Question

Nếu \(x^2+x\left(6-2x\right)=3x\left(x+1\right)-4\left(x^2-1\right)\) thì \(x=\)

Answer 1

Ta có: x² + x(6 - 2x) = 3x(x + 1) - 4(x² - 1)

=> x² + 6x - 2x² = 3x² + 3x - 4x² + 4

=> 3x = 4

=> x = 4/3

Answer 2

\(x^2+x\left(6-2x\right)=3x\left(x+1\right)-4\left(x^2-1\right)\)

\(\Leftrightarrow x^2+6x-2x^2=3x^2+3x-4x^2+4\)

\(\Leftrightarrow3x=4\)

\(\Leftrightarrow x=\frac{4}{3}\)

Vậy phương trình có nghiệm \(x=\frac{4}{3}\)