a,\(\left(x-15\right):50+22=24\)

\(< =>\frac{\left(x-15\right)}{50}=2< =>x-15=100\)

\(< =>x=100+15=115\)

b,\(42-\left(2x+32\right)+12:2=6\)

\(< =>42-2x-32=0\)

\(< =>10-2x=0< =>x=\frac{10}{2}=5\)

Làm nốt :

c) \(134-2\left\{156-6\cdot\left[54-2\cdot\left(9+6\right)\right]\right\}\cdot x=86\)

=> 134 - 2{156 - 6 . [54 - 2 . 15]} . x = 86

=> 134 - 2{156 - 6 . [54 - 30]} . x = 86

=> 134 - 2{156 - 6. 24} . x = 86

=> 134 - 2{156 - 144} . x = 86

=> 134 - 2.12 . x = 86

=> 134 - 24 . x = 86

=> 24.x = 48

=> x = 2

Bài 2 : a) 120 : [21 - (4x - 4)] = 2³.3

=> 120 : [21 - (4x - 4)] = 8.3

=> 120 : [21 - (4x - 4)] = 24

=> 21 - (4x - 4) = 5

=> 4x - 4 = 16

=> 4x = 20

=> x = 5

b) 3.[205 - (x - 9)] - 486 = 0

=> 3.[205 - (x - 9)] = 486

=> 205 - (x - 9) = 162

=> x - 9 = 205 - 162 = 43

=> x = 43 + 9 = 52

c) 204 - 2{200 - 5.[64 - 2.(11 + 6)]} . x = 4

=> 204 - 2{200 - 5.[64 - 2.17]} . x = 4

=> 204 - 2{200 - 5 .[64 - 34]}.x = 4

=> 204 - 2{200 - 5.30} . x = 4

=> 204 - 2{200 - 150}.x = 4

=> 204 - 2.50 . x = 4

=> 2.50.x = 200

=> 100.x = 200

=> x = 2

nhiều quá :((

\(a,2\left(x-5\right)-3\left(x+7\right)=14\)

\(2x-10-3x-21=14\)

\(-x-31=14\)

\(-x=45\)

\(x=45\)

\(b,5\left(x-6\right)-2\left(x+3\right)=12\)

\(5x-30-2x-6=12\)

\(3x-36==12\)

\(3x=48\)

\(x=16\)

\(c,3\left(x-4\right)-\left(8-x\right)=12\)

\(3x-12-8+x=0\)

\(4x-20=0\)

\(4x=20\)

\(x=5\)

Cố nốt nha bn ! 

cảm ơn, bn nha:)))

mà hình như bạn TOP 3 trả lời câu hỏi pải ko nhỉ???

d, \(-7\left(3x-5\right)+2\left(7x-14\right)=28\)

\(\Leftrightarrow-21x+35+14x-28=28\Leftrightarrow-7x=21\Leftrightarrow x=-3\)

e, \(5\left(3-2x\right)+5\left(x-4\right)=6-4x\)

\(\Leftrightarrow15-6x+5x-20=6-4x\Leftrightarrow-5-x=6-4x\)

\(\Leftrightarrow-11+3x=0\Leftrightarrow x=\frac{11}{3}\)

f, \(-5\left(2-x\right)+4\left(x-3\right)=10x-15\)

\(\Leftrightarrow-10+5x+4x-12=10x-15\Leftrightarrow-22+9x=10x-15\)

\(\Leftrightarrow-7-x=0\Leftrightarrow x=-7\)

các bạn giúp m với =(((((

a) x=7

b)x=2/3

c)x=-13

d)x=3

e)x=-1

1) |x + 2| = 4

\(\Leftrightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)

2) 3 – |2x + 1| = (-5)

\(\Leftrightarrow\left|2x+1\right|=8\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}\)

3) 12 + |3 – x| = 9

\(\Leftrightarrow\left|3-x\right|=-3\)(vô lí)

=>\(x=\varnothing\) 

1) I x+2 I=4

\(\Rightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-6\end{cases}}}\)

2) \(3-|2x+1|=-5\)

\(\Leftrightarrow|2x+1|=8\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}}\)

3) \(12+|3-x|=9\)

\(\Leftrightarrow|3-x|=-3\)(vô lí vì I 3-x I \(\ge\)0)

1; |\(x\) + 2| = 4

    \(\left[{}\begin{matrix}x+2=-4\\x+2=4\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=-4-2\\x=4-2\end{matrix}\right.\)

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

Vậy \(x\) \(\in\) {- 6; 2}

a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)

\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)

\(\Leftrightarrow24x=-13\)

hay \(x=-\dfrac{13}{24}\)

a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3

=>x=-1/3+3/4=-4/12+9/12=5/12

b: =>x(1/2-5/6)=7/2

=>-1/3x=7/2

hay x=-21/2

c: (4-x)(3x+5)=0

=>4-x=0 hoặc 3x+5=0

=>x=4 hoặc x=-5/3

d: x/16=50/32

=>x/16=25/16

hay x=25

e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4

=>2x=-7/4+3=5/4

hay x=5/8