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`@` `\text {Ans}`

`\downarrow`

`1)`

\(2x+\dfrac{1}{2}=\dfrac{5}{3}\)

`\Rightarrow`\(2x=\dfrac{5}{3}-\dfrac{1}{2}\)

`\Rightarrow`\(2x=\dfrac{7}{6}\)

`\Rightarrow`\(x=\dfrac{7}{6}\div2\)

`\Rightarrow`\(x=\dfrac{7}{12}\)

Vậy, `x = 7/12`

`2)`

\(\dfrac{1}{7}+\dfrac{4}{5}x=\dfrac{5}{3}\)

`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{5}{3}-\dfrac{1}{7}\)

`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{32}{21}\)

`\Rightarrow`\(x=\dfrac{32}{21}\div\dfrac{4}{5}\)

`\Rightarrow`\(x=\dfrac{40}{21}\)

Vậy, `x = 40/21`

`3)`

\(\dfrac{3}{5}-\dfrac{3}{5}x=\dfrac{1}{7}\)

`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{3}{5}-\dfrac{1}{7}\)

`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{16}{35}\)

`\Rightarrow`\(x=\dfrac{16}{35}\div\dfrac{3}{5}\)

`\Rightarrow`\(x=\dfrac{16}{21}\)

Vậy, `x = 16/21`

`4)`

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

`\Rightarrow`\(3x=\dfrac{5}{6}-\dfrac{3}{4}\)

`\Rightarrow`\(3x=\dfrac{1}{12}\)

`\Rightarrow`\(x=\dfrac{1}{12}\div3\)

`\Rightarrow`\(x=\dfrac{1}{36}\)

Vậy, `x  = 1/36`

`5)`

\(\dfrac{5}{3}-\dfrac{1}{2}x=\dfrac{3}{7}\)

`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{5}{3}-\dfrac{3}{7}\)

`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{26}{21}\)

`\Rightarrow`\(x=\dfrac{26}{21}\div\dfrac{1}{2}\)

`\Rightarrow`\(x=\dfrac{52}{21}\)

Vậy, `x = 52/21`

`6)`

\(5x+\dfrac{1}{2}=\dfrac{2}{3}\)

`\Rightarrow`\(5x=\dfrac{2}{3}-\dfrac{1}{2}\)

`\Rightarrow`\(5x=\dfrac{1}{6}\)

`\Rightarrow`\(x=\dfrac{1}{6}\div5\)

`\Rightarrow`\(x=\dfrac{1}{30}\)

Vậy, `x = 1/30.`

18 tháng 1 2019

1)6x-8=3x+1

6x-3x=1+8

3x=9

x=3

Vậy x=3

2: 12-10x=25-30x

=>20x=13

=>x=13/20

3: \(3\left(2x+3\right)-2\left(4x-5\right)=10x+21\)

=>6x+9-8x+10=10x+21

=>10x+21=-2x+19

=>12x=-2

=>x=-1/6

4: \(\Leftrightarrow25x-15-6x+12=11-5x\)

=>19x-3=11-5x

=>24x=14

=>x=7/12

5: \(\Leftrightarrow8-12x-5+10x=4-6x\)

=>4-6x=-2x+3

=>-4x=-1

=>x=1/4

6: \(\Leftrightarrow32x-24-6+9x=13-40x\)

=>41x-30=13-40x

=>81x=43

=>x=43/81

7: \(\Leftrightarrow10x-5+20x=5x-11\)

=>30x-5=5x-11

=>25x=-6

=>x=-6/25

1: =>25x-15-6x+12=11-5x

=>19x-3=11-5x

=>24x=14

=>x=7/12

2: =>8-12x-5+10x=4-6x

=>4-6x=-2x+3

=>-4x=-1

=>x=1/4

AH
Akai Haruma
Giáo viên
14 tháng 11 2023

Lời giải:

1. $(x+2)-2=0$

$x+2=2$

$x=0$

2.

$(x+3)+1=7$

$x+3=7-1=6$

$x=6-3=3$

3.

$(3x-4)+4=12$

$3x-4+4=12$

$3x=12$

$x=12:3=4$

4.

$(5x+4)-1=13$

$5x+4=13+1=14$

$5x=14-4=10$

$x=10:5=2$

5.

$(4x-8)-3=5$

$4x-8=5+3=8$

$4x=8+8=16$

$x=16:4=4$

6.

$3+(x-5)=7$

$x-5=7-3=4$

$x=4+5=9$

7.

$8-(2x-4)=2$

$2x-4=8-2=6$

$2x=6+4=10$

$x=10:2=5$

8.

$7+(5x+2)=14$

$5x+2=14-7=7$

$5x=7-2=5$

$x=5:5=1$

9.

$5-(3x-11)=1$

$3x-11=5-1=4$

$3x=11+4=15$

$x=15:3=5$

10.

$16-(8x+2)=6$

$8x+2=16-6=10$

$8x=10-2=8$

$x=8:8=1$

a: =>x/27+1=-2/3

=>x/27=-5/3

=>x=-45

b: \(\Leftrightarrow x-4=\dfrac{2}{5}:\dfrac{20}{21}=\dfrac{2}{5}\cdot\dfrac{21}{20}=\dfrac{42}{100}=\dfrac{21}{50}\)

=>x=221/50

c: \(\Leftrightarrow x+\dfrac{2}{3}=\dfrac{4}{60}=\dfrac{1}{15}\)

=>x=1/15-2/3=1/15-10/15=-9/15=-3/5

d: \(\Leftrightarrow x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{15}{14}\cdot\dfrac{21}{20}\)

=>\(x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{3}{2}\cdot\dfrac{3}{4}=\dfrac{1}{5}-\dfrac{9}{8}=\dfrac{-37}{40}\)

=>x=-37/24

e: =>-3/7x=84/45

=>x=-196/45

f: =>11/10x=-2/3

=>x=-20/33

1) Ta có: \(3\left(x-1\right)-5\left(x-2\right)=4\left(x+1\right)\)

\(\Leftrightarrow3x-5-5x+10-4x-4=0\)

\(\Leftrightarrow-6x+1=0\)

\(\Leftrightarrow-6x=-1\)

hay \(x=\dfrac{1}{6}\)

2) Ta có: \(-2\left(x-2\right)-4\left(x+1\right)=-3\left(x+3\right)\)

\(\Leftrightarrow-2x+4-4x-4+3x+9=0\)

\(\Leftrightarrow-3x=-9\)

hay x=3

3) Ta có: \(3x^2+2x=0\)

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

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

4) Ta có: \(x^2-5x=0\)

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

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

5) Ta có: \(\left(2x-3\right)^2=36\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

6) Ta có: \(\left(5x-1\right)^3=125\)

\(\Leftrightarrow5x-1=5\)

\(\Leftrightarrow5x=6\)

hay \(x=\dfrac{6}{5}\)

7) Ta có: \(3^{x+1}=27\)

\(\Leftrightarrow x+1=3\)

hay x=2

4 tháng 2 2023

\(-\dfrac{3}{5}\times x+\dfrac{1}{2}=\dfrac{4}{5}\)

\(-\dfrac{3}{5}\times x=\dfrac{4}{5}-\dfrac{1}{2}\)

\(-\dfrac{3}{5}\times x=\dfrac{3}{10}\)

\(x=\dfrac{3}{10}:-\dfrac{3}{5}\)

\(x=-\dfrac{1}{2}\)

Vậy \(x=-\dfrac{1}{2}\)

\(\dfrac{1}{2}\times x+\dfrac{3}{8}=\dfrac{7}{16}\)

\(\dfrac{1}{2}\times x=\dfrac{7}{16}-\dfrac{3}{8}\)

\(\dfrac{1}{2}\times x=\dfrac{1}{16}\)

\(x=\dfrac{1}{8}\)

Vậy \(x=\dfrac{1}{8}\)

\(\dfrac{2}{3}x-\dfrac{5}{6}x=-\dfrac{1}{2}\)

\(x\times\left(\dfrac{2}{3}-\dfrac{5}{6}\right)=-\dfrac{1}{2}\)

\(x\times\dfrac{-1}{6}=-\dfrac{1}{2}\)

\(x=3\)

Vậy \(x=3\)

\(\dfrac{2}{5}x+\dfrac{3}{10}x=-\dfrac{1}{5}\)

\(x\times\left(\dfrac{2}{5}+\dfrac{3}{10}\right)=-\dfrac{1}{5}\)

\(x\times\dfrac{7}{10}=-\dfrac{1}{5}\)

\(x=-\dfrac{2}{7}\)

Vậy \(x=-\dfrac{2}{7}\).