Question

Tìm x,biết

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

b, x+(a+b)\(^2\)=\(\dfrac{a^4+b^4}{\left(a-b\right)^2}\)

Answer 1

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

\(\Leftrightarrow x=\dfrac{2a\left(a-b\right)}{b\left(b-a\right)}+\dfrac{3a+b}{b}\)

\(\Leftrightarrow x=-2a+\dfrac{3a+b}{b}=\dfrac{-2ab+3a+b}{b}\)

b: \(x+\left(a+b\right)^2=\dfrac{a^4+b^4}{\left(a-b\right)^2}\)

\(\Leftrightarrow x=\dfrac{a^4+b^4}{\left(a-b\right)^2}-\left(a+b\right)^2\)

\(\Leftrightarrow x=\dfrac{a^4+b^4-\left(a^2-b^2\right)^2}{\left(a-b\right)^2}=\dfrac{a^4+b^4-a^4+2a^2b^2-b^4}{\left(a-b\right)^2}=\dfrac{2a^2b^2}{\left(a-b\right)^2}\)