a, \(M=7.\left(x-y\right)+4a.\left(x-y\right)-5\)

Theo bài ra ta có: x-y=0

=> \(M=0+0-5\)

\(\Rightarrow M=-5\)

b,

\(N=\left(x^2+y^2\right).\left(x-y\right)+3\)

\(\Rightarrow N=0+3=3\)

lớp 7 lên 8 à làm quen nhá :)

a) \(M=7x-7y+4ax-4ay-5\)

\(M=7\left(x-y\right)+4a\left(x-y\right)-5\)

\(M=0+0-5=-5\)

b) \(N=x\left(x^2+y^2\right)-y\left(x^2+y^2\right)+3\)

\(N=\left(x-y\right)\left(x^2+y^2\right)+3\)

\(N=0+3=3\)

  Ta có   \(x-y=0\) 

a)    M = 7x - 7y +4ax - 4ay - 5

           = 7( x - y )+ 4a( x - y ) - 5

           = 7.0 + 4a . 0 - 5

           = -5

b)  N = x( x²+y² ) - y( x² + y² ) + 3

         =  ( x - y ).( x² + y² ) +3

         = 0. ( x² + y² ) + 3

         = 3

a: \(M=7\left(x-y\right)+4a\left(x-y\right)-5=-5\)

b: \(N=x^3+xy^2-x^2y-y^3+3\)

\(=\left(x-y\right)^3+3xy\left(x-y\right)+xy\left(y-x\right)+3\)

=3

\(Ta có: x-y=0 => x=y \)

\(Khi đó, ta có: \)

\(a) M=7x-7y+4ax-4ay-5 =(7x-7x) +(4ax-4ax)-5 \)

\(M=0+0-5=-5 \)

\(Vậy M=-5\)

\(b) N=x(x^2+y^2)-y(x^2+y^2)+3\)

\(=> N=[x(x^2+x^2)-x(x^2+x^2)]+3\)

\(=> N=0+3=3\)

\(Vậy N=3\)

M=7x-7y+4ax-4ay-5

=7.(x-y)+4a(x-y)-5 (1)

thay x-y=0 vào (1) ta được:

7.0+4a.0-5

=-5

vậy với x-y=0 thì M=-5

câu 2 đề sai đúng là

N=x(x²+y²)-y(x²+y²)+3

=(x²+y²)(x-y)+3 (2)

thay x-y=0 vào (2) ta được:

(x²+y²).0+3

=3

vậy với x-y=0 thì N=3

\(M=7x-7y+4ax-4ay-5=\left(7x-7y\right)+\left(4ax-4ay\right)-5\)=7(x-y)+4a(x-y)-5

Do x-y=0

=>M=0+0-5=-5

N=\(x\left(x^2+y^2\right)-y\left(x^2+y^2\right)+3=\left(x^2+y^2\right)\left(x-y\right)+3\)

Do x-y=0

=>N=0+3=3

\(ĐK:x\ne y;x\ne-y;x^2+xy+y^2\ne0;x^2-xy+y^2\ne0\)

\(A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\left[1:\dfrac{\left(x^3+y^3\right)\left(x^2+y^2\right)}{\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2+y^2\right)}\right]\\ A=\dfrac{x^2-xy+y^2}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)\left(x^2+y^2\right)}\\ A=x-y=B\)

\(x=0;y=0\Leftrightarrow B=0\)

Giá trị của A không xác định vì \(x=y\) trái với ĐK:\(x\ne y\)

Vậy \(A\ne B\)

bài 11

a) \(x^2-xy+x\\ =x\left(x-y+1\right)\)

b)

\(x^2-2xy-4+y^2\\ =\left(x^2-2xy+y^2\right)-4\\ =\left(x-y\right)^2-4\\ =\left(x-y-2\right)\left(x-y+2\right)\)

c)

\(x^3-x^2-16x+16\\ =x^2\left(x-1\right)-16\left(x-1\right)\\ =\left(x-1\right)\left(x-4\right)\left(x+4\right)\)

bài 12

\(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(2x^2-10x-3x-2x^2=26\)

\(-13x=26\\ x=-2\)

b)

\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

M=7x-7y+4ax-4ay-5

M=7(x-y)+4a(x-y)-5

M=0+0-5=-5.

N=x(x^2+y^2)-y(x^2+y^2)+3

N=(x-y)(x^2+y^2)+3=0+3=3

a, M = 7x - 7y + 4ax - 4ay - 5

=> M = 7(x - y) + 4a(x - y) - 5

=> M = 0 + 0 - 5 = -5

b, N = x (x² + y²) - y ( x² + y²) + 3 

=> N = (x - y)(x² + y²) + 3 = 0 + 3 = 3