Tìm x, biết: a) -3/5x = 12/25; b) 3/5x - 3/4 = -1 1/2; c) 2/5 + 3/5 : x = 0,5;

* Hướng dẫn giải

a) $-\frac{3}{5}x=\frac{12}{25}$

$x=\frac{12}{25}:\left(-\frac{3}{5}\right)$

$x=\frac{12}{25}.\frac{-5}{3}$

$x=\frac{-4}{5}$

Vậy $x=\frac{-4}{5}$

b) $\frac{3}{5}x-\frac{3}{4}=-1\frac{1}{2}$

$\frac{3}{5}x-\frac{3}{4}=\frac{-3}{2}$

$\frac{3}{5}x=\frac{-3}{2}+\frac{3}{4}$

$\frac{3}{5}x=\frac{-6}{4}+\frac{3}{4}$

$\frac{3}{5}x=\frac{-3}{4}$

$x=\frac{-3}{4}:\frac{3}{5}$

$x=\frac{-3}{4}.\frac{5}{3}$

$x=\frac{-5}{4}$

Vậy $x=\frac{-5}{4}$

c) $\frac{2}{5}+\frac{3}{5}:x=0,5$

$\frac{2}{5}+\frac{3}{5}:x=\frac{1}{2}$

$\frac{3}{5}:x=\frac{1}{2}-\frac{2}{5}$

$\frac{3}{5}:x=\frac{5}{10}-\frac{4}{10}$

$\frac{3}{5}:x=\frac{1}{10}$

$x=\frac{3}{5}:\frac{1}{10}$

$x=\frac{3}{5}.\frac{10}{1}$

x = 6

Vậy x = 6

d) $\frac{3}{4}-\left(x-\frac{1}{2}\right)=1\frac{2}{3}$

$\frac{3}{4}-\left(x-\frac{1}{2}\right)=\frac{5}{3}$

$$\begin{gathered} x-\frac{1}{2}=\frac{3}{4}-\frac{5}{3} \\ x-\frac{1}{2}=\frac{9}{12}-\frac{20}{12} \\ x-\frac{1}{2}=\frac{-11}{12} \\ x=\frac{-11}{12}+\frac{1}{2} \\ x=\frac{-11}{12}+\frac{6}{12} \\ x=\frac{-5}{12} \end{gathered}$$

Vậy $x=\frac{-5}{12}$

e) $2\frac{2}{15}:\left(\frac{1}{3}-5x\right)=-2\frac{2}{5}$

$$\begin{gathered} \frac{32}{15}:\left(\frac{1}{3}-5x\right)=\frac{-12}{5} \\ \frac{1}{3}-5x=\frac{32}{15}:\frac{-12}{5} \\ \frac{1}{3}-5x=\frac{32}{15}.\frac{5}{-12} \\ \frac{1}{3}-5x=\frac{-8}{9} \\ 5x=\frac{1}{3}-\frac{-8}{9} \\ 5x=\frac{3}{9}+\frac{8}{9} \\ 5x=\frac{11}{9} \\ x=\frac{11}{9}:5 \\ x=\frac{11}{45} \end{gathered}$$

Vậy $x=\frac{11}{45}$

g) $x^2+\frac{1}{9}=\frac{5}{3}:3$

$$\begin{gathered} x^2+\frac{1}{9}=\frac{5}{9} \\ {x}^2=\frac{5}{9}-\frac{1}{9} \\ {x}^2=\frac{4}{9} \\ {x}^2=\frac{4}{9} \\ {x}^2={\left(\frac{2}{3}\right)}^2={\left(\frac{-2}{3}\right)}^2 \\ +) x=\frac{-2}{3} \\ +) x=\frac{2}{3} \\ Vậy x=\frac{2}{3} hoặc x=\frac{-2}{3} \end{gathered}$$