a) x² - 5x - y² -5y

= ( x² - y² ) + ( -5x - 5y)

= ( x - y ) ( x + y) - 5( x + y )

= ( x + y ) ( x - y -5)

b) x³ + 2x² - 4x - 8

= x² ( x + 2 ) - 4 ( x + 2 )

= ( x +2 ) ( x² -4 )

= ( x+2)² ( x-2)

Bai 2 : 

a, \(A=\left(x+3\right)^2+\left(x-2\right)^2-2\left(x+3\right)\left(x-2\right)\)

\(=x^2+6x+9+x^2-4x+4-2\left(x^2-2x+3x-6\right)\)

\(=2x^2+2x+13-2x^2-2x+12=25\)

b, \(B=\left(x-2\right)^2-x\left(x-1\right)\left(x-3\right)+3x^2-9x+8\)

\(=x^2-4x+4-x\left(x^2-3x-x+3\right)+3x^2-9x+8\)

\(=4x^2-13x+12-x^3+4x^2-3x=-16x+12-x^3\)

\(M=\left(\frac{1}{a^2-a}+\frac{1}{a-1}\right):\frac{a+1}{a^2-2a+1}\)

\(M=\left(\frac{1}{a\left(a-1\right)}+\frac{1}{a-1}\right):\frac{a+1}{\left(a-1\right)^2}\)ĐKXĐ : a khác 0, a khác 1

\(M=\frac{1+a}{a\left(a-1\right)}.\frac{\left(a-1\right)^2}{a+1}\)

\(M=\frac{a-1}{a}\)

\(M=\left(\frac{1}{a^2-a}+\frac{1}{a-1}\right):\frac{a+1}{a^2-2a+1}\)DK : \(x\ne0;\pm1\)

\(=\left(\frac{1}{a\left(a-1\right)}+\frac{1}{a-1}\right):\frac{a+1}{\left(a-1\right)^2}=\left(\frac{1}{a\left(a-1\right)}+\frac{a}{a\left(a-1\right)}\right):\frac{a+1}{\left(a-1\right)^2}\)

\(=\frac{a+1}{a\left(a-1\right)}.\frac{\left(a-1\right)^2}{a+1}=\frac{a-1}{a}\)

x² + 5y² - 4xy + 6x - 14y + 10 = 0

=> (x² - 4xy + 4y²) + (6x - 12y) + 9 + (y² - 2y + 1) = 0

=> (x - 2y)² + 6(x - 2y) + 9 + (y - 1)² = 0

=> (x - 2y + 3)² + (y - 1)² = 0

=> \(\hept{\begin{cases}x-2y+3=0\\y-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}\)

Vậy x = 1 ; y = - 1 là giá trị cần tìm

làm nhanh giùm mình nha ! đang cần gấp <:)

Ta có : \(4x^2+2y^2+2z^2-4xy-4zx+2yz-6y-10z+34=0\)

\(\Rightarrow\left(4x^2+y^2+z^2-4xy-4zx+2yz\right)+\left(y^2-6y+9\right)+\left(z^2-10z+25\right)=0\)

\(\Rightarrow\left(2x-y-z\right)^2+\left(y-3\right)^2+\left(z-5\right)^2=0\)

Vì \(\hept{\begin{cases}\left(2x-y-z\right)^2\ge0\forall x,y,z\\\left(y-3\right)^2\ge0\forall y\\\left(z-5\right)^2\ge0\forall z\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\left(2x-y-z\right)^2=0\\\left(y-3\right)^2=0\\\left(z-5\right)^2=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x-y-z=0\\y-3=0\\z-5=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x-3-5=0\\y=3\\z=5\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x=8\\y=3\\z=5\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=4\\y=3\\z=5\end{cases}}\left(1\right)\)

Lại có : \(S=\left(x-4\right)^{2017}+\left(y-4\right)^{2017}+\left(z-4\right)^{2017}\)

Thay \(\left(1\right)\)vào \(S\),ta được :

\(S=0^{2017}+\left(-1\right)^{2017}+1^{2017}\)

    \(=0-1+1=0\)

Vậy \(S=0\)