Question

tìm x

a)\(5x\left(\frac{1}{5}x-2\right)+3\left(6-\frac{1}{3}x^2\right)=12\)

b)\(7x\left(x-2\right)-5\left(x-1\right)=7x^2+3\)

Answer 1

a) Ta có: \(5x\left(\frac{1}{5}x-2\right)+3\left(6-\frac{1}{3}x^2\right)=12\)

\(\Leftrightarrow x^2-10x+18-x^2=12\)

\(\Leftrightarrow-10x+18=12\)

\(\Leftrightarrow-10x=-6\)

hay \(x=\frac{3}{5}\)

Vậy: \(x=\frac{3}{5}\)

b) Ta có: \(7x\left(x-2\right)-5\left(x-1\right)=7x^2+3\)

\(\Leftrightarrow7x^2-14x-5x+5-7x^2-3=0\)

\(\Leftrightarrow-19x+2=0\)

\(\Leftrightarrow-19x=-2\)

hay \(x=\frac{2}{19}\)

Vậy: \(x=\frac{2}{19}\)