a) x+7=-12

        x=(-12)-7

        x=-19

b)x-15=-21

        x=(-21)+15

        x=-6

c)13-x=20

        x=13-20

        x=-7

d)17-(2+x)=3

      x=17-3

      x=14

      x=14-2

      x=12

a,x+7=-12

=>x= -12-7

=>x= -19

b,x-15= -21

=>x= -21+15

=>x= -6

c,13-x=20

=>x=13-20

=>x= -7

d, 17-(2+x)=3

=>2+x=17-3

=>2+x=14

=>x=14-2

=>x=12

a) \(x+7=-12\) 

    \(x=-12-7\) 

    \(x=-19\) 

b) \(x-15=-21\)  

    \(x=-21+15\) 

    \(x=-6\)  

c) \(13-x=20\) 

            \(x=13-20\)  

            \(x=-7\) 

d) \(17-\left(2+x\right)=3\)  

             \(2+x=17-3\)  

             \(2+x=14\)   

                   \(x=14-2\)  

                   \(x=12\)

b: \(\dfrac{5}{7}-\dfrac{2}{3}\cdot x=\dfrac{4}{5}\)

=>\(\dfrac{2}{3}x=\dfrac{5}{7}-\dfrac{4}{5}=\dfrac{25-28}{35}=\dfrac{-3}{35}\)

=>\(x=-\dfrac{3}{35}:\dfrac{2}{3}=\dfrac{-3}{35}\cdot\dfrac{3}{2}=-\dfrac{9}{70}\)
c: \(\dfrac{1}{2}x+\dfrac{3}{5}x=-\dfrac{2}{3}\)

=>\(x\left(\dfrac{1}{2}+\dfrac{3}{5}\right)=-\dfrac{2}{3}\)

=>\(x\cdot\dfrac{5+6}{10}=\dfrac{-2}{3}\)

=>\(x\cdot\dfrac{11}{10}=-\dfrac{2}{3}\)

=>\(x=-\dfrac{2}{3}:\dfrac{11}{10}=-\dfrac{2}{3}\cdot\dfrac{10}{11}=\dfrac{-20}{33}\)

d: \(\dfrac{4}{7}\cdot x-x=-\dfrac{9}{14}\)

=>\(\dfrac{-3}{7}\cdot x=\dfrac{-9}{14}\)

=>\(\dfrac{3}{7}\cdot x=\dfrac{9}{14}\)

=>\(x=\dfrac{9}{14}:\dfrac{3}{7}=\dfrac{9}{14}\cdot\dfrac{7}{3}=\dfrac{3}{2}\)

a)

\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)                        

Vậy \(x = \frac{{ - 2}}{3}\).

b)

\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)

Vậy\(x = \frac{1}{12}\).

c)

\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)               

Vậy \(x = \frac{7}{3}\).

d)

\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)

Vậy \(x = \frac{{ - 9}}{{10}}\).

a, \(\left(\dfrac{1}{2}+\dfrac{4}{7}\right):x=\dfrac{-3}{4}\)

\(\dfrac{15}{14}:x=\dfrac{-3}{4}\)

=> x= \(\dfrac{-7}{10}\)

b, 0,5:x-\(1\dfrac{3}{4}\)= 25%

0,5:x-\(\dfrac{7}{4}=\dfrac{1}{4}\)

0,5:x = 2

=> x = \(\dfrac{1}{4}\)

a) Ta có: \(\left(x-3\right)=\left(3-x\right)^2\)

\(\Leftrightarrow\left(x-3\right)^2-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)

\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)

\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)

hay \(x=-\dfrac{1}{4}\)

c) Ta có: \(8x^3-50x=0\)

\(\Leftrightarrow2x\left(4x^2-25\right)=0\)

\(\Leftrightarrow x\left(2x-5\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

e) Ta có: \(x\left(x+3\right)-x^2-3x=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)

\(a.x+\dfrac{1}{6}=-\dfrac{3}{8}\)

\(\Leftrightarrow x=-\dfrac{13}{24}\)

\(b.2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)

\(\Leftrightarrow2-\dfrac{3}{4}+x=\dfrac{7}{12}\)

\(\Leftrightarrow x=-\dfrac{2}{3}\)

\(c.\dfrac{1}{2}x+\dfrac{1}{8}x=\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{5}{8}x=\dfrac{3}{4}\)

\(\Leftrightarrow x=\dfrac{6}{5}\)

\(d.75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}-\left(\dfrac{-1}{2}\right)^2\)

\(=\dfrac{75}{100}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}-\dfrac{1}{4}\)

\(=-\dfrac{3}{4}+\dfrac{6}{5}-\dfrac{1}{4}\)

\(=\dfrac{1}{5}\)

a) \(x+\dfrac{1}{6}=\dfrac{-3}{8}\)

            \(x=\dfrac{-3}{8}-\dfrac{1}{6}\)

           \(x=\dfrac{-13}{24}\)

vậy x =....

b) \(2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)

             \(\dfrac{3}{4}-x=2-\dfrac{7}{12}\)

             \(\dfrac{3}{4}-x=\dfrac{17}{12}\)

                    \(x=\dfrac{3}{4}-\dfrac{17}{12}\)

                   \(x=\dfrac{-2}{3}\)

vậy x =....

a: \(\left(\dfrac{3}{4}x+2\dfrac{1}{2}\right)\cdot\dfrac{-2}{3}=\dfrac{1}{8}\)

=>\(\left(\dfrac{3}{4}x+\dfrac{5}{2}\right)=\dfrac{1}{8}:\dfrac{-2}{3}=\dfrac{-3}{16}\)

=>\(\dfrac{3}{4}x=-\dfrac{3}{16}-\dfrac{5}{2}=-\dfrac{3}{16}-\dfrac{40}{16}=-\dfrac{43}{16}\)

=>\(x=-\dfrac{43}{16}:\dfrac{3}{4}=\dfrac{-43}{16}\cdot\dfrac{4}{3}=\dfrac{-43}{12}\)

b: \(\dfrac{1}{3}\cdot x-0,5x=0,75\)

=>\(x\left(\dfrac{1}{3}-\dfrac{1}{2}\right)=0,75\)

=>\(x\cdot\dfrac{-1}{6}=0,75\)

=>\(x=-0,75\cdot6=-4,5\)

\(a,=x^2-4-x^2-2x-1=-2x-5\\ b,=8x^3-1-8x^3-1=-2\\ 3,\\ a,\Rightarrow x^3+8-x^3+2x=15\\ \Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\\ b,\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\\ \Rightarrow7x=14\Rightarrow x=2\)

Bài 2:

a) \(=x^2-4-x^2-2x-1=-2x-5\)

b) \(=8x^3-1-8x^3-1=-2\)

Bài 3:

a) \(\Rightarrow x^3+8-x^3+2x=15\)

\(\Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)

b) \(\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\)

\(\Rightarrow7x=14\Rightarrow x=2\)

`@` ` \text {Ans}`

`\downarrow`

`a,`

`1/4+3/4*x=3/2-x`

`=> 1/4 + 3/4x - 3/2 + x = 0`

`=> (1/4 - 3/2) + (3/4x + x) = 0`

`=> -5/4 + 7/4x = 0`

`=> 7/4x = 5/4`

`=> x = 5/4 \div 7/4`

`=> x = 5/7`

Vậy, `x=5/7`

`b,`

`3/5*x-1/4=1/10*x-1/2`

`=> 3/5x - 1/4 - 1/10x + 1/2 = 0`

`=> (3/5x - 1/10x) + (-1/4 + 1/2)=0`

`=> 1/2x + 1/4 = 0`

`=> 1/2x = -1/4`

`=> x = -1/4 \div 1/2`

`=> x = -1/2`

Vậy, `x=-1/2`

`c,`

`3x-3/5=x-1/4`

`=> 3x - 3/5 - x + 1/4 = 0`

`=> (3x - x) - (3/5 - 1/4) = 0`

`=> 2x - 7/20 = 0`

`=> 2x = 0,35`

`=> x = 0,35 \div 2`

`=> x = 7/40`

Vậy, `x=7/40`

`d,`

`3/2*x-2/5=1/3*x-1/4`

`=>  3/2x - 2/5 - 1/3x + 1/4 = 0`

`=> (3/2x - 1/3x) - (2/5 - 1/4) = 0`

`=> 7/6x - 3/20 = 0`

`=> 7/6x = 3/20`

`=> x = 3/20 \div 7/6`

`=> x = 9/70`

Vậy, `x=9/70`

`@` `\text {Kaizuu lv uuu}`