Question

bài 1.Rút gọn :

A=\(\frac{2a^3b^5}{3a^3b^2}\)

B=\(\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}\)

Answer 1

Ta có: \(A=\frac{2a^3b^5}{3a^3b^2}=\frac{2b^3}{3}\)

Ta có:

\(B=\frac{x^2+y^2-z^2+2xy}{x^2-y^2+z^2+2xz}\)

\(=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}\)

\(=\frac{\left(x+y-z\right)\left(x+y+z\right)}{\left(x-y+z\right)\left(x+y+z\right)}\)

\(=\frac{x+y-z}{x-y+z}\)

Answer 2

A= \(\frac{2b^3}{3}\)

B= \(\frac{\left(x^2+2xy+y^2\right)-z^2}{\left(x^2+2xz+z^2\right)-y^2}=\frac{\left(x+y\right)^2-z^2}{\left(x+z\right)^2-y^2}=\frac{\left(x+y+z\right)\left(x+y-z\right)}{\left(x+z+y\right)\left(x+z-y\right)}=\frac{x+y-z}{x+z-y}\)