Question

[Yến BTC Nguyễn]

làm hộ tớ bài 10 với

Answer 1

[Đào Tùng Dương]

\(1,x-\dfrac{2}{7}=\dfrac{4}{3};x=\dfrac{4}{3}+\dfrac{2}{7}=\dfrac{34}{21}\)

\(2,x+\dfrac{3}{4}=\dfrac{6}{8};x=\dfrac{6}{8}-\dfrac{3}{4}=0\)

\(3,\left|x\right|-\dfrac{3}{4}=\dfrac{2}{3}+\dfrac{1}{4}=\dfrac{11}{12};\left|x\right|=\dfrac{11}{12}+\dfrac{3}{4};\left|x\right|=\dfrac{5}{3}\Rightarrow x=\left[{}\begin{matrix}\dfrac{-5}{3}\\\dfrac{5}{3}\end{matrix}\right.\)

\(4,\left(x+1\right).3=4.5=20;x+1=\dfrac{20}{3}\Leftrightarrow x=\dfrac{17}{3}\)

Answer 2

[Yến BTC Nguyễn]

thanks