1/x+1/y+1/z và 1/x^2+1/y^2+1/z^2=1.CMR x+y+z=4xyz
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3 tháng 12 2021
\(x^2+y^2+z^2+4xyz=2\left(xy+yz+zx\right)\\ \Leftrightarrow\left(x-y-z\right)^2=\left(1-x\right)4yz\ge0\\ \Leftrightarrow1-x\ge0\Leftrightarrow0< x\le1\\ \Leftrightarrow\left(x-y-z\right)^2=\left(1-x\right)4yz\le\left(1-x\right)\left(y+z\right)^2\\ \Leftrightarrow x^2-2x\left(y+z\right)+\left(y+z\right)^2\le\left(1-x\right)\left(y+z\right)^2\\ \Leftrightarrow x^2-2x\left(y+z\right)\le\left(y+z\right)^2\left(1-x-1\right)=-x\left(y+z\right)^2\\ \Leftrightarrow x-2\left(y+z\right)\le-\left(y+z\right)^2\\ \Leftrightarrow x\le\left(y+z\right)\left[2-\left(y+z\right)\right]\)
Đặt \(2-\left(y+z\right)=t\)
\(P=x\left(1-y\right)\left(1-z\right)\le x\left(\dfrac{1-y+1-z}{2}\right)^2=\dfrac{x\left[2-\left(y+z\right)\right]^2}{4}\\ \Leftrightarrow4P\le x\left[2-\left(y+z\right)\right]^2\le\left(y+z\right)\left[2-\left(y+z\right)\right]^3\\ \Leftrightarrow4P\le t^3\left(2-t\right)=\dfrac{27}{16}-\dfrac{\left(4t^2+4t+3\right)\left(2t-3\right)^2}{16}\)
Mà \(-\dfrac{\left(4t^2+4t+3\right)\left(2t-3\right)^2}{16}\le0\Leftrightarrow4P\le\dfrac{27}{16}\Leftrightarrow P\le\dfrac{27}{64}\)
Dấu \("="\Leftrightarrow x=\dfrac{3}{4};y=z=\dfrac{1}{4}\)
17 tháng 12 2023
a: \(2x^2+3xy-14y^2\)
\(=2x^2+7xy-4xy-14y^2\)
\(=\left(2x^2+7xy\right)-\left(4xy+14y^2\right)\)
\(=x\left(2x+7y\right)-2y\left(2x+7y\right)\)
\(=\left(2x+7y\right)\left(x-2y\right)\)
b: \(\left(x-7\right)\left(x-5\right)\left(x-3\right)\left(x-1\right)+7\)
\(=\left(x-7\right)\left(x-1\right)\left(x-5\right)\left(x-3\right)+7\)
\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)+7\)
\(=\left(x^2-8x\right)^2+15\left(x^2-8x\right)+7\left(x^2-8x\right)+105+7\)
\(=\left(x^2-8x\right)^2+22\left(x^2-8x\right)+112\)
\(=\left(x^2-8x\right)^2+8\left(x^2-8x\right)+14\left(x^2-8x\right)+112\)
\(=\left(x^2-8x\right)\left(x^2-8x+8\right)+14\left(x^2-8x+8\right)\)
\(=\left(x^2-8x+8\right)\left(x^2-8x+14\right)\)
c: \(\left(x-3\right)^2+\left(x-3\right)\left(3x-1\right)-2\left(3x-1\right)^2\)
\(=\left(x-3\right)^2+2\left(x-3\right)\left(3x-1\right)-\left(x-3\right)\left(3x-1\right)-2\left(3x-1\right)^2\)
\(=\left(x-3\right)\left[\left(x-3\right)+2\left(3x-1\right)\right]-\left(3x-1\right)\left[\left(x-3\right)+2\left(3x-1\right)\right]\)
\(=\left(x-3+6x-2\right)\left(x-3-3x+1\right)\)
\(=\left(7x-5\right)\left(-2x-2\right)\)
\(=-2\left(x+1\right)\left(7x-5\right)\)
d: \(xy\left(x-y\right)+yz\left(y-z\right)+zx\left(z-x\right)\)
\(=x^2y-xy^2+y^2z-yz^2+zx\left(z-x\right)\)
\(=\left(x^2y-yz^2\right)-\left(xy^2-y^2z\right)+xz\left(z-x\right)\)
\(=y\left(x^2-z^2\right)-y^2\left(x-z\right)-xz\left(x-z\right)\)
\(=y\cdot\left(x-z\right)\left(x+z\right)-\left(x-z\right)\left(y^2+xz\right)\)
\(=\left(x-z\right)\left(xy+zy-y^2-xz\right)\)
\(=\left(x-z\right)\left[\left(xy-y^2\right)+\left(zy-zx\right)\right]\)
\(=\left(x-z\right)\left[y\cdot\left(x-y\right)-z\left(x-y\right)\right]\)
\(=\left(x-z\right)\left(x-y\right)\left(y-z\right)\)
AH
Akai Haruma
Giáo viên
7 tháng 9 2021
Lời giải:
$2\text{VT}=2(x+y+z)-4(xy+yz+xz)+8xyz$
$=(2x-1)(2y-1)(2z-1)+1$
Do $x,y,z\in [0;1]$ nên $-1\leq 2x-1, 2y-1, 2z-1\leq 1$
$\Rightarrow (2x-1)(2y-1)(2z-1)\leq 1$
$\Rightarrow 2\text{VT}\leq 2$
$\Rightarrow \text{VT}\leq 1$
Ta có đpcm.
Dấu "=" xảy ra khi $(x,y,z)=(1,1,1), (0,0,1)$ và hoán vị.