Question

1) | x+4\5 | + 7\5 = 3\5

2) | x + 2 |= x-3

3) 5- | x - 2\3| - 2= x

4) |x- 4 | = | 2x+3 |

5) (x - 1\3 )^3 = -1\27

6) (x + 1\2)^2 = 5\5

7) | (- 2\2\3)^2 - x | - 1\3 = 0

Answer 1

1) \(\left|x+\frac{4}{5}\right|+\frac{7}{5}=\frac{3}{5}\)

\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{3}{5}-\frac{7}{5}\)

\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{-4}{5}\)

\(x+\frac{4}{5}=\pm\frac{4}{5}\)

\(TH1:x+\frac{4}{5}=\frac{4}{5}\)

\(\Rightarrow x=\frac{4}{5}-\frac{4}{5}=0\)

\(TH2:x+\frac{4}{5}=\frac{-4}{5}\)

\(\Rightarrow x=\frac{-4}{5}-\frac{4}{5}=\frac{-8}{5}\)

Vậy x ∈ {0; \(\frac{-8}{5}\)}

Answer 2

Hai câu cuối khó nhìn nên ko giải

Answer 3

5,\(\left(x-\frac{1}{3}\right)^3=-\frac{1}{27}\)

Có \(-\frac{1}{27}=\left(-\frac{1}{3}\right)^3\)

\(\Rightarrow\left(x-\frac{1}{3}\right)^3=\left(-\frac{1}{3}\right)^3\)

TH1: \(\left(x-\frac{1}{3}\right)^3=\left(-\frac{1}{3}\right)^3\)

\(\Rightarrow x-\frac{1}{3}=-\frac{1}{3}\)

\(x=-\frac{1}{3}+\frac{1}{3}\Rightarrow x=0\)

Vậy x = 0