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a: Ta có: \(25x^2\left(x-y\right)-x+y\)
\(=\left(x-y\right)\left(25x^2-1\right)\)
\(=\left(x-y\right)\left(5x-1\right)\left(5x+1\right)\)
b: Ta có: \(16x^2\left(z^2-y^2\right)-z^2+y^2\)
\(=\left(z^2-y^2\right)\left(16x^2-1\right)\)
\(=\left(z-y\right)\left(z+y\right)\left(4x-1\right)\left(4x+1\right)\)
c: Ta có: \(x^3+x^2y-x^2z-xyz\)
\(=x^2\left(x+y\right)-xz\left(x+y\right)\)
\(=x\left(x+y\right)\left(x-z\right)\)
x2+y2−z22xy−y2+z2−x22yz+z2+x2−y22xz=1x2+y2−z22xy−y2+z2−x22yz+z2+x2−y22xz=1
Tính P = x + y + z
\(=x^2y-x^2z+y^2\left(z-x\right)+xz^2-yz^2\\ =\left(x^2y-yz^2\right)-\left(x^2z-xz^2\right)-y^2\left(x-z\right)\\ =y\left(x-z\right)\left(x+z\right)-xz\left(x-z\right)-y^2\left(x-z\right)\\ =\left(x-z\right)\left(xy+yz-xz-y^2\right)\\ =\left(x-z\right)\left[x\left(y-z\right)+y\left(z-y\right)\right]\\ =\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
\(x^2+y^2+z^2=x\left(y+z\right)\Rightarrow2x^2+2y^2+2z^2=2xy+2xz\)
\(\Rightarrow2x^2+2y^2+2z^2-2xy-2xz=0\)
\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)+y^2+z^2=0\)
\(\Rightarrow\left(x-y\right)^2+\left(x-z\right)^2+y^2+z^2=0\)
Vì \(\left(x-y\right)^2\ge0\forall x,y\)
\(\left(x-z\right)^2\ge0\forall x,z\)
\(y^2\ge0\forall y\)
\(z^2\ge0\forall z\)
\(\Rightarrow\left(x-y\right)^2+\left(x-z\right)^2+y^2+z^2\ge0\forall x,y,z\)
Dấu = xảy ra <=>\(\hept{\begin{cases}x=y\\x=z\\y=0;z=0\end{cases}}\)
=> x=y=z=0 là nghiệm của pt