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giải đáp
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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 |
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giải đáp
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mọi người giúp em nhanh đi ạ! 195
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Ta co: $(1+x)^{n}=C^{0}_{n}+C^{1}_{n}.x+C^{2}_{n}.x^{2}+...+C^{k}_{n}.x^{k}+C^{n}_{n}.x^{n}$Cho x=-1 ta duoc:$0=C^{0}_{n}-C^{1}_{n}+C^{2}_{n}-C^{3}_{n}+...+C^{2k}_{n}-C^{2k+1}_{n}+...+(-1)^{n}C^{n}_{n}$ $\Rightarrow$$C^{0}_{n}+C^{2}_{n}+C^{4}_{n}+...=C^{1}_{n}+C^{3}_{n}+C^{5}_{n}+...$ (1) Cho x=1 ta co:$2^{n}=C^{0}_{n}+C^{1}_{n}+C^{2}_{n}+C^{3}_{n}+....C^{2k}_{n}+C^{2k+1}_{n}+...+C^{n-1}_{n}+C^{n}_{n}$ Ap dung (1) ta co : $2^{n}=2(C^{0}_{n}+C^{2}_{n}+C^{4}_{n}+...)$ $\Rightarrow$$\frac{2^{n}}{2}=C^{0}_{n}+C^{2}_{n}+C^{4}_{n}+...$hay$2^{n-1}=C^{0}_{n}+C^{2}_{n}+C^{4}_{n}+...$(2) Tu (1) va(2)$\Rightarrow$dpcm
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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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