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2: 12-10x=25-30x
=>20x=13
=>x=13/20
3: \(3\left(2x+3\right)-2\left(4x-5\right)=10x+21\)
=>6x+9-8x+10=10x+21
=>10x+21=-2x+19
=>12x=-2
=>x=-1/6
4: \(\Leftrightarrow25x-15-6x+12=11-5x\)
=>19x-3=11-5x
=>24x=14
=>x=7/12
5: \(\Leftrightarrow8-12x-5+10x=4-6x\)
=>4-6x=-2x+3
=>-4x=-1
=>x=1/4
6: \(\Leftrightarrow32x-24-6+9x=13-40x\)
=>41x-30=13-40x
=>81x=43
=>x=43/81
7: \(\Leftrightarrow10x-5+20x=5x-11\)
=>30x-5=5x-11
=>25x=-6
=>x=-6/25
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3) \(...\Rightarrow2^x\left(2^3+1\right)=36\)
\(\Rightarrow2^x.9=36\)
\(\Rightarrow2^x=4\)
\(\Rightarrow2^x=2^2\Rightarrow x=2\)
4) \(...\Rightarrow4^{x+1}-4^x=12\)
\(\Rightarrow4^x\left(4-1\right)=12\)
\(\Rightarrow4^x.3=12\)
\(\Rightarrow4^x=4=4^1\Rightarrow x=1\)
5) \(...\Rightarrow5^{x+1}\left(5^2-1\right)=3000\)
\(\Rightarrow5^{x+1}.24=3000\)
\(\Rightarrow5^{x+1}=125\)
\(\Rightarrow5^{x+1}=5^3\)
\(\Rightarrow x+1=3\)
\(\Rightarrow x=2\)
6) Bạn xem lại đề
a. \(2^x.2^3+2^x=36\)
\(2^x\left(2^3+1\right)=36\)
\(2^x.9=36\)
\(2^x=4\Rightarrow x=2\)
b. \(4^x.4^1-\left(2^2\right)^x=12\)
\(4^x.4-4^x=12\)
\(4^x\left(4-1\right)=12\)
\(4^x.3=12\)
\(4^x=4\)
x = 1
c. \(5^x.5^3-5^x.5^1=3000\)
\(5^x\left(5^3-5^1\right)=3000\)
\(5^x.120=3000\)
\(5^x=25\)
x = 2
d. \(4^{x+1}=2^{2x}\)
\(4^x.4=\left(2^2\right)^x\)
\(4^x.4=4^x\)
Có vẻ như câu 4 này để bài thiếu
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1)8(4x-3)-3(2-3x)=13-40x
(32x-24)-(6-9x)=13-40x
32x-24-6+9x=13-40x
41x-30=13-40x
41x+40x=13+30
81x=43
x=43/81
Vậy x=43/81
2)10x-5(1-4x)=5x-11
10x-(5-20x)=5x-11
10x-5+20x=5x-11
30x-5=5x-11
30x-5x=5-11
25x=-6
x=-6/25
Vậy x=-6/25
![](https://rs.olm.vn/images/avt/0.png?1311)
5x-(3+4x)=5
=>5x-3-4x=5
=>5x-4x-3=5
=>x-3=5
=>x=5+3
=>x=8
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a) (x-3)100-1=0
\(\Leftrightarrow\left(x-3\right)^{100}=0+1=1\)
\(\Leftrightarrow\left(x-3\right)^{100}=1^{100}=\left(-1\right)^{100}\)
\(\Rightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1+3=4\\x=\left(-1\right)+3=2\end{matrix}\right.\)
Vậy: x\(\in\left\{4;2\right\}\)
b) (9-5x)2019+1=0
\(\Leftrightarrow\left(9-5x\right)^{2019}=0-1=-1=\left(-1\right)^{2019}\)
\(\Rightarrow9-5x=-1\)
\(\Leftrightarrow5x=9-\left(-1\right)=10\)
\(\Leftrightarrow x=10:5=2\)
Vậy: x=2
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\(\left(2x+3\right)\cdot\left(10x+2\right)=\left(5x+2\right)\cdot\left(4x+5\right)\)
\(\Leftrightarrow20x^2+4x+30x+6-20x^2-25x-8x-10=0\)
\(\Leftrightarrow x-4=0\)
\(\Leftrightarrow x=4\)
\(< =>20x^2+4x+30x+6=20x^2+25x+8x+10\)
\(< =>34x+6-25x-8x-10=0\)
\(< =>x-4=0< =>x=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : 5x-(3+4x) =5
5x-3-4x =5
(5x-4x)-3=5
x-3 =5
x =8
Vậy x=8
\(5x-\left(3+4x\right)=5\\ \Leftrightarrow5x-3-4x=5\\ \Leftrightarrow x=8\)
Vậy $x=8$ là nghiệm phương trình