a)Đk:\(x^2-y^2\ne0\Rightarrow\left(x-y\right)\left(x+y\right)\ne0\)\(\Rightarrow\left\{\begin{matrix}x\ne y\\x\ne-y\end{matrix}\right.\)
b)\(A=\frac{x^2+2x-y^2-2y}{x^2-y^2}=\frac{x^2+xy+2x-xy-y^2-2y}{x^2-y^2}\)
\(\frac{x\left(x+y+2\right)-y\left(x+y+2\right)}{\left(x-y\right)\left(x+y\right)}=\frac{\left(x-y\right)\left(x+y+2\right)}{\left(x-y\right)\left(x+y\right)}=\frac{x+y+2}{x+y}\)
c)Khi \(\left\{\begin{matrix}x=5\\y=6\end{matrix}\right.\) thay vào A ta có:
\(A=\frac{x+y+2}{x+y}=\frac{5+6+2}{5+6}=\frac{13}{11}\)
cacs bn làm ra giấy rồi chụp cũng ddcj,lm ơn đó , tối nay mk đi học thêm rồi
a)ĐKXĐ : x \(\ne\)\(\pm\)y
b)Rút gọn\(\dfrac{x^2+2x-y^2-2y}{x^2-y^2}=\dfrac{x^2+xy+2x-xy-y^2-2y}{\left(x-y\right)\left(x+y\right)}=\dfrac{x\left(x+y+2\right)-y\left(x+y+2\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x+y+2\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{x+y+2}{x+y}\)
c)Thay x=5, y=6 vào BT A ta có :
A=\(\dfrac{5+6+2}{5+6}=\dfrac{13}{11}\)
