Cho (x+ y +z)( xy + yz +xz) = xyz
CMR x\(^{2017}\)+ y\(^{2017}\)+z \(^{2017}\)= (x+ y + z)\(^{2017}\)
\(Cho\left(x+y+z\right)\left(xy+yz+xz\right)=xyz\)
\(CMR:x^{2017}+y^{2017}+z^{2017}=\left(x+y+z\right)^{2017}\)
cho ( x +y +z)( xy +yz +xz) =xyz. CHỨNg minh rằng:
x2017 + y2017 +z2017 = (x+y+z)2017
THanks for your help!!!!!~~~~~
cho (x+y+z) (xy+yz+zx)=xyz .CMR:
x^2017+y^2017+z^2017= (x+y+z)^2017
\(\left(x+y+z\right)\left(xy+yz+zx\right)=xyz\\ \Leftrightarrow\left(x+y+z\right)\left(xy+yz+zx\right)-xyz=0\\ \Leftrightarrow3\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-y\\y=-z\\z=-x\end{matrix}\right.\)
\(\forall x=-y\Leftrightarrow VT=-y^{2017}+y^{2017}+z^{2017}=z^{2017}=\left(-y+y+z\right)^{2017}=VP\\ \forall y=-z\Leftrightarrow VT=x^{2017}-z^{2017}+z^{2017}=x^{2017}=\left(x-z+z\right)^{2017}=VP\\ \forall z=-x\Leftrightarrow VT=x^{2017}+y^{2017}-x^{2017}=y^{2017}=\left(x+y-x\right)^{2017}=VP\)
Vậy ta đc đpcm
cho (x+y+z)(xy+yz+zx)= xyz
chứng minh: x2017+y2017+z2017= (x+y+z)2017
1)Tìm MinB
B = x2 - 4xy + 5y2+10x - 2xy + 2042
2)Chứng minh
x2017+y2017+z2017=(x+y+z)2017
Biết (x+ y + z)(xy + yz +xz) = xyz
Cho (x+y+z)(xy+yz+zx)=xyz CM: x2017 + y2017 + z2017
Cho ba số thực dương x, y, z thỏa mãn: xy+yz+zx=2017. chứng minh : \(\sqrt{\dfrac{yz}{x^2+2017}}+\sqrt{\dfrac{zx}{y^2+2017}}+\sqrt{\dfrac{xy}{z^2+2017}}\le\dfrac{3}{2}\)
Ta có:\(\sqrt{\dfrac{yz}{x^2+2017}}=\sqrt{\dfrac{yz}{x^2+xy+yz+zx}}=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}\)
\(=\sqrt{\dfrac{y}{x+y}\cdot\dfrac{z}{x+z}}\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}}{2}\)
Tương tự ta có:\(\sqrt{\dfrac{zx}{y^2+2017}}\le\dfrac{\dfrac{x}{x+y}+\dfrac{z}{y+z}}{2}\)
\(\sqrt{\dfrac{xy}{z^2+2017}}\le\dfrac{\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)
Cộng vế với vế ta có:
\(\sqrt{\dfrac{yz}{x^2+2017}}+\sqrt{\dfrac{zx}{y^2+2017}}+\sqrt{\dfrac{xy}{z^2+2017}}\)
\(\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}+\dfrac{z}{z+y}+\dfrac{x}{x+y}+\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)
\(=\dfrac{\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{z+x}{z+x}}{2}=\dfrac{1+1+1}{2}=\dfrac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{\sqrt{2017}}{\sqrt{3}}\)
Cho \(\left(x+y+z\right)\left(xy+yz+z\right)=xyz\)
Chứng minh rằng: \(x^{2017}+y^{2017}+z^{2017}=\left(x+y+z\right)^{2017}\)
Ta có:
\(\left(x+y+z\right)\left(xy+yz+zx\right)=xyz\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow x=-y;y=-z;z=-x\)
Với \(x=-y\)
\(\Rightarrow x^{2017}+y^{2017}+z^{2017}=z^{2017}=\left(x+y+z\right)^{2017}\)
Tương tự cho 2 trường hợp còn lại
Cho x + y + z = 0 và xy + yz + xz = 0. Tính S= (x - 1)^2015 + (y - 1)^2016 + ( z + 1)^2017
(x+y+z)^2=0
x^2+y^2+z^2+2xy +2yz+2xz=0
x^2+y^2+z^2+2(xy+yz+xz)=0
Vì xy + yz +xz=0 nên x^2+y^2+z^2=0.
Vì x^2, y^2, z^2 luôn lớn hơn hoặc bằng 0 mà x^2+y^2+z^2=0.Vì vậy:
x^2=0, y^2=0, z^2=0
x=y=z=0
Thay x=y=z=o vào S ta được: S=1