Với \(a\ne0,a+b\ne0\), hãy giải phương trình \(\frac{3abc}{a+b}+\frac{a^2b^2}{\left(a+b\right)^3}+\frac{\left(2a+b\right)b^2x}{a\left(a+b\right)^2}=3cx+\frac{bx}{a}\) .
\(x=\frac{a+b}{ab}\) \(x=\frac{ab}{a+b}\) \(x=\frac{a-b}{ab}\) \(x=\frac{ab}{a-b}\) Hướng dẫn giải:Biến đổi tương đương phương trình đã cho:
\(\dfrac{3ac}{a+b}+\dfrac{a^2b^2}{\left(a+b\right)^2}+\dfrac{\left(2a+b\right)b^2x}{a\left(a+b\right)^2}=3cx+\dfrac{bx}{a}\) \(\Leftrightarrow\dfrac{3ac\left(a+b\right)^2+b\left(a+b\right)^2-\left(2a+b\right)b^2}{a\left(a+b\right)^2}x=\dfrac{ab[3c\left(a+b\right)^2+ab]}{\left(a+b\right)^3}\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2\left(3ac+b\right)-\left(2a+b\right)b^2}{a}x=\dfrac{ab[3c\left(a+b\right)^2+ab]}{\left(a+b\right)}\)\(\Leftrightarrow[3c\left(a+b\right)^2+ab]x=\dfrac{ab[3c\left(a+b\right)^2+ab]}{\left(a+b\right)}\)\(\Leftrightarrow x=\dfrac{ab}{a+b}\).
Đáp số: \(x=\dfrac{ab}{a+b}\)