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1: Ta có: \(20-2\left(x+4\right)=4\)
\(\Leftrightarrow2\left(x+4\right)=16\)
\(\Leftrightarrow x+4=8\)
hay x=4
5: Ta có: \(\left(x+1\right)^3=27\)
\(\Leftrightarrow x+1=3\)
hay x=2
c: \(=\dfrac{4}{9}\left(\dfrac{5}{7}+\dfrac{2}{5}+\dfrac{2}{7}-\dfrac{7}{5}\right)=\dfrac{4}{9}\cdot\left(1-1\right)=0\)
d: \(=\dfrac{4-3-1}{12}\cdot\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)=0\cdot\left(\dfrac{67}{111}+\dfrac{2}{33}-\dfrac{15}{117}\right)=0\)
a: \(\left(-120\right):15+12\left(2x-1\right)=52\)
=>\(12\left(2x-1\right)-8=52\)
=>\(12\left(2x-1\right)=60\)
=>\(2x-1=\dfrac{60}{12}=5\)
=>2x=5+1=6
=>\(x=\dfrac{6}{2}=3\)
c: \(x+4⋮x+1\)
=>\(x+1+3⋮x+1\)
=>\(3⋮x+1\)
=>\(x+1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{0;-2;2;-4\right\}\)
d: \(2x+7⋮x+2\)
=>\(2x+4+3⋮x+2\)
=>\(3⋮x+2\)
=>\(x+2\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{-1;-3;1;-5\right\}\)
e: \(3x⋮x-1\)
=>\(3x-3+3⋮x-1\)
=>\(3⋮x-1\)
=>\(x-1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{2;0;4;-2\right\}\)
b) 157-(x-124)=-483
x-124=157-(-483)
x-124=640
x=640+124
x=764
bạn đã kiểm tra kĩ chưa vậy?mình đọc đề câu B mà loạn não luôn á;-;
a,(5.9.x-25):12=0
(45.x-25):12=0
(45.x-25) =0x12
(45.x-25)=0
45.x=0+25
45.x=25
x=25:45
x=5/9
b,30.(x-28)=120
(x-28)=120:30
(x-28)=4
x=4+28
x=32
c'84-4.(2.x+1)=48
4.(2.x+1)=84-48
4.(2.x+1)=36
(2.x+1)=36:4
(2.x+1)=9
2.x=9-1
2.x=8
x=8:2
x=4
d,[(x+32)-15].2=60
[(x+32)-15]=60:2
[(x+32)-15]=30
(x+32)=30+15
(x+32)=45
x=45-32
x=13
(67/111+2/33-15/117) . (1/3-1/4-1/12)
= (67/111+2/33-15/117) . 0
= 2836/5291 . 0
= 0
`(12.15-x). 1/4= 120 . 1/4`
`(12.15-x). 1/4=30`
`12.15-x=30:1/4`
`12.15-x=120`
`180-x=120`
`x=180-120`
`x=60`
Vậy `x=60`