Bài 1:

a. $(-20)+x=-30$

$x-20=-30$

$x=-30+20=-(30-20)=-10$

b.

$(-10)-x=-20$

$x=(-10)-(-20)=-10+20=20-10=10$

c. Đề sai. Bạn xem lại.

d.

$x+(-3)=-7$

$x=-7-(-3)=-7+3=-(7-3)=-4$

e.

$x-(-5)=-9$

$x=(-9)+(-5)=-14$

f.

$x(-11)=12$

$x=\frac{12}{-11}=\frac{-12}{11}$

h.

$2x-10=20$

$2x=20+10=30$

$x=30:2=15$

l.

$4x-8=-8$
$4x=-8+8=0$

$x=0:4=0$

k.

$-12-(-2)x=-8$

$(-2)x=-12-(-8)=-12+8=-(12-8)=-4$

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

Bài 2:

a. $-20-(10-x)=-3$

$10-x=-20-(-3)=-20+3=-(20-3)=-17$

$x=10-(-17)=10+17=27$

b.

$14+(14-x)=-2$
$14-x=-2-14=-16$

$x=14-(-16)=14+16=30$

c.

$-15-(x-3)=-7$

$x-3=-15-(-7)=-15+7=-8$

x=-8+3=-5$

d.

$(x+4)+(-20)=-8$

$x+4=-8-(-20)=-8+20=12$
$x=12-4=8$

e.

$-2x-2=-4$

$-2x=-4+2=-2$

$x=(-2):(-2)=1$

f.

$-2x+4=-4$

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

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

l.

$-12-(-2)x=-2-4=-6$

$(-2)x=-12-(-6)=-12+6=-6$

$x=(-6):(-2)=3$

|x|=20

=> x= +2 hoặc x=-2

37-|x|=12

|x|=37-12

|x|=25

=>x=\(\pm25\)

|x|-12=8

|x|=8+12

|x|=20

=>x=\(\pm20\)

1, 

|x| = 20

=> x = + 20

37 - |x| = 12

=> |x| = 25

=> x = + 25

|x| - 12 = 8

=> |x| = 20

=> x = + 20

|x| + 8 = |-20| - 12

=> |x| + 8 = 20 - 12

=> |x| + 8 = 8

=> |x| = 0

=> x = 0

1.8100

2. 34 x 18 + 18 x 66

= 18 x ( 34 + 66)

= 18 x 100 = 1800

3. X × ( 8 +  12) =  160 + 20 × 12

= X x 20 = 160 + 240

= X x 20 = 400

X = 400 : 20 = 20

4. X x 12 - X x 2 = 2020

(12 - 2) x X = 2020

10 x X = 2020

X = 2020 : 10 = 202

1.8100

2. 34 x 18 + 18 x 66

= 18 x ( 34 + 66)

= 18 x 100 = 1800

3. X × ( 8 +  12) =  160 + 20 × 12

= X x 20 = 160 + 240

= X x 20 = 400

X = 400 : 20 = 20

4. X x 12 - X x 2 = 2020

(12 - 2) x X = 2020

10 x X = 2020

X = 2020 : 10 = 202

1) +) \(\left|x\right|=20\Rightarrow\orbr{\begin{cases}x=20\\x=-20\end{cases}}\)

+)  \(37-\left|x\right|=12\)

\(\Rightarrow\left|x\right|=37-12=25\)

\(\Rightarrow\orbr{\begin{cases}x=25\\x=-25\end{cases}}\)

+)  \(\left|x\right|-12=8\)

\(\Rightarrow\left|x\right|=8+12=20\Rightarrow\orbr{\begin{cases}x=20\\x=-20\end{cases}}\)

+)  \(\left|x\right|+8=\left|-20\right|-12\)

\(\Rightarrow\left|x\right|+8=20-12=8\)

\(\Rightarrow\left|x\right|=8-8=0\)

\(\Rightarrow x=0\)

2) Ta có: \(\left|x\right|\ge0\)

               \(\left|y\right|\ge0\)

Mà \(\left|x\right|+\left|y\right|\le0\)

\(\Rightarrow\left|x\right|=\left|y\right|=0\)

\(\Rightarrow x=y=0\)

\(a,\dfrac{3}{4}x-\dfrac{7}{12}=\dfrac{5}{6}-\dfrac{2}{3}\\ \Rightarrow\dfrac{3}{4}x-\dfrac{7}{12}=\dfrac{1}{6}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{6}+\dfrac{7}{12}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{3}{4}\\ \Rightarrow x=\dfrac{3}{4}:\dfrac{3}{4}\\ \Rightarrow x=1\\ b,\dfrac{-5}{x}=\dfrac{20}{28}\\ \Rightarrow\dfrac{-5}{x}=\dfrac{5}{7}\\ \Rightarrow\dfrac{-5}{x}=\dfrac{-5}{-7}\\ \Rightarrow x=-7\\ c,2\dfrac{1}{3}:x=7\\ \Rightarrow\dfrac{7}{3}:x=7\\ \Rightarrow x=\dfrac{7}{3}:7\\ \Rightarrow x=\dfrac{1}{3}\)

\(d,\dfrac{-105}{12}< x< \dfrac{20}{7}\Rightarrow x\in\left\{-8;-7;...;2\right\}\)

a: \(\Leftrightarrow x\cdot\dfrac{3}{4}=\dfrac{3}{4}\)

hay x=1

b: \(\Leftrightarrow x=\dfrac{-28\cdot5}{20}=-7\)

c: \(\Leftrightarrow x=\dfrac{7}{3}:7=\dfrac{1}{3}\)

d: \(\Leftrightarrow-8< x< 3\)

hay \(x\in\left\{-7;-6;-5;-4;-3;-2;-1;0;1;2\right\}\)

\(a)\dfrac{3}{4}x-\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{3}{4}x=\dfrac{1}{6}+\dfrac{7}{12}\\ \dfrac{3}{4}x=\dfrac{2}{3}\\ x=\dfrac{2}{3}:\dfrac{3}{4}\\ x=\dfrac{8}{9}\\ b)-\dfrac{5}{x}=\dfrac{20}{28}\\ -5\cdot28=x\cdot20=-140\\ x=-140:20\\ x=-7\\ c)\dfrac{7}{3}:x=7\\ x=\dfrac{7}{3}:7\\ x=\dfrac{1}{3}\)