Question

giải phương trình sau

a) \(\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)

Answer 1

\(\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)

=> \(\dfrac{3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)}{15}\) \(=\dfrac{7x^2-14x-5}{15}\)

=> \(\dfrac{7x^2+22x-2}{15}=\dfrac{7x^2-14x-5}{15}\)

=> \(\dfrac{7x^2+22x-2-7x^2+14x+5}{15}=0\)

=>\(\dfrac{36x+3}{15}=0\)

=> 36x=-3

=>\(x=\dfrac{-1}{12}\)

Vay \(S=\left\{\dfrac{-1}{12}\right\}\)

Answer 2

Phương trình tương đương: \(3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)=7x^2-14x-5\\ \Leftrightarrow7x^2+22x+3=7x^2-14x\\ \Leftrightarrow22x+3=-14x\\ \Leftrightarrow22x+14x=-3\\ \Leftrightarrow36x=-3\\ \Leftrightarrow x=-\dfrac{1}{12}\)

Vậy \(S=\left\{-\dfrac{1}{2}\right\}\)