a: =>(x+10)(x-1)=0

=>x=-10 hoặc x=1

b: \(A=x^3-1-\left(x+5\right)\left(x^2-3\right)-5x^2-10x-5\)

\(=x^3-5x^2-10x-6-x^3+3x-5x^2+15\)

=-7x+9

=110/13

Lời giải:

a. 

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

$x=\frac{2}{3}+\frac{1}{2}=\frac{7}{6}$

b. 

$x\times \frac{5}{6}=\frac{1}{2}$
$x=\frac{1}{2}: \frac{5}{6}=\frac{3}{5}$

c.

$x:\frac{2}{5}=10$

$x=10\times \frac{2}{5}=4$

d.

$\frac{7}{5}-x=\frac{12}{10}$

$x=\frac{7}{5}-\frac{12}{10}=\frac{1}{5}$

a,

(2x-3)-(x-5)=(x+2)-(x-1)

2x-3-x+5=x+2-x+1

x+2=3

x=1

b,

2(x-1)-5(x+2)=-10

2x-2-5x+10=-10

-3x+8=-10

-3x=-18

x=6

a) ( 2x - 3 ) - ( x - 5 ) = ( x + 2 ) - ( x - 1 )

    2x - 3 - x + 5        = x + 2 - x + 1

    ( 2x - x ) + ( 5 - 3 ) = ( x - x ) + ( 2 + 1 )

    x + 2                    = 3

    x                          = 3 - 2

    x                          = 1

b) 2( x - 1 ) - 5( x + 2 ) = -10

    2x - 2.1 - 5x + 5.2   = -10

    2x - 2 - 5x + 10       = -10

    ( 2x - 5x ) + ( 10 - 2 ) = -10

    -3x + 8                    = -10

    -3x                          = -10 - 8

    -3x                          = -18

       x                         = -18 : -3

       x                         = 6

\(a,\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=0\\ \Rightarrow\left(x^3-27\right)+x\left(4-x^2\right)=0\\ \Rightarrow x^3-27+4x-x^3=0\\ \Rightarrow4x-27=0\\ \Rightarrow4x=27\\ \Rightarrow x=\dfrac{27}{4}\)

\(b,\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\\ \Rightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)=-10\\ \Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0\)

\(\Rightarrow12x+6=0\\ \Rightarrow12x=-6\\ \Rightarrow x=-\dfrac{1}{2}\)

câu a x = 7

Câu b x = -0,5

k nha

\(\text{a , (x-3).(x^2+3x+9)+x(x+2).(2-x)=1 }\)

=(x³-3³)+x(4-x²)=1

=x³-27+4x-x³=1

4x-27=1

4x=28

x=7

\(\text{b, (x+1)^3-(x-1)^3-6.(x-1)^2=-10}\)

=-0,5

trả lo nhanh nhen mik con` lamf bt nữa nha

a) (2x -3) - (x- 5) = (x +2 ) - (x - 1)

( 2x - 3) - (x - 5) - (x + 2) + (x - 1) =0

     2x - 3 - x +5 - x - 2 + x - 1      =0

 (2x - x - x + x) + (-3 + 5 - 1)        = 0

           x          +       1               = 0

                     x                          = 0 - 1

                     x                          = -1 

b) 2(x - 1) - 5(x + 2) = -10

   2x - 2 - 5x + 10    = -10

  (2x - 5x) + (-2 + 10) = -10

  -3x + 8 = -10

  -3x = -10 - 8

  -3x = -18

    x = -18 : (-3)

    x = 6

k mình điiiiiii

bn dưới kia ơi sao lại nhân 2 vậy

a, \(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)ĐK : \(x\ne\pm2\)

\(=\left(\frac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\right)\)

\(=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{x^2-4+10-x^2}{x+2}=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}=\frac{-1}{x-2}\)

b, Ta có : \(2\left|x\right|=1\Leftrightarrow\left|x\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)

TH1 : Thay x = 1/2 vào biểu thức A ta được : \(-\frac{1}{\frac{1}{2}+2}=-\frac{1}{\frac{5}{2}}=-\frac{2}{5}\)

TH2 : Thay x = -1/2 vào biểu thức A ta được : \(\frac{-1}{-\frac{1}{2}-2}=-\frac{1}{-\frac{5}{2}}=-1.\left(-\frac{2}{5}\right)=\frac{2}{5}\)

c, Ta có A < 0 hay \(\frac{-1}{x-2}< 0\Rightarrow x-2>0\)do - 1 < 0 

\(\Leftrightarrow x>2\)

d, Ta có A = x hay \(\frac{-1}{x-2}=x\Rightarrow x^2-2x=-1\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)