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sửa đổi
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Hệ trục tọa độ (toán 10)
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$MA+MB$ lon nhat khi va chi khi $M=AB'\cap d$, trong do $B'$ la diem doi xung voi $B$ qua $d$. Dat $B'=(a,b)$, ta co $\vec{BB'}=(a-2, b-5)'\bot d$. Cho nen $$\frac{a-2}{1}=\frac{b-5}{-2}$$Do vay, dat $a=t+2, b=-2t+5$. Goi trung diem cua $BB'$ la $X$ thi $X=(\frac{a+2}{2},\frac{b+5}{2})=(2+\frac{t}{2}, -t+5)$ va $X\in d$. Do vay $$(2+\frac{t}{2})-2(-t+5)+2=0\Rightarrow t=12/5$$Nhu vay $B'=(22/5, 1/5)$. Phuong trinh duong thang $AB'$ la $$\frac{x-22/5}{0-22/5}=\frac{y-1/5}{6-1/5}$$Vi $M\in AB'$ nen dat $M=(\frac{22}{5}(-t+1),\frac{1}{5}(39t+1)) \quad (*)$. Do $M\in d$ nen $$\frac{22}{5}(-t+1)-\frac{2}{5}(39t+1)+2=0\Rightarrow t=\frac{3}{10}$$Thay $t$ vao $(*)$ ta duoc toa do cua $M$
$MA+MB$ lon nhat khi va chi khi $M=AB'\cap d$, trong do $B'$ la diem doi xung voi $B$ qua $d$. Dat $B'=(a,b)$, ta co $\vec{BB'}=(a-2, b-5)\bot d$. Cho nen $$\frac{a-2}{1}=\frac{b-5}{-2}$$Do vay, dat $a=t+2, b=-2t+5$. Goi trung diem cua $BB'$ la $X$ thi $X=(\frac{a+2}{2},\frac{b+5}{2})=(2+\frac{t}{2}, -t+5)$ va $X\in d$. Do vay $$(2+\frac{t}{2})-2(-t+5)+2=0\Rightarrow t=12/5$$Nhu vay $B'=(22/5, 1/5)$. Phuong trinh duong thang $AB'$ la $$\frac{x-22/5}{0-22/5}=\frac{y-1/5}{6-1/5}$$Vi $M\in AB'$ nen dat $M=(\frac{22}{5}(-t+1),\frac{1}{5}(39t+1)) \quad (*)$. Do $M\in d$ nen $$\frac{22}{5}(-t+1)-\frac{2}{5}(39t+1)+2=0\Rightarrow t=\frac{3}{10}$$Thay $t$ vao $(*)$ ta duoc toa do cua $M$
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sửa đổi
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Phương trình đại số
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Ta co $a^b=b^a$ suy ra $blna=alnb\Rightarrow \frac{lna}{a}=\frac{lnb}{b}$. Xet ham so $f(t)=\frac{lnt}{t}$ co $f'(t)=\frac{1-lnt}{t^2}$. Ta thay $f'(t)<0$ voi $t>2$ va $f'(t)>0$ voi $t\leq2$. Neu co $a\ne b, a,b\in \mathbb{Z}$ ma $f(a)=f(b)$ thi mot trong hai so phai nho hon hoac bang $2$, so con lai lon hon $2$. Thu cho $a=1,2$ dan toi ko co $b$ thoa man. Vay vo nghiem
Ta co $a^b=b^a$ suy ra $blna=alnb\Rightarrow \frac{lna}{a}=\frac{lnb}{b}$. Xet ham so $f(t)=\frac{lnt}{t}$ co $f'(t)=\frac{1-lnt}{t^2}$. Ta thay $f'(t)<0$ voi $t>2$ va $f'(t)>0$ voi $t\leq2$. Neu co $a\ne b, a,b\in \mathbb{Z}$ ma $f(a)=f(b)$ thi mot trong hai so phai nho hon hoac bang $2$, so con lai lon hon $2$. Thu cho $a=1,2$ dan toi ko co $b=4$ thoa man ( xet $2^b=b^2$ thi $b=2^h$ voi $h$ la so nguyen nao do. Suy ra $2^{2^h}=2^{2h}$ nen $h=2$). Vay nghiem la $a=2,b=4$ hoac $b=4,a=2$
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sửa đổi
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Phương trình đại số
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Khong ton tai $a,b$ thoa man tinh chat nhu vay. Vi neu $c$ la uoc nguyen to cua $a$ thi tu $a^b=b^a$, dan toi $c$ cung la uoc cua $b$ va nguoc lai. Do vay, ta co the dat dat $$a=p_1^{n_1}\cdots p_k^{n_k}\\ b=p_1^{m_1}\cdots p_k^{m_k}$$ trong do $p_k$ la cac so nguyen to khac nhau. Vi $a^b=b^a$ nen $p_i^{b}=p_i^a$, nhu vay $a=b$ ( mau thuan).
Ta co $a^b=b^a$ suy ra $blna=alnb\Rightarrow \frac{lna}{a}=\frac{lnb}{b}$. Xet ham so $f(t)=\frac{lnt}{t}$ co $f'(t)=\frac{1-lnt}{t^2}$. Ta thay $f'(t)<0$ voi $t>2$ va $f'(t)>0$ voi $t\leq2$. Neu co $a\ne b, a,b\in \mathbb{Z}$ ma $f(a)=f(b)$ thi mot trong hai so phai nho hon hoac bang $2$, so con lai lon hon $2$. Thu cho $a=1,2$ dan toi ko co $b$ thoa man. Vay vo nghiem
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sửa đổi
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[toán 9]
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Bai 3) Do $xy+yx +zx =1$ nen $$1+y^2=xy+yz+zx+y^2=(z+y)(x+y)\\ 1+x^2 =xy+yz+zx+x^2 = (z+x)(y+x)\\ 1+z^2 =xy+yz+zx +z^2=(x+z)(y+z)$$Do vay $$ \frac{(1+y^2)(1+z^2)}{1+x^2}=(y+z)^2\\ \frac{(1+x^2)(1+z^2)}{1+y^2}=(z+x)^2\\ \frac{(1+y^2)(1+x^2)}{1+z^2}=(x+y)^2 $$ Nen $P= x(y+z)+y(z+x)+z(x+y)=2(xy+yz+zx)=2 $
Bai 3) Do $xy+yx +zx =1$ nen $$1+y^2=xy+yz+zx+y^2=(z+y)(x+y)\\ 1+x^2 =xy+yz+zx+x^2 = (z+x)(y+x)\\ 1+z^2 =xy+yz+zx +z^2=(x+z)(y+z)$$Do vay$$\frac{(1+y^2)(1+z^2)}{1+x^2}=(y+z)^2\\ \frac{(1+x^2)(1+z^2)}{1+y^2} =(x+z)^2\\ \frac{(1+x^2)(1+y^2)}{1+z^2}=(x+y)^2 $$Nen $P= x(y+z)+y(z+x)+z(x+y)=2(xy+yz+zx)=2 $
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sửa đổi
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Tính nguyên hàm (2)
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8) $I_8=\int \sin(ln x)dx = x\sin(ln x)-\int x d(\sin(ln x))= x\sin(ln x)-\int \cos (lnx)dx$Do $\int \cos(lnx)dx= x \cos(ln x)-\int x d(\cos(ln x))=x\cos(ln x)+\int \sin(ln x)$nên $I_8=x\sin(ln x)-x\cos(ln x)-I_8$. Suy ra $I_8=\frac{x \sin(ln x)-x\cos(lnx) }{2}$
8) $I_8=\int \sin(ln x)dx = x\sin(ln x)-\int x d(\sin(ln x))= x\sin(ln x)-\int \cos (lnx)dx$Do $\int \cos(lnx)dx= x \cos(ln x)-\int x d(\cos(ln x))=x\cos(ln x)+\int \sin(ln x)$nên $I_8=x\sin(ln x)-x\cos(ln x)-I_8$. Suy ra $I_8=\frac{x \sin(ln x)-x\cos(lnx) }{2}+C$
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sửa đổi
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Tính nguyên hàm (2)
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9) $I_9 =\int e^{-2x}\sin x dx=\int e^{-2x} d(-\cos x)=-\cos xe^{-2x}+\int \cos x d(e^{-2x})$$=-\cos x e^{-2x}-2\int \cos x e^{-2x} dx $Do $\int \cos x e^{-2x}dx=\int e^{-2x}d(\sin x)=e^{-2x}\sin x -\int \sin x d(e^{-2x})$$=e^{-2x} \sin x +2\int e^{-2x}\sin x dx$nên$I_9=-\cos x e^{-2x}-2e^{-2x}\sin x -4I_9$. Suy ra $I_9=\frac{-\cos x e^{-2x}-2e^{-2x}\sin x}{5}$
9) $I_9 =\int e^{-2x}\sin x dx=\int e^{-2x} d(-\cos x)=-\cos xe^{-2x}+\int \cos x d(e^{-2x})$$=-\cos x e^{-2x}-2\int \cos x e^{-2x} dx $Do $\int \cos x e^{-2x}dx=\int e^{-2x}d(\sin x)=e^{-2x}\sin x -\int \sin x d(e^{-2x})$$=e^{-2x} \sin x +2\int e^{-2x}\sin x dx$nên$I_9=-\cos x e^{-2x}-2e^{-2x}\sin x -4I_9$. Suy ra $I_9=\frac{-\cos x e^{-2x}-2e^{-2x}\sin x}{5}+C$
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sửa đổi
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Tính nguyên hàm (2)
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5) Đặt $t=\sqrt{x}$, suy ra $$I_5=2\int t^2\cos t dt \\ =2\int t^2 d(\sin t)\\ = 2(t^3 \sin t - 3\int t^2 \sin t dt ) \\=2t^3\sin t + 6 \int t^2 d(\cos t)\\=2t^3\sint t + 6( t^2 \cos t - 2\int t \cos t dt )\\=2t^3\sin t + 6 t^2\cos t -12 \int t d(\sin t )\\=2t^3\sin t + 6t^2 \cos t - 12 ( t\sin t -\int \sin t dt) \\=2t^3 \sin t + 6t^2 \cos t -12 t\sin t +12 \cos t +C $$
5) Đặt $t=\sqrt{x}$, suy ra $I_5 = 2\int t^3 \cos t dt = 2\int t^3 d(\sin t)=2(t^3 \sin t-3\int t^2 \sin t dt)$. $=2t^3 \sin t + 6 \int t^2 d(\cos t )= 2t^3 \sin t + 6(t^2\cos t -2\int t\cos t dt)$.$=2t^3 \sin t + 6 t^2 \cos t -12 \int t d(\sin t)$$=2t^3 \sin t + 6 t^2 \cos t -12 (t \sin t - \int \sin t dt)$$= 2t^3 \sin t + 6t^2 \cos t -12 t \sin t + \cos t +C$6) $I_6= \int xln(2x)dx = \int ln(2x) d(\frac{x^2}{2})=\frac{x^2}{2} ln(2x)-\int \frac{x^2}{2}.\frac{2}{2x}dx$$=\frac{x^2}{2}ln(x)-\int\frac{x}{2}dx=\frac{x^2}{2}ln(x)-\frac{x^2}{4}+C$7) $I_7=\int (x^2-2x)ln x dx=\int ln x d(x^3/3-x^2)=(x^3/3-x^2)ln x-\int (x^3/3-x^2)\frac{1}{x} dx$$=(x^3/3-x^2)ln x-\int (x^2/3-x)dx=(x^3/3-x^2)ln x -(x^3/9-x^2/2)+C$
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sửa đổi
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Cần gấp ạ
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Đặt $A(x)=\sum\limits_{k \, lẻ \in [1,2n] } x^{2n-k}$ và $B(x)=\sum\limits_{k\,chẵn\in[1,2n] }x^{2n-k}$, ta có $$A(x)+B(x)=\sum\limits_{k=1}^{2n} x^{2n-k}=(x+1)^{2n}$$$$-A(x)+B(x)=\sum\limits_{k=1}^{2n}(-1)^k x^{2n-k}=(x-1)^n$$Ta có $A(x)=\frac{(x+1)^{2n}-(x-1)^{2n}}{2}$ và $A(1)=C^{1}_{2n}+\cdots+C^{2n-1}_{2n}=2048$. Như vậy $2^{2n-1}=2048$, suy ra $2n-1=11$. Nên $n=6$
Đặt $A(x)=\sum\limits_{k \, lẻ \in [1,2n] } x^{2n-k}$ và $B(x)=\sum\limits_{k\,chẵn\in[1,2n] }x^{2n-k}$, ta có $$A(x)+B(x)=\sum\limits_{k=1}^{2n} x^{2n-k}=(x+1)^{2n}$$$$-A(x)+B(x)=\sum\limits_{k=1}^{2n}(-1)^k x^{2n-k}=(x-1)^{2n}$$Ta có $A(x)=\frac{(x+1)^{2n}-(x-1)^{2n}}{2}$ và $A(1)=C^{1}_{2n}+\cdots+C^{2n-1}_{2n}=2048$. Như vậy $2^{2n-1}=2048$, suy ra $2n-1=11$. Nên $n=6$
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