Question

Answer 1

Lời giải:

a. 

$\frac{1}{3}+\frac{2}{3}:x=-7$

$\frac{2}{3}:x=-7-\frac{1}{3}=\frac{-22}{3}$

$x=\frac{2}{3}: \frac{-22}{3}=\frac{-1}{11}$

b.

$x: \frac{3}{4}=\frac{-2}{3}-\frac{1}{4}=\frac{-11}{12}$

$x=\frac{-11}{12}.\frac{3}{4}=\frac{-11}{16}$
c.

$\frac{1}{3}x+\frac{2}{5}(x-1)=0$

$\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0$

$\frac{11}{15}x=\frac{2}{5}$

$x=\frac{2}{5}: \frac{11}{15}=\frac{6}{11}$
d.

$(3x+\frac{1}{2})^2=\frac{1}{16}=(\frac{1}{4})^2=(\frac{-1}{4})^2$

$\Rightarrow 3x+\frac{1}{2}=\frac{1}{4}$ hoặc $3x+\frac{1}{2}=\frac{-1}{4}$

$\Rightarrow 3x=\frac{1}{4}-\frac{1}{2}$ hoặc $3x=\frac{-1}{4}-\frac{1}{2}$

$\Rightarrow 3x=\frac{-1}{4}$ hoặc $3x=\frac{-3}{4}$

$\Rightarrow x=\frac{-1}{12}$ hoặc $x=\frac{-1}{4}$

Answer 2

e.

$(x+\frac{1}{3})^4=\frac{16}{81}=(\frac{2}{3})^4=(\frac{-2}{3})^4$
$\Rightarrow x+\frac{1}{3}=\frac{2}{3}$ hoặc $x+\frac{1}{3}=\frac{-2}{3}$

$\Rightarrow x=\frac{2}{3}-\frac{1}{3}$ hoặc $x=\frac{-2}{3}-\frac{1}{3}$
$\Rightarrow x=\frac{1}{3}$ hoặc $x=-1$

f.

$(\frac{3}{7})^5x=(\frac{3}{7})^7$

$x=(\frac{3}{7})^7:(\frac{3}{7})^5=(\frac{3}{7})^2=\frac{9}{49}$

g.

$(3x+1)^3=27=3^3$

$\Rightarrow 3x+1=3$

$\Rightarrow 3x=2$

$\Rightarrow x=\frac{2}{3}$
h.

$(2x-\frac{1}{3})^3=\frac{1}{125}=(\frac{1}{5})^3$
$\Rightarrow 2x-\frac{1}{3}=\frac{1}{5}$

$\Rightarrow 3x=\frac{1}{5}+\frac{1}{3}=\frac{8}{15}$

$\Rightarrow x=\frac{8}{15}:3 = \frac{8}{45}$