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a) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}\\ \dfrac{x}{5}+\dfrac{1}{2}=\dfrac{3}{5}\\ \dfrac{x}{5}=\dfrac{3}{5}-\dfrac{1}{2}\\ \dfrac{x}{5}=\dfrac{6}{10}-\dfrac{5}{10}\\ \dfrac{x}{5}=\dfrac{1}{10}\\ \dfrac{2x}{10}=\dfrac{1}{10}\\ \Rightarrow2x=1\\ x=1:2\\ x=0,5=\dfrac{1}{2}\)

b) \(x+\dfrac{3}{15}=\dfrac{1}{3}\\ x=\dfrac{1}{3}-\dfrac{3}{15}\\ x=\dfrac{5}{15}-\dfrac{3}{15}\\ x=\dfrac{2}{15}\)

c) \(x-\dfrac{12}{4}=\dfrac{1}{2}\\ x-3=\dfrac{1}{2}\\ x=\dfrac{1}{2}+3\\ x=\dfrac{1}{2}+\dfrac{6}{2}\\ x=\dfrac{7}{2}\)

d) \(\dfrac{1}{2}x+\dfrac{1}{2}=\dfrac{5}{2}\\ \dfrac{1}{2}x=\dfrac{5}{2}-\dfrac{1}{2}\\ \dfrac{1}{2}x=2\\ x=2:\dfrac{1}{2}\\ x=4\)

a. \(\dfrac{2x+5}{10}=\dfrac{6}{10}\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)

b. \(\dfrac{15x+3}{15}=\dfrac{5}{15}\Leftrightarrow15x=2\Leftrightarrow x=\dfrac{2}{15}\)

c. \(\dfrac{4x-12}{4}=\dfrac{2}{4}\Leftrightarrow4x=14\Leftrightarrow x=\dfrac{7}{2}\)

d. \(\dfrac{1+x}{2x}=\dfrac{5x}{2x}\Leftrightarrow-4x=-1\Leftrightarrow x=\dfrac{1}{4}\)

e. \(\dfrac{-4\left(2x-5\right)}{6\left(2x-5\right)}-\dfrac{2}{6\left(2x-5\right)}=\dfrac{9\left(2x-5\right)}{6\left(2x-5\right)}\)

\(\Leftrightarrow-8x+20-2=18x-45\)

\(\Leftrightarrow-26x=-63\Leftrightarrow x=\dfrac{63}{26}\)

Câu 1: 

a: =>-2x-x+17=34+x-25

=>-3x+17=x+9

=>-4x=-8

hay x=2

b: =>17x+16x+27=2x+43

=>33x+27=2x+43

=>31x=16

hay x=16/31

c: =>-2x-3x+51=34+2x-50

=>-5x+51=2x-16

=>-7x=-67

hay x=67/7

e: 3x-32>-5x+1

=>8x>33

hay x>33/8

Bài 1

a, 32.(-64) - 64. 68

= -64.(32 + 68)

= -64 .100

= - 6400

b, -54.76 + 12.(-76) - 76.34

= -76.(54 + 12 + 34)

= -76.100

= -7600

19- 42.(-19) + 38.5

= 19 + 42.19 + 19.2.5

= 19.(1 + 42 + 10)

= 19.53

= 1007

a/ \(51-(-12+3x)=27\)

\(\Leftrightarrow51+12-27-3x=0\Leftrightarrow36=3x\Leftrightarrow x=\frac{36}{3}=12\)

KL:........

b/ $-x + 21=15+ 2x$

\(\Leftrightarrow2x+x=21-15\Leftrightarrow2x=6\Leftrightarrow x=3\)

KL: ...........

c) $7.(x-9)-5(6-x)=-6+11.x$

\(\Leftrightarrow7x-63-30+5x=-6+11x\Leftrightarrow7x+5x-11x=-6+63+30\Leftrightarrow x=87\)

KL:............

d) $(x-3).(x^2 + 2)=0$

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

e) $|2x-7|-22=-13$

\(\Leftrightarrow\left|2x-7\right|=9\Leftrightarrow\left[{}\begin{matrix}2x-7=9\\2x-7=-9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-1\end{matrix}\right.\)

KL: ...........

f) $(2x - 1)^3=-125$

\(\Leftrightarrow\left(2x-1\right)^3=\left(-5\right)^3\Leftrightarrow2x-1=-5\Leftrightarrow x=-2\)

KL: ...........

d) |2x – 5| –7 = 22

| 2x -5 | = 22+7

| 2x -5 | = 29

TH1: 2x-5 = -29

2x = -29+5

2x= -24

x= -24:2

x= -12

TH2: 2x -5 =29

2x = 29+5

2x= 34

x= 34:2

x= 17 

Vậy...

a) x – 14 = 3x + 18

x -14 - 3x = 18

-2x -14 =18

-2x = 18-14

-2x = 4

x = 4:( -2) 

x= -2

Vậy...

g. \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)

Vậy \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)

f. \(\frac{2}{3}x-\frac{1}{2}x=\frac{5}{12}\)

\(\Leftrightarrow x\left(\frac{2}{3}-\frac{1}{2}\right)=\frac{5}{12}\)

\(\Leftrightarrow x\left(\frac{4}{6}-\frac{3}{6}\right)=\frac{5}{12}\)

\(\Leftrightarrow\frac{1}{6}x=\frac{5}{12}\)

\(\Leftrightarrow x=\frac{5}{12}\div\frac{1}{6}\)

\(\Leftrightarrow x=\frac{30}{12}=\frac{5}{2}\)

Bài 1 Tìm x biết:

a)65-(29-x)=32

65 -29+x=31

x=31-65+29

x=-5

b)(x+5)-(x+23)=x-34

x+5 -x +23 = x-34

(x-x)+ (23+5)=x-34

0+28=x-34

28=x-34

28+34=x

62=x

=>x=62

c)(16-x)+(x-38)=x+44

16-x+x-38=x+44

-x+x-x=44-16+38

-x=36

=>x=-36

d)-12+3(-x+7)=-18

3(-x+7)=-18+12

3(-x+7)=-6

-x+7=-6:3

-x+7=-2

-x=-2-7

-x=-9

=>x=9

Baif 2

d)|7-x|=10

=> \(\left[{}\begin{matrix}7-x=10\\7-x=-10\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=7-10\\x=-10-7\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=-3\\x=-17\end{matrix}\right.\)

e)(x-6).(7-2x)=0

\(\Rightarrow\)\(\left[{}\begin{matrix}x-6=0\\7-2x=0\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=0+6\\2x=7\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=6\\x=7:2\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=6\\x=3,5\end{matrix}\right.\)

f)(9-x).(2x+8)=0

\(\Rightarrow\)\(\left[{}\begin{matrix}9-x=0\\2x+8=0\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=0+9\\2x=-8\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)

g)x(-x+8).(-3x-18)=0

\(\Rightarrow\) \(\left[{}\begin{matrix}x=0\\-x+8=0\\-3x-18=0\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\-x=0+8\\-3x=0+18\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\-x=8\\-3x=18\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=-8\\x=18:\left(-3\right)\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=0\\x=-8\\x=-6\end{matrix}\right.\)

h)(-x+8).(x-54).(-24-x)=0

\(\Rightarrow\)\(\left[{}\begin{matrix}-x+8=0\\x-54=0\\-24-x=0\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}-x=8\\x=0+54\\-x=0+24\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\-x=24\end{matrix}\right.\)

\(\Rightarrow\)\(\left[{}\begin{matrix}x=8\\x=54\\x=-24\end{matrix}\right.\)