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AH
Akai Haruma
Giáo viên
7 tháng 10 2021
Lời giải:
$P=(xy+yz+xz)^2+(x^2-yz)^2+(y^2-zx)^2+(z^2-xy)^2$
$=x^2y^2+y^2z^2+z^2x^2+2x^2yz+2xy^2z+2xyz^2+x^4+y^2z^2-2x^2yz+y^4+z^2x^2-2xzy^2+z^4+x^2y^2-2xyz^2$
$=x^4+y^4+z^4+2x^2y^2+2y^2z^2+2z^2x^2$
$=(x^2+y^2+z^2)^2=10^2=100$
5 tháng 7 2019
Xét tích : \(\left[x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)\right]\left(x+y+z\right)\)
=\(x^3\left(z-y\right)+x^2\left(z-y\right)\left(z+y\right)+y^3\left(x-z\right)+y^2\left(x-z\right)\left(x+z\right)\)
\(+z^3\left(y-x\right)+z^2\left(y-x\right)\left(y+x\right)\)
\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)+x^2\left(z^2-y^2\right)+y^2\left(x^2-z^2\right)+z^2\left(y^2-x^2\right)\)
\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)+x^2z^2-x^2y^2+y^2x^2-y^2z^2+z^2y^2-z^2x^2\)
\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)\)
Như vậy:
\(\left[x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)\right]\left(x+y+z\right)\)\(=x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)\)
<=> \(\frac{x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)}{x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)}=x+y+z\)
Ta có: \(\frac{\frac{x^2\left(z-y\right)}{yz}+\frac{y^2\left(x-z\right)}{xz}+\frac{z^2\left(y-x\right)}{xy}}{\frac{x\left(z-y\right)}{yz}+\frac{y\left(x-z\right)}{xz}+\frac{z\left(y-x\right)}{xy}}\)
\(=\frac{\frac{x^3\left(z-y\right)}{xyz}+\frac{y^3\left(x-z\right)}{xyz}+\frac{z^3\left(y-x\right)}{xyz}}{\frac{x^2\left(z-y\right)}{xyz}+\frac{y^2\left(x-z\right)}{xyz}+\frac{z^2\left(y-x\right)}{xyz}}\)
\(=\frac{x^3\left(z-y\right)+y^3\left(x-z\right)+z^3\left(y-x\right)}{x^2\left(z-y\right)+y^2\left(x-z\right)+z^2\left(y-x\right)}=x+y+z\)
2 tháng 7 2021
a) xy(x + y) + yz(y + z) + xz(z + x) + 3xyz
= xy(X + y + z) + yz(x + y + z) + xz(X + y + z)
= (x + y +z)(xy + yz+ xz)
b) xy(x + y) - yz(y + z) - xz(z - x)
= x2y + xy2 - y2z - yz2 - xz2 + x2z
= x2(y + z) - yz(y + z) + x(y2 - z2)
= x2(y + z) - yz(y + z) + x(y + z)(y - z)
= (y + z)(x2 - yz + xy - xz)
= (y + z)[x(x + y) - z(x + y)]
= (y + z)(x + y)(x - z)
c) x(y2 - z2) + y(z2 - x2) + z(x2 - y2)
= x(y - z)(y + z) + yz2 - yx2 + x2z - y2z
= x(y - z)(y + z) - yz(y - z) - x2(y - z)
= (y - z)((xy + xz - yz - x2)
= (y - z)[x(y - x) - z(y - x)]
= (y - z)(x - z)(y -x)
17 tháng 7 2020
\(\frac{x^2-yz}{\left(x+y\right)\left(x+z\right)}=\frac{x^2+xy}{\left(x+y\right)\left(x+z\right)}-\frac{xy+yz}{\left(x+y\right)\left(x+z\right)}=\frac{x}{x+z}-\frac{y}{x+y}\)
Tương tự:\(\frac{y^2-zx}{\left(y+z\right)\left(y+x\right)}=\frac{y}{x+y}-\frac{z}{y+z};\frac{z^2-xy}{\left(z+x\right)\left(z+y\right)}=\frac{z}{z+y}-\frac{x}{z+x}\)
Khi đó:
\(\frac{x^2-yz}{\left(x+y\right)\left(x+z\right)}+\frac{y^2-zx}{\left(y+z\right)\left(y+x\right)}+\frac{z^2-xy}{\left(z+x\right)\left(z+y\right)}=0\)
18 tháng 11 2017
\(M=\dfrac{x\left(yz-x^2\right)+y\left(zx-y^2\right)+z\left(xy-z^2\right)}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{xyz-x^3+xyz-y^3+xyz-z^3}{x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2}\)
\(=\dfrac{-\left(x^3+y^3+z^3-3xyz\right)}{2x^2+2y^2+2z^2-2xy-2yz-2zx}\)
\(=\dfrac{-\left(x^3+3x^2y+3xy^2+y^3+z^3-3x^2y-3xy^2-3xyz\right)}{2\left(x^2+y^2+z^2-xy-yz-zx\right)}\)
\(=\dfrac{-\left[\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)\right]}{2\left(x^2+y^2+z^2-xy-yz-zx\right)}\)
\(=\dfrac{-\left\{\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\right\}}{2\left(x^2+y^2+z^2-xy-yz-zx\right)}\)
\(=\dfrac{-\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)}{2\left(x^2+y^2+z^2-xy-yz-zx\right)}\)
\(=\dfrac{-\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)}{2\left(x^2+y^2+z^2-xy-yz-zx\right)}=\dfrac{-x-y-z}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 1 2019
\(A=\dfrac{x^2-yz}{\left(x+y\right)\left(x+z\right)}+1+\dfrac{y^2-xz}{\left(y+z\right)\left(y+x\right)}+1+\dfrac{z^2-xy}{\left(z+x\right)\left(z+y\right)}+1-3\)
Xét \(\dfrac{x^2-yz}{\left(x+y\right)\left(x+z\right)}+1=\dfrac{x^2-yz+x^2+xz+xy+yz}{\left(x+y\right)\left(x+z\right)}\)
\(=\dfrac{x^2+xy+x^2+xz}{\left(x+y\right)\left(x+z\right)}=\dfrac{x\left(x+y\right)+x\left(x+z\right)}{\left(x+y\right)\left(x+z\right)}=\dfrac{x}{x+y}+\dfrac{x}{x+z}\)
Tương tự: \(\left\{{}\begin{matrix}\dfrac{y^2-zx}{\left(y+z\right)\left(y+x\right)}+1=\dfrac{y}{y+z}+\dfrac{y}{y+x}\\\dfrac{z^2-xy}{\left(z+y\right)\left(z+x\right)}+1=\dfrac{z}{z+y}+\dfrac{z}{z+x}\end{matrix}\right.\)
Cộng vế với vế ta được:
\(A=\dfrac{x}{x+y}+\dfrac{x}{x+z}+\dfrac{y}{y+x}+\dfrac{y}{y+z}+\dfrac{z}{z+x}+\dfrac{z}{z+y}-3\)
\(A=\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{z+x}{z+x}-3=1+1+1-3=0\)