Question

\(\frac{-22}{15}\)x + \(\frac{1}{3}\)= l \(\frac{-2}{3}\) + \(\frac{1}{5}\)|

Tìm x, dấu l là trị tuyệt đối nha

Answer 1

\(\frac{-22}{15}x+\frac{1}{3}=\left|\frac{-2}{3}+\frac{1}{5}\right|\)

=>\(\frac{-22}{15}x+\frac{1}{3}=\left|\frac{-7}{15}\right|\)

=>\(\frac{-22}{15}x+\frac{1}{3}=\frac{7}{15}\)

=>\(\frac{-22}{15}x=\frac{7}{15}-\frac{1}{3}\)

=>\(\frac{-22}{15}x=\frac{2}{15}\)

=>\(x=\frac{2}{15}:\frac{-22}{15}\)

=>\(x=\frac{1}{-11}\)

Answer 2

có 2 trường hợp:

*\(\frac{-22}{15}\)x +\(\frac{1}{3}\)=\(\frac{-7}{15}\)  HOẶC     \(\frac{-22}{15}\)x +\(\frac{1}{3}\)=\(\frac{7}{15}\)

Trường hợp 1: x=\(\frac{6}{11}\)                     Trường hợp 2: x=\(\frac{-1}{11}\)