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sửa đổi
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lượng giác
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lượng giác 1)Chứng minhg rằng nếu $a$ là góc nhọn thì $(1+\frac{1}{sina})(1+\frac{1}{cosa )}>5$2)Giải pt:$cos^2x-sinxcos4x-cos^4x=\frac{1}{4}$
lượng giác 1)Chứng minhg rằng nếu $a$ là góc nhọn thì $(1+\frac{1}{sina})(1+\frac{1}{cosa} )>5$2)Giải pt:$cos^2x-sinxcos4x-cos^4x=\frac{1}{4}$
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sửa đổi
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giải pt lượng giác 11
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2)$cos3x-cos2x-cosx=sin2x-sinx-1=0$$\Leftrightarrow -2sin2xsinx-1+2sin^2x-sin2x+sinx+1=0$$\Leftrightarrow 2sin^2x+sinx-(2sin2x.sinx+sin2x)=0$$\Leftrightarrow sinx(2sinx+1)-sin2x(2sinx+1)=0$$\Leftrightarrow sinx(2sinx+1)(1-2cosx)=0$
2)$cos3x-cos2x-cosx=sin2x-sinx-1$$\Leftrightarrow -2sin2xsinx-1+2sin^2x-sin2x+sinx+1=0$$\Leftrightarrow 2sin^2x+sinx-(2sin2x.sinx+sin2x)=0$$\Leftrightarrow sinx(2sinx+1)-sin2x(2sinx+1)=0$$\Leftrightarrow sinx(2sinx+1)(1-2cosx)=0$
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HELP ME
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Ptdt $AB$:$x+2y-5=0$Ta co:$\frac{-1}{-2}=\frac{2}{4}\Rightarrow \Delta _1//\Delta _2$Goi $M=AB\cap \Delta _1$ Toa do cua M la nghiem cua he:$\left\{ \begin{array}{l} x+2y-5=0\\ -x+2y-7=0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} x=-1\\ y=3 \end{array} \right.$$\Rightarrow M(-1;3)$Goi $N=AB\cap \Delta _2$ toa do cua $N$la nghiem cua he:$\left\{ \begin{array}{l} x+2y-5=0\\ -2x+4y-5=0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} x=5/4\\ y=15/8 \end{array} \right.$$\Rightarrow N(5/4;15/8)$$C $ cach deu $\Delta_1 ,\Delta _2 \Rightarrow C$ la trung diem cua $MN:\left\{ \begin{array}{l} x_C=\frac{-1+5/4}{2}\\ y_C=\frac{3+15/4}{2} \end{array} \right.$Vay$ C(\frac{1}{8};\frac{27}{8})$
Ptdt $AB$:$x+2y-5=0$Ta co:$\frac{-1}{-2}=\frac{2}{4}\Rightarrow \Delta _1//\Delta _2$Goi $M=AB\cap \Delta _1$ Toa do cua M la nghiem cua he:$\left\{ \begin{array}{l} x+2y-5=0\\ -x+2y-7=0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} x=-1\\ y=3 \end{array} \right.$$\Rightarrow M(-1;3)$Goi $N=AB\cap \Delta _2$ toa do cua $N$la nghiem cua he:$\left\{ \begin{array}{l} x+2y-5=0\\ -2x+4y-5=0 \end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} x=5/4\\ y=15/8 \end{array} \right.$$\Rightarrow N(5/4;15/8)$$C $ cach deu $\Delta_1 ,\Delta _2 \Rightarrow C$ la trung diem cua $MN:\left\{ \begin{array}{l} x_C=\frac{-1+5/4}{2}\\ y_C=\frac{3+15/8}{2} \end{array} \right.$Vay$ C(\frac{1}{8};\frac{39}{16})$
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sửa đổi
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hình học 11 nâng cao
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ban chiu kho tu ve hinh nhe!a,Trong mp$(ABN)$: $ME\cap BN=J\Rightarrow \left\{ \begin{array}{l} J\in ME\\ J\in BN\subset (BCD) \end{array} \right.\Rightarrow J=ME\in (BCD)$b,$*$ta co $Q\in (EMQ)\cap ( BCD) (1)$ Trong mp $(BCD):CD\cap QJ=P\Rightarrow P\in (EMQ)\cap (BCD) (2)$Tu $(1)$ va $(2)$$:(EMQ)\cap (ACD)=PQ$$*M\in (EMQ)\cap (ABD) $Trong mp$(ABD):EP\cap AD=F\Rightarrow F\in (EMQ)\cap (ABD)$$\Rightarrow (EMQ)\cap (ABD)=FM$c,Ta co:$\left\{ \begin{array}{l} (QME)\cap (ABC)=QM\\(QME)\cap (BCD)=QP\\(QME)\cap(ACD)=FP\\ (QME)\cap(ABD)=MF \end{array} \right.$Vay thiet dien cat boi mp$(QME)$la tu giac $FMQP$
ban chiu kho tu ve hinh nhe!a,Trong mp$(ABN)$: $ME\cap BN=J\Rightarrow \left\{ \begin{array}{l} J\in ME\\ J\in BN\subset (BCD) \end{array} \right.\Rightarrow J=ME\in (BCD)$b,$*$ta co $Q\in (EMQ)\cap ( BCD) (1)$ Trong mp $(BCD):CD\cap QJ=P\Rightarrow P\in (EMQ)\cap (BCD) (2)$Tu $(1)$ va $(2)$$:(EMQ)\cap (BCD)=PQ$$*M\in (EMQ)\cap (ABD) $Trong mp$(ACD):EP\cap AD=F\Rightarrow F\in (EMQ)\cap (ABD)$$\Rightarrow (EMQ)\cap (ABD)=FM$c,Ta co:$\left\{ \begin{array}{l} (QME)\cap (ABC)=QM\\(QME)\cap (BCD)=QP\\(QME)\cap(ACD)=FP\\ (QME)\cap(ABD)=MF \end{array} \right.$Vay thiet dien cat boi mp$(QME)$la tu giac $FMQP$
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sửa đổi
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Tổ hợp
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Ta co:$(1+x)^{r}=C^{0}_{r}+C^{1}_{r}x+C^{2}_{r}x^{2}+...+C^{r}_{r}x^{r}$ $(1+x)^{q}=C^{0}_{q}+C^{1}_{q}x+C^{2}_{q}x^{2}+...+C^{q}_{q}x^{q}$He so $x^{p}$ trong tich $(1+x)^{r}(1+x)^{q}la:$ $C^{0}_{r}.C^{p}_{q}+C^{1}_{0}.C^{p-1}_{q}+...+C^{p}_{r}.C^{0}_{q}$ (1)ta lai co:$(1+x)^{r+p}=C^{0}_{r+q}+...+C^{p}_{r+q}x^{p}+...+C^{r+q}_{r+q}.x^{r+q}$He so $x^{p}$ trong $(1+x)^{r+q} $ la $C^{p}_{r+q} (2)ma $(1+x)^{r}.(1+x)^{q}=(1+x)^{q+r} nen tu $(1)$ va $(2)$ $\Rightarrow $dpcm
Ta co:$(1+x)^{r}=C^{0}_{r}+C^{1}_{r}x+C^{2}_{r}x^{2}+...+C^{r}_{r}x^{r}$ $(1+x)^{q}=C^{0}_{q}+C^{1}_{q}x+C^{2}_{q}x^{2}+...+C^{q}_{q}x^{q}$He so $x^{p}$ trong tich $(1+x)^{r}(1+x)^{q}la:$ $C^{0}_{r}.C^{p}_{q}+C^{1}_{0}.C^{p-1}_{q}+...+C^{p}_{r}.C^{0}_{q}$ (1)ta lai co:$(1+x)^{r+p}=C^{0}_{r+q}+...+C^{p}_{r+q}x^{p}+...+C^{r+q}_{r+q}.x^{r+q}$He so $x^{p}$ trong $(1+x)^{r+q} $ la $C^{p}_{r+q} (2)$ma $(1+x)^{r}.(1+x)^{q}=(1+x)^{q+r}$nen tu $(1)$ va $(2)$ $\Rightarrow $dpcm
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sửa đổi
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giúp em nhị thức niuton
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$(x-\frac{y}{2})^{30}=a_{0}x^{30}+a_{2}x^{29}(-\frac{y}{2})+a_{3}x^{28}(-\frac{y}{2})^{2}+....+a_{30}(-\frac{y}{2})^{30}$Taco : $S=f(1;1)=(1-\frac{1}{2})^{30}=a_{0}+a_{1}+a_{2}+...+a_{30}=(\frac{1}{2})^{30}$
$(x-\frac{y}{2})^{30}=a_{0}x^{30}+a_{1}x^{29}(-\frac{y}{2})+a_{2}x^{28}(-\frac{y}{2})^{2}+....+a_{30}(-\frac{y}{2})^{30}$Taco : $S=f(1;1)=(1-\frac{1}{2})^{30}=a_{0}+a_{1}(-\frac{1}{2})+a_{2}(-\frac{1}{2})^{2}+...+a_{30}(-\frac{1}{2})^{30}=(\frac{1}{2})^{30}$
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sửa đổi
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Ai giúp mình bài toán với
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$\sqrt[]{3}sin2x(1+2cosx)=1+2cosx+cos2x-cos3x$$\Leftrightarrow$$2\sqrt{3}sinx cosx(1+2cosx)=1+2cosx+2cos^2x-1-4cos^3x+cos3x$$\Leftrightarrow$$cosx(2\sqrt{3}sinx+2\sqrt{3}sin2x-2cosx+2cos2x+2-5)=0$m chi lam duoc den day ^^
$\sqrt[]{3}sin2x(1+2cosx)=1+2cosx+cos2x-cos3x$$\Leftrightarrow$$2\sqrt{3}sinx cosx(1+2cosx)=1+2cosx+2cos^2x-1-4cos^3x+3cosx$$\Leftrightarrow$$cosx(2\sqrt{3}sinx+2\sqrt{3}sin2x-2cosx+2cos2x+2-5)=0$m chi lam duoc den day ^^
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sửa đổi
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giải giùm em câu lượng giác này với.
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$\frac{1}{2}sinx .sin4x+ cosx.cos2x=1$ $\Leftrightarrow$ $2sin^{2}x.cosx.cos2x+cosx.cos2x=cos2x-2sin^{2}x$$\Leftrightarrow$$2sin^{2}x(cosx.cos2x+1)+cos2x(cosx-1)=0$$\Leftrightarrow$$2(1-cosx)(1+cosx)(cosx.cos2x+1)+cos2x(cosx-1)=0$$\Leftrightarrow$$(1-cosx)[(2+2cosx)(cosx.cos2x+1)-cos2x]=0$$\Leftrightarrow$$cosx=1$ nhan tu con lai thi chiu
$\frac{1}{2}sinx .sin4x+ cosx.cos2x=1$ $\Leftrightarrow$ $2sin^{2}x.cosx.cos2x+cosx.cos2x=cos2x+2sin^{2}x$$\Leftrightarrow$$2sin^{2}x(cosx.cos2x-1)+cos2x(cosx-1)=0$$\Leftrightarrow$$2(1-cosx)(1+cosx)(cosx.cos2x-1)+cos2x(cosx-1)=0$$\Leftrightarrow$$(1-cosx)[(2+2cosx)(cosx.cos2x-1)-cos2x]=0$$\Leftrightarrow$$cosx=1$ nhan tu con lai thi chiu
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sửa đổi
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giải giùm em câu lượng giác này với.
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$\frac{1}{2}sinx .sin4x+ cosx.cos2x=1$ $\Leftrightarrow$ $2sin^{2}x.cosx.cos2x+cosx.cos2x=cos2x-2sin^{2}x$$\Leftrightarrow$$2sin^{2}x(cosx.sinx-1)+cos2x(cosx-1)=0$$\Leftrightarrow$$2(1-cosx)(1+cosx)(cosx.sinx-1)+cos2x(cosx-1)=0$$\Leftrightarrow$$(1-cosx)[(2+2cosx)(cosx.sinx-1)-cos2x]=0$$\Leftrightarrow$$cosx=1$ nhan tu con lai thi chiu
$\frac{1}{2}sinx .sin4x+ cosx.cos2x=1$ $\Leftrightarrow$ $2sin^{2}x.cosx.cos2x+cosx.cos2x=cos2x-2sin^{2}x$$\Leftrightarrow$$2sin^{2}x(cosx.cos2x+1)+cos2x(cosx-1)=0$$\Leftrightarrow$$2(1-cosx)(1+cosx)(cosx.cos2x+1)+cos2x(cosx-1)=0$$\Leftrightarrow$$(1-cosx)[(2+2cosx)(cosx.cos2x+1)-cos2x]=0$$\Leftrightarrow$$cosx=1$ nhan tu con lai thi chiu
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sửa đổi
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Bài tập về giải HPT bậc 2 ai giúp em với help!!! :'(
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b,$\begin{cases}2x^{2}-3x=y^{2} -2 (1)\\ 2y^{2}-3y=x^{2}-2(2) \end{cases}$lay (1) -(2) ta duoc:$3(x-y)(x+y-1)=0$$\Leftrightarrow$$x=y $ hoac $ x=1-y$ ket hop voi (2)giai ra duoc 2nghiem $(x;y)$la$(2;2), (1;1),$c.$\Leftrightarrow $$\left\{ \begin{array}{l} 2x^{2}+3xy+3y^{2}=3\\ 3x^{2}+3xy+3y^{2}=3 \end{array} \right.$$\Leftrightarrow$$\left\{ \begin{array}{l} 2x^{2}+3xy+3y^{2}\\ x=0 \end{array} \right.$$\Leftrightarrow$$\left\{ \begin{array}{l} x=0\\ y=1 \end{array} \right.$d.$\Leftrightarrow$$\left\{ \begin{array}{l} \sqrt{x}+\sqrt{y}=4\\ \sqrt{xy}=3 \end{array} \right.$ de roi tu giai nhae.Dat$\left\{ \begin{array}{l}a= \sqrt[3]{x}+\sqrt[3]{y}\\ b=\sqrt[3]{xy} \end{array} \right.$ he tro thanh :$\left\{ \begin{array}{l} a+b=3\\ a^{2} -3b=3\end{array} \right.$
b,$\begin{cases}2x^{2}-3x=y^{2} -2 (1)\\ 2y^{2}-3y=x^{2}-2(2) \end{cases}$lay (1) -(2) ta duoc:$3(x-y)(x+y-1)=0$$\Leftrightarrow$$x=y $ hoac $ x=1-y$ ket hop voi (2)giai ra duoc 2nghiem $(x;y)$la$(2;2), (1;1),$c.$\Leftrightarrow $$\left\{ \begin{array}{l} 2x^{2}+3xy+3y^{2}=3\\ 3x^{2}+3xy+3y^{2}=3 \end{array} \right.$$\Leftrightarrow$$\left\{ \begin{array}{l} 2x^{2}+3xy+3y^{2}\\ x=0 \end{array} \right.$$\Leftrightarrow$$\left\{ \begin{array}{l} x=0\\ y=1 \end{array} \right.$d.$\Leftrightarrow$$\left\{ \begin{array}{l} \sqrt{x}+\sqrt{y}=4\\ \sqrt{xy}=3 \end{array} \right.$ de roi tu giai nhae.Dat$\left\{ \begin{array}{l}a= \sqrt[3]{x}+\sqrt[3]{y}\\ b=\sqrt[3]{xy} \end{array} \right.$ he tro thanh :$\left\{ \begin{array}{l} a=3\\ a^{2} -3b=3\end{array} \right.$
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sửa đổi
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Bài tập về giải HPT bậc 2 ai giúp em với help!!! :'(
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b,$\begin{cases}2x^{2}-3x=y^{2} -2 (1)\\ 2y^{2}-3y=x^{2}-2(2) \end{cases}$lay (1) -(2) ta duoc:$3(x-y)(x+y-1)=0$$\Leftrightarrow$$x=y $ hoac $ x=1-y$ ket hop voi (2)giai ra duoc 4nghiem $(x;y)$la$(2;2), (1;1), (\frac{1+\sqrt{5}}{2};\frac{1-\sqrt{5}}{2}), (\frac{1-\sqrt{5}}{2};\frac{1+\sqrt{5}}{2})$c.$\Leftrightarrow $$\left\{ \begin{array}{l} 2x^{2}+3xy+3y^{2}=3\\ 3x^{2}+3xy+3y^{2}=3 \end{array} \right.$$\Leftrightarrow$$\left\{ \begin{array}{l} 2x^{2}+3xy+3y^{2}\\ x=0 \end{array} \right.$$\Leftrightarrow$$\left\{ \begin{array}{l} x=0\\ y=1 \end{array} \right.$d.$\Leftrightarrow$$\left\{ \begin{array}{l} \sqrt{x}+\sqrt{y}=4\\ \sqrt{xy}=3 \end{array} \right.$ de roi tu giai nhae.Dat$\left\{ \begin{array}{l}a= \sqrt[3]{x}+\sqrt[3]{y}\\ b=\sqrt[3]{xy} \end{array} \right.$ he tro thanh :$\left\{ \begin{array}{l} a+b=3\\ a^{2} -3b=3\end{array} \right.$
b,$\begin{cases}2x^{2}-3x=y^{2} -2 (1)\\ 2y^{2}-3y=x^{2}-2(2) \end{cases}$lay (1) -(2) ta duoc:$3(x-y)(x+y-1)=0$$\Leftrightarrow$$x=y $ hoac $ x=1-y$ ket hop voi (2)giai ra duoc 2nghiem $(x;y)$la$(2;2), (1;1),$c.$\Leftrightarrow $$\left\{ \begin{array}{l} 2x^{2}+3xy+3y^{2}=3\\ 3x^{2}+3xy+3y^{2}=3 \end{array} \right.$$\Leftrightarrow$$\left\{ \begin{array}{l} 2x^{2}+3xy+3y^{2}\\ x=0 \end{array} \right.$$\Leftrightarrow$$\left\{ \begin{array}{l} x=0\\ y=1 \end{array} \right.$d.$\Leftrightarrow$$\left\{ \begin{array}{l} \sqrt{x}+\sqrt{y}=4\\ \sqrt{xy}=3 \end{array} \right.$ de roi tu giai nhae.Dat$\left\{ \begin{array}{l}a= \sqrt[3]{x}+\sqrt[3]{y}\\ b=\sqrt[3]{xy} \end{array} \right.$ he tro thanh :$\left\{ \begin{array}{l} a+b=3\\ a^{2} -3b=3\end{array} \right.$
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sửa đổi
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Bài tập về giải HPT bậc 2 ai giúp em với help!!! :'(
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a, Dat $\left\{ \begin{array}{l} a=u+v\\ b=\sqrt{uv} \end{array} \right.$ he pt tro thanh:$\left\{ \begin{array}{l} a^{2}-3b^{2}=13\\ a-b=3\end{array} \right.$giai ra duoc:$\left\{ \begin{array}{l} a=4\\ b=1\end{array} \right.$ hoac $\left\{ \begin{array}{l} a=5\\ b=2 \end{array} \right.$$\Rightarrow $$\left\{ \begin{array}{l} u=2-\sqrt{3}\\ v=2+\sqrt{3} \end{array} \right.$hoac $\left\{ \begin{array}{l} u=2+\sqrt{3}\\ v=2-\sqrt{3} \end{array} \right.$ hoac$\left\{ \begin{array}{l} u=1\\ v=4 \end{array} \right.$ hoac $\left\{ \begin{array}{l} u=4\\ v=1 \end{array} \right.$
a, Dat $\left\{ \begin{array}{l} a=u+v\\ b=\sqrt{uv} \end{array} \right.$ he pt tro thanh:$\left\{ \begin{array}{l} a^{2}-3b^{2}=13\\ a-b=3\end{array} \right.$giai ra duoc:$\left\{ \begin{array}{l} a=4\\ b=1\end{array} \right.$ hoac $\left\{ \begin{array}{l} a=5\\ b=2 \end{array} \right.$.$\left\{ \begin{array}{l} a=4\\ b=1 \end{array} \right.$: $ \left\{ \begin{array}{l} u+v=4\\ \sqrt{uv}=1 \end{array} \right.$ $\Leftrightarrow$$\left\{ \begin{array}{l} u+v=4\\ uv=1 \end{array} \right.$ $\Rightarrow $$\left\{ \begin{array}{l} u=2-\sqrt{3}\\ v=2+\sqrt{3} \end{array} \right.$hoac $\left\{ \begin{array}{l} u=2+\sqrt{3}\\ v=2-\sqrt{3} \end{array} \right.$ .$\left\{ \begin{array}{l} a=5\\ b=2 \end{array} \right.$$\Rightarrow $$\left\{ \begin{array}{l} u+v=5\\ \sqrt{uv}=2 \end{array} \right.$ $\Leftrightarrow $$\left\{ \begin{array}{l} u+v=5\\ uv=4\end{array} \right.$$\Leftrightarrow $$\left\{ \begin{array}{l} u=1\\ v=4 \end{array} \right.$ hoac $\left\{ \begin{array}{l} u=4\\ v=1 \end{array} \right.$
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sửa đổi
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ai giúp em tí đê cần gấp !!!!
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minh lam cach nay ban xem thu xem sao:$x^{2}$-3x+k-1=0$\triangle $=9-4(k-1)=13-4kpt co nghiem$\Leftrightarrow$ $\triangle$$\geq$0 $\Leftrightarrow$k$\leq$$\frac{13}{4}$Ta co :$ \left\{ \begin{array}{l} x_{1}+x_{2}=3 (1)\\ x_{1} .x_{2}=k-1 (2)\end{array} \right.$ket hop voi yeu cau bai ra , giai he$\Rightarrow$ka, theo bai ra ta co 2x1-5x2=-8 ket hop voi (1) ta duoc he:$\left\{ \begin{array}{l} x_{1}+x_{2}=3\\ 2x_{1}-5x_{2} =-8\end{array} \right.$$\Leftrightarrow$$\left\{ \begin{array}{l} x_{1}=1\\ x_{2}=2 \end{array} \right.$thay vao (2) ta duoc k=3b.theo bai ra ta co: $x^{2}_{1}$-$x^{2}_{2}$=-8 ket hop voi (1)ta co he:$\left\{ \begin{array}{l} (x_{1}-x_{2})(x_{1}+x_{2})=-8\\ x_{1}+x_{2}=3 \end{array} \right.$$\Leftrightarrow$$\left\{ \begin{array}{l} x_{1}=1/6\\ x_{2} =17/6\end{array} \right.$ thay vao (2)$\Rightarrow$k=53/36cau c lam tuong tu
minh lam cach nay ban xem thu xem sao:$x^{2}$-3x+k-1=0$\triangle $=9-4(k-1)=13-4kpt co nghiem$\Leftrightarrow$ $\triangle$$\geq$0 $\Leftrightarrow$k$\leq$$\frac{13}{4}$Ta co :$ \left\{ \begin{array}{l} x_{1}+x_{2}=3 (1)\\ x_{1} .x_{2}=k-1 (2)\end{array} \right.$ket hop voi yeu cau bai ra , giai he$\Rightarrow$ka, theo bai ra ta co 2x1-5x2=-8 ket hop voi (1) ta duoc he:$\left\{ \begin{array}{l} x_{1}+x_{2}=3\\ 2x_{1}-5x_{2} =-8\end{array} \right.$$\Leftrightarrow$$\left\{ \begin{array}{l} x_{1}=1\\ x_{2}=2 \end{array} \right.$thay vao (2) ta duoc k=3b.theo bai ra ta co: $x^{2}_{1}$-$x^{2}_{2}$=-8 ket hop voi (1)ta co he:$\left\{ \begin{array}{l} (x_{1}-x_{2})(x_{1}+x_{2})=15\\ x_{1}+x_{2}=3 \end{array} \right.$$\Leftrightarrow$$\left\{ \begin{array}{l} x_{1}=4\\ x_{2} =-1\end{array} \right.$ thay vao (2)$\Rightarrow$k=-3cau c lam tuong tu
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sửa đổi
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AI GIUP TOI VOI
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Cau 2:goi hai so tu nhien lien tiep do la x va x+1theo bai ra ta co:$(x+1)^{2}$ -$x^{2}$=15 $\Leftrightarrow$2x=14$\Leftrightarrow$=7vay hai so do la 7 va 8
Cau 2:goi hai so tu nhien lien tiep do la x va x+1theo bai ra ta co:$(x+1)^{2}$ -$x^{2}$=15 $\Leftrightarrow$2x=14$\Leftrightarrow$x=7vay hai so do la 7 va 8
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giai he phuong trinh
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giai he phuong trinh giai he:$ \left\{ \begin{array}{l} x^{4}+y^{2}=\frac{698}{81}\\ x^{2}+y^{2} +xy-3x-4y+4\end{array} \right.$
giai he phuong trinh giai he:$ \left\{ \begin{array}{l} x^{4}+y^{2}=\frac{698}{81}\\ x^{2}+y^{2} +xy-3x-4y+4 =0\end{array} \right.$
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