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![](https://rs.olm.vn/images/avt/0.png?1311)
\(2,\)
\(a,20-\left[4^2+\left(x-6\right)\right]=90\)
\(\Rightarrow20-16-x+6=90\)
\(\Rightarrow10-x=90\)
\(\Rightarrow x=-80\)
Vậy: \(x=-80\)
\(b,\left(x+3\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\2x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;2\right\}\)
\(c,1000:\left[30+\left(2^x-6\right)\right]=3^2+4^2\left(x\in N\right)\)
\(\Rightarrow1000:\left(30+2^x-6\right)=25\)
\(\Rightarrow24+2^x=40\)
\(\Rightarrow2^x=16\)
\(\Rightarrow x=4\)
Vậy: \(x=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1000 : [30 + (2x - 6)] = 32 + 42
= 1000 : [ 30 + (2x-6)] = 9 + 16 = 25
= 30 + (2x - 6) = 1000 : 25 = 40
= 2x - 6 = 40 - 30 = 10
2x = 10 + 6 = 16
⇒ 2x = 24
⇒ x = 4
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1000:\text{[}30+\left(2^x-6\right)\text{]}=3^2+4^2\)
\(1000:\text{[}30+\left(2^x-6\right)\text{]}=9+16\)
\(1000:\text{[}30+\left(2^x-6\right)\text{]}=25\)
\(\text{ }30+\left(2^x-6\right)\text{ }=40\)
\(2^x-6=10\)
\(2^x=16\)
\(=>2^x=2^4\)
\(=>x=4\)
\(1000:\left[30+\left(2^x-6\right)\right]=3^2+4^2\\ 1000:\left[30+\left(2^x-6\right)\right]=9+16\\ 1000:\left[30+\left(2^x-6\right)\right]=25\\ 30+\left(2^x-6\right)=1000:25\\ 30+\left(2^x-6\right)=40\\ 2^x-6=40-30\\ 2^x-6=10\\ 2^x=10+6\\ 2^x=16\\ 2^x=2^4\\ x=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
1: =>5(2x+6)=40
=>2x+6=8
=>2x=2
=>x=1
2: =>12-(x+3)=256:64=4
=>(x+3)=8
=>x=5
3: =>2x-1=3 hoặc 2x-1=-3
=>x=2 hoặc x=-1
4: \(\Leftrightarrow3^{x+2017}=3^{2015}\)
=>x+2017=2015
=>x=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Rightarrow42-2x-32+6=6\\ \Rightarrow16-2x=6\\ \Rightarrow2x=10\\ \Rightarrow x=5\)
\(\Rightarrow42-2x-32+6=6\\ \Rightarrow16-2x=6\\ \Rightarrow2x=10\\ \Rightarrow x=5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
\(2,\)
\(a,20-\left[42+\left(x-6\right)\right]=90\)
\(\Rightarrow20-42-x+6-90=0\)
\(\Rightarrow x=-106\)
Vậy: \(x=-106\)
\(b,\left(x+3\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\2x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;2\right\}\)
\(c,1000:\left[30+\left(2x-6\right)\right]=32+42\left(x\in N\right)\)
\(\Rightarrow1000:\left(30+2x-6\right)=74\)
\(\Rightarrow1000:\left(24+2x\right)=74\)
\(\Rightarrow24+2x=\dfrac{500}{37}\)
\(\Rightarrow2x=-\dfrac{388}{37}\)
\(\Rightarrow x=-\dfrac{194}{37}\)
Mà \(x\in N\)
\(\Rightarrow x\in\varnothing\)
Vậy: \(x\in\varnothing\)
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