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\(120-4\left(1-x\right)=106-3x\)
\(120-4+4x+3x=106\)
\(116+7x=106\)
\(7x=-10\)
\(x=\frac{-10}{7}\)
120 - 4(1-x) = 106 - 3x
120 - 4 - x = 106 - 3x
116 - x = 106 - 3x
x - 3x = 116 - 106
-2x = 10
x = 10 : ( -2 )
x = -5
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 5:
a: x(x-4)=0
=>\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
b: Đề thiếu vế phải rồi bạn
Bài 6:
a: \(\left(-5\right)\cdot\left(-6\right)\cdot\left(-4\right)\cdot2\)
\(=-\left(2\cdot5\right)\cdot\left(4\cdot6\right)\)
\(=-24\cdot10=-240\)
b: \(\left(-3\right)\cdot2\cdot\left(-8\right)\cdot5\)
\(=3\cdot2\cdot8\cdot5\)
\(=\left(3\cdot8\right)\cdot\left(2\cdot5\right)\)
\(=24\cdot10=240\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a,(5-x)(x-1) < 0`
`<=>5-x<0` hoặc `x-1<0`
`<=>5 <x` hoặc `x<1`
Vậy `S={x|5<x;x<1}`
`b,(x-4)(x+1/2) >= 0`
`<=>TH1 : {(x-4>=0),(x+1/2 >=0):}<=>{(x>=4(TM)),(x>= -1/2(L)):}`
`<=>TH2 :{(x-4<=0),(x+1/2 <= 0):} <=>{(x<=4(L)),(x<=-1/2(TM)):}`
`=>x<= -1/2` hoặc `x>=4`
Vậy `S={x|x<= -1/2 ; x>=4}`
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(3x⋮x-1\)
=>\(3x-3+3⋮x-1\)
=>\(3⋮x-1\)
=>\(x-1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{2;0;4;-2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. 2x+\(\dfrac{4}{5}\)=0 hoặc 3x-\(\dfrac{1}{2}\)=0
2x=- 4/5 hoặc 3x=1/2
x=-2/5 hoặc x=\(\dfrac{1}{6}\)
b. x-\(\dfrac{2}{5}\)=0 hoặc x+\(\dfrac{4}{7}\)=0
x=2/5 hoặc x=-\(\dfrac{4}{7}\)
d. x(1+5/8-12/16)=1
\(\dfrac{7}{8}\)x=1=> x=8/7