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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a;80x-64=160
80x=224
x=14/5
b;x-4=42:6
x-4=7
x=11
c;40-x=16*2
40-x=32
x=8
d;20-x=8:4
20-x=2
x=18
`# \text {Ryo}`
`a)`
`80x - 2^2 . 2^4 = 160`
`\Rightarrow 80x - 2^6 = 160`
`\Rightarrow 80x - 64 = 160`
`\Rightarrow 80x = 224`
`\Rightarrow x = 224 \div 80`
`\Rightarrow x = 2,8`
Vậy, `x = 2,8`
`b)`
`(x - 4).6 = 42`
`\Rightarrow x - 4 = 42 \div 6`
`\Rightarrow x - 4 = 7`
`\Rightarrow x = 7 + 4`
`\Rightarrow x = 11`
Vậy, `x = 11`
`c)`
`(40 - x) \div 2 = 16`
`\Rightarrow 40 - x = 16 . 2`
`\Rightarrow 40 - x = 32`
`\Rightarrow x = 40 - 32`
`\Rightarrow x = 8`
Vậy, `x = 8`
`d)`
`(20 - x) . 4 = 8`
`\Rightarrow 20 - x = 8 \div 4`
`\Rightarrow 20 - x = 2`
`\Rightarrow x = 20 - 2`
`\Rightarrow x = 18`
Vậy, `x = 18.`
![](https://rs.olm.vn/images/avt/0.png?1311)
a) -32 -4.(x-5) = 0
<=>4.(x-5)=-32
<=>x-5=(-32):4
<=>x+5=-8
<=>x=-8+5
<=>x=-3
Vậy x=-3
b) 13.(x-5)=-169
<=>x-5=(-169):13
<=>x-5=-13
<=>x=-13+5
<=>x=-8
vậy x=-8
c) (-2).x+5 = (-3).(-3)+8
<=>(-2).x+5=17
<=>(-2).x=17-5
<=>(-2).x=12
<=>x=12:(-2)
<=>x=-6
Vậy x=-6
d) (-8).x = (-10).(-2)-4
<=>(-8).x=16
<=>x=16:(-8)
<=>x=-2
vậy x=-2
e) (-9).x+3 = (-2).(-7)+16
<=>(-9).x+3=30
<=>(-9).x=30-3
<=>(-9).x=27
<=>x=27:(-9)
<=>x=-3
Vậy x=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
Tính ngoặc tròn
\(4\cdot8-16\cdot2\)
\(=32-32\)
\(=0\)
Vậy tích trên bằng 0 ( vì có 1 thừa số = 0 )
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(\frac{4}{13}.\frac{6}{5}+\frac{4}{13}.\frac{2}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(\frac{4}{13}.\frac{8}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\frac{32}{65}.\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(2x+1\right)^2=\frac{10}{13}\div\frac{32}{65}\)
\(\left(2x+1\right)^2=\frac{25}{16}\)
\(\Rightarrow2x+1\in\left\{\frac{5}{4};-\frac{5}{4}\right\}\)
\(\hept{\begin{cases}2x+1=\frac{5}{4}\\2x+1=-\frac{5}{4}\end{cases}\Rightarrow\hept{\begin{cases}2x=\frac{1}{4}\\2x=-\frac{9}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{8}\\x=-\frac{9}{8}\end{cases}}}\)
Vậy \(x\in\left\{\frac{1}{8};-\frac{9}{8}\right\}\)
\(x^3-\frac{9}{16}.x=0\)
\(x\left(x^2-\frac{9}{16}\right)=0\)
\(\hept{\begin{cases}x=0\\x^2-\frac{9}{16}=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=\frac{9}{16}\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=\pm\frac{3}{4}\end{cases}}}\)
Vậy \(x\in\left\{0;\frac{3}{4};-\frac{3}{4}\right\}\)
a) 2.x - 16 = 40 + x
=> 2.x - x = 40 + 16
x = 56
b) 16 - 2.x = 40 - 3.x
=> -2x + 3.x = 40 -16
x = 24
c) 6.(x-1) - 5.x = 15
6x -6 - 5.x = 15
6x-5x - 6 = 15
x - 6 = 15
x = 21
d) 2.(4.x-2) - 7.x = 16
8x - 4 - 7x = 16
8x-7x - 4 = 16
x - 4 = 16
x = 20
\(a,2x-16=40+x\)
\(\Rightarrow2x-x=40+16\)
\(\Rightarrow x=56\)
\(b,16-2x=40-3x\)
\(\Rightarrow-2x+3x=40-16\)
\(\Rightarrow x=24\)
\(c,6.\left(x-1\right)-5.x=15\)
\(\Rightarrow6x-6-5x=15\)
\(\Rightarrow x=21\)
\(d,2.\left(4x-2\right)-7.x=16\)
\(\Rightarrow8x-4-7x=16\)
\(\Rightarrow x=20\)