a/ \(A=x.\left(x+y\right)-y.\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y\right)\)
\(=x^2-y^2\)
Thế x = 0 , y = 10 vào A ta có:
\(A=0^2-10^2=-100\)
b/ \(B=\left(x+2\right)^2+\left(x-2\right)^2-2\left(x+2\right)\left(x-2\right)\)
\(=\left[\left(x+2\right)-\left(x-2\right)\right]^2\)
\(=\left(x+2-x+2\right)^2\)
\(=4^2=16\)
a, \(A=x\left(x+y\right)-y\left(x+y\right)=\left(x-y\right)\left(x+y\right)=x^2-y^2\)
Thay \(x=0;y=10\) vào A ta được:
\(x^2-y^2=0^2-10^2=0-100=-100\)
b, \(B=\left(x+2\right)^2+\left(x-2\right)^2-2\left(x+2\right)\left(x-2\right)\)
\(=\left[x+2-\left(x-2\right)\right]^2=\left(0x+4\right)^2=0x+16\)
Thay x = -4 vào đa thức ta được:
\(0x+16=-4.0+16=16\)