\(\Leftrightarrow\left|x-1\right|=5\Leftrightarrow\left[{}\begin{matrix}x=5+1=6\\x=-5+1=-4\end{matrix}\right.\)

1) = \(\frac{3}{5}\)

2) =\(\frac{6}{7}\)

3)\(\frac{9}{13}\)

4)\(\frac{4}{13}\)

a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)

=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)

=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)

b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)

=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)

=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)

TThế giới oi oi oi 

b) Ta có: \(9x^4+8x^2-1=0\)

\(\Leftrightarrow9x^4+9x^2-x^2-1=0\)

\(\Leftrightarrow9x^2\left(x^2+1\right)-\left(x^2+1\right)=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(9x^2-1\right)=0\)

mà \(x^2+1>0\forall x\)

nên \(9x^2-1=0\)

\(\Leftrightarrow9x^2=1\)

\(\Leftrightarrow x^2=\dfrac{1}{9}\)

hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3}\right\}\)

Vậy: \(S=\left\{\dfrac{1}{3};-\dfrac{1}{3}\right\}\)

\(-\frac{1}{7}+\frac{5}{3}+\frac{5}{4}+\frac{1}{3}-\frac{3}{2}\)

\(=\left(-\frac{1}{7}+\frac{5}{3}-\frac{3}{2}\right)+\left(\frac{5}{3}+\frac{1}{3}\right)\)

\(=\frac{-6}{42}+\frac{70}{42}-\frac{63}{42}+\frac{6}{3}\)

\(=\frac{-6+70-63}{42}+2\)

\(=\frac{1}{42}+\frac{84}{42}\)

\(=\frac{85}{42}\)

a,(-12).x= 60-12

(-12).x=48

x=48:(-12)

x = -4

b,(-5).x + 5 = -24+6

(-5).x + 5 = 30

(-5).x      = 30-5

(-5).x      = 25

x = 25 : (-5)

x = -5

c,

$3x+36=-7x-64$

$10x=-100$

$x=-10$

$b,-5x-178=14x+145$

$-19x=323$

$x=-17$

\(\dfrac{3}{x+2}\)=\(\dfrac{5}{2x+1}\)(x khác -1/2 và x khác -2)

=>6x+3=5x+10

<=>x=7 tm 

A,-3/-7+5/19+-4/7

=(-3/-7+-4/7)+5/19

=-1/7+5/19=16/133

B,-13/24+-5/24+7/21

=-3/4+7/21=-5/12

C,-5/13+(-8/13+1)

=(-5/13+-8/13)+1

=-1+1=0

D,2/3+(3/8+-2/3)

=(2/3+-2/3)+3/8

=0+3/8=3/8

E,(-3/4+5/8)+-1/8

=-1/8+1/8=0