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