Question

Chứng minh đăng thức:

\(\dfrac{x^2+ax+ab+bx}{3bx-a^2-ax+3ab}=\dfrac{x+b}{3b-a}\)

Answer 1

- Xét tử:

\(x^2+ax+ab+bx\)

\(=x\left(x+a\right)+b\left(x+a\right)\)

\(=\left(x+a\right)\left(x+b\right)\)

- Xét mẫu:

\(3bx-a^2-ax+3ab\)

\(=3bx+3ab-a^2-ax\)

\(=3b\left(x+a\right)-a\left(a+x\right)\)

\(=\left(x+a\right)\left(3b-a\right)\)

Vậy \(\dfrac{x^2+ax+ab+bx}{3bx-a^2-ax+3ab}=\dfrac{\left(x+a\right)\left(x+b\right)}{\left(x+a\right)\left(3b-a\right)}=\dfrac{x+b}{3b-a}\) với \(x\ne-a\)

Answer 2

Ta có: \(VT=\dfrac{x^2+ax+ab+bx}{3bx-a^2-ax+3ab}\)

\(=\dfrac{x\left(a+x\right)+b\left(a+x\right)}{3b\left(x+a\right)-a\left(a+x\right)}\)

\(=\dfrac{\left(b+x\right)\left(a+x\right)}{\left(3b-a\right)\left(a+x\right)}=\dfrac{b+x}{3b-a}=VP\)

\(\Rightarrowđpcm\)