a) \(-x-14+32=-26\)

\(-x=-26-32+14\)

\(-x=-44\)

\(x=44\)

b) \(-152-\left(3x+1\right)=\left(-2\right)\left(-27\right)\)

\(-152-3x-1=54\)

\(-3x=96\)

\(x=-32\)

c) \(x+x+x+82=-2-x\)

\(3x+82=2+x\)

\(2x=-80\)

\(x=-40\)

d) \(\left(-1\right)\left(-3\right)\left(-6\right)x=36\)

\(-18x=36\)

\(x=-2\)

1)

\(-x-14+32=-26\)

\(\Rightarrow-x+18=-26\)

\(\Rightarrow-x=-26-18\)

\(\Rightarrow-x=-44\)

\(\Rightarrow x=44\)

Vậy \(x=44\)

2)

\(-152-\left(3x+1\right)=\left(-2\right)\left(-27\right)\)

\(\Rightarrow-152-3x-1=54\)

\(\Rightarrow-153-3x=54\)

\(\Rightarrow-3x=54+163\)

\(\Rightarrow-3x=207\)

\(\Rightarrow x=207:\left(-3\right)\)

\(\Rightarrow x=-69\)

Vậy \(x=-69\)

3)

\(x+x+x+82=-2-x\)

\(\Rightarrow3x+82=-2-x\)

\(\Rightarrow3x+x=-2-82\)

\(\Rightarrow4x=-84\)

\(\Rightarrow x=-21\)

Vậy \(x=-21\)

4)

\(\left(-1\right)\left(-3\right)\left(-6\right).x=36\)

\(\Rightarrow-18x=36\)

\(\Rightarrow x=-2\)

Vậy \(x=-2\)

14  x \(x\) + \(x\) x 26 = 25 

( 14 + 26 ) x \(x\) = 25 

40 x \(x\) = 25 

        \(x\) = 25 : 40

       \(x\) = \(\dfrac{5}{8}\)

Ta có : \(\frac{3}{2x+1}+\frac{10}{4x+2}-\frac{6}{6x+3}=\frac{12}{26}\)

\(\Rightarrow\frac{3}{2x+1}+\frac{5.2}{2\left(2x+1\right)}-\frac{3.2}{3\left(2x+1\right)}=\frac{6}{13}\)

=> \(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)

=> \(\frac{3+5-2}{2x+1}=\frac{6}{13}\)

=> \(\frac{6}{2x+1}=\frac{6}{13}\)

=> 2x + 1 = 13

=> 2x = 12

=> x = 6

Vậy x = 6

\(\frac{3}{2x+1}+\frac{10}{2\left(2x+1\right)}-\frac{6}{3\left(2x+1\right)}=\frac{6}{13}\)                

\(\frac{3}{2x+1}+\frac{5}{2x+1}-\frac{2}{2x+1}=\frac{6}{13}\)                

\(\frac{6}{2x+1}=\frac{6}{13}\)  

\(\Rightarrow2x+1=13\left(6=6\right)\)         

\(2x=12\)    

\(x=6\)     

\(a,\left|x-\dfrac{1}{2}\right|+\dfrac{1}{3}=\dfrac{2}{3}\\ \Rightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{1}{3}\\x-\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\\x=-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{1}{6}\end{matrix}\right.\\ b,\dfrac{x}{-2}=\dfrac{y}{5}=\dfrac{x-y}{-2-5}=\dfrac{14}{-7}=-2\\ \Rightarrow x=-2.\left(-2\right)=4;y=-2.5=-10\)

x-y=-14 nhé

a)

\(\left|x-\dfrac{1}{2}\right|+\dfrac{1}{3}=\dfrac{2}{3}\\ \left|x-\dfrac{1}{2}\right|=\dfrac{1}{3}\\ \left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{1}{3}\\x-\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{1}{3}+\dfrac{1}{2}\\x=-\dfrac{1}{3}+\dfrac{1}{2}\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{5}{6}\\x=\dfrac{1}{6}\end{matrix}\right.\)

b)

\(\dfrac{x}{-2}=\dfrac{y}{5}\)

mà `x-y=14` nên áp dụng tính chất dãy tỉ số bằng nhau ta có

\(\dfrac{x}{-2}=\dfrac{y}{5}=\dfrac{x-y}{-2-5}=\dfrac{14}{-7}=-2\\ =>\left\{{}\begin{matrix}x=-2\cdot\left(-2\right)=4\\y=-2\cdot5=-10\end{matrix}\right.\)

`a, x/7 =-4/14`

`=> 14x=7.(-4)`

`=>14x=-28`

`=>x=-28:14`

`=>x=-2`

`b,x/2=-2/-x`

`=>x/2=2/x`

`=>x.x=2.2`

`=>x^2=4`

`=>x= +-2`

`c,(x-1)/5=5/(x-1)`

`=>(x-1)^2 = 5.5`

`=>(x-1)^2=25`

`=>(x-1)^2=5^2`

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

`d,x+3/2=-12/16`

`=>x=-12/16 -3/2`

`=>x= -12/16 - 24/16`

`=>x= -36/16`

`=>x=-9/4`

a: \(\Leftrightarrow3x+9=-2x+6\)

=>5x=-3

hay x=-3/5

b: =>3/x=y/35=3/7

=>x=7; y=15

c: =>9x/5=-3/5

=>9x=-3

hay x=-1/3

d: =>x+2/26=-1/4

=>x+2=-13/2

hay x=-17/2

(-8)-[(-5)+8]

=(-8)-3

=-11

25.134+25.(-34)

=25.[134+(-34)]

=25.100

=2500

210+[46+(-210)+(-26)]

=210+46+(-210)+(-26)

=[210+(-210)]+[46+(-26)]

=0+20

=20

x+10=-14

x     =(-14)-10

x     =-24

2x-(-17)=15

2x        =15+(-17)

2x       =-2

 x       =(-2):2

 x      =-1