a)  2x-3=-x+6

2x - 3 + x - 6 =0

3x -9 = 0

3x = 9

x = 9 : 3

x= 3

c/ \(\left(-12x-4^3\right).8^3=4.8^4\)

\(\left(-12x-64\right).512=16384\)

\(-12x-64=\dfrac{16384}{512}=32\)

\(-12x=32+64=96\)

\(x=\dfrac{96}{\left(-12\right)}=-8\)

a,   2x-3=-x+6

\(\Leftrightarrow\)2x-3+x-6=0

\(\Leftrightarrow\)3x-9=0

\(\Leftrightarrow\)3x=9

\(\Leftrightarrow\)x=3

a) Ta có: \(2x-3=-x+6\)

\(\Leftrightarrow2x+x=6+3\)

\(\Leftrightarrow3x=9\)

hay x=3

Vậy: x=3

b) Ta có: \(\left|2x-7\right|+2=13\)

\(\Leftrightarrow\left|2x-7\right|=11\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=11\\2x-7=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=18\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{9;-2\right\}\)

c) Ta có: \(\left(-12x-4^3\right)\cdot8^3=4\cdot8^4\)

\(\Leftrightarrow\left(-12x-64\right)=\dfrac{4\cdot8^4}{8^3}=32\)

\(\Leftrightarrow-12x=96\)

hay x=-8

Vậy: x=-8

\(a,\Rightarrow x=30-18=12\\ b,\Rightarrow x+6=45:5=9\\ \Rightarrow x=9-6=3\\ c,\Rightarrow38-3x=4^2=16\\ \Rightarrow3x=38-16=22\\ \Rightarrow x=\dfrac{22}{3}\)

a: =>2x^3=58-4=54

=>x^3=27

=>x=3

b; =>(5-x)^5=2^5

=>5-x=2

=>x=3

c: =>(5x-6)^3=4^3

=>5x-6=4

=>5x=10

=>x=2

d: (3x)^3=(2x+1)^3

=>3x=2x+1

=>x=1

1=>2x³=54
=>x³=27  =>x=3
2=>(5-x)⁵=2⁵
=>5-x=2
=>x=3
3=>(5x-6)³=4³
=>5x-6=4
=>5x=10=>x=2
4=>3x=2x+1
=>x=1

Bài 9,

6²x73+36x3³=36x73+36x27=36(73+27)=36x100=3600.

197-\([\)6x(5-1)²+2022⁰\(]\):5=197-\([\)6x16+1\(]\):5=197-97:5=197-97/5=888/5.

Bài 10,

21-4x=13

=>4x=21-13=8

=>x=8:4=2.

30:(x-3)+1=4⁵:4³=4²=16

=>30:(x-3)=16-1=15

=>x-3=30:15=2

=>x=2+3=5.

(x-1)³+5x6=38

=>(x-1)³+30=38

=>(x-1)³=38-30=8=2³

=>x-1=2

=>x=3.

Bài 1

a) \(x=x^5\)

\(x^5-x=0\)

\(x\left(x^4-1\right)=0\)

\(x=0\) hoặc \(x^4-1=0\)

* \(x^4-1=0\)

\(x^4=1\)

\(x=1\)

Vậy x = 0; x = 1

b) \(x^4=x^2\)

\(x^4-x^2=0\)

\(x^2\left(x^2-1\right)=0\)

\(x^2=0\) hoặc \(x^2-1=0\)

*) \(x^2=0\)

\(x=0\)

*) \(x^2-1=0\)

\(x^2=1\)

\(x=1\)

Vậy \(x=0\); \(x=1\)

c) \(\left(x-1\right)^3=x-1\)

\(\left(x-1\right)^3-\left(x-1\right)=0\)

\(\left(x-1\right)\left[\left(x-1\right)^2-1\right]=0\)

\(x-1=0\) hoặc \(\left(x-1\right)^2-1=0\)

*) \(x-1=0\)

\(x=1\)

*) \(\left(x-1\right)^2-1=0\)

\(\left(x-1\right)^2=1\)

\(x-1=1\) hoặc \(x-1=-1\)

**) \(x-1=1\)

\(x=2\)

**) \(x-1=-1\)

\(x=0\)

Vậy \(x=0\); \(x=1\); \(x=2\)

a) 128 - 3(x + 4) = 23

             3(x + 4) = 128 -23

             3(x + 4) = 105

                x + 4  = 105 : 3

                x + 4  = 35

                x        = 35 - 4

                x        = 31

\(a.128-3\left(x+4\right)=23.\)

\(3\left(x+4\right)=128-23\)

\(3\left(x+4\right)=105\)

\(x+4=105:3=35\)

\(x=35-4=31\)

\(b.\left[\left(4x+28\right)\cdot3+55\right]:5=35\)

\(\left(4x+28\right)\cdot3+55=35\cdot5=175\)

\(\left(4x+28\right)\cdot3=175-55=120\)

\(4x+28=120:3=40\)

\(4x=40-28=12\)

\(x=12:4=3\)

a, \(128-3.\left(x+4\right)=23\)

\(\Rightarrow\) \(3.\left(x+4\right)=105\)

\(\Rightarrow\) \(x+4=35\)

\(\Rightarrow\) \(x=31\)

b, \(\left[\left(4x+28\right).3+55\right]:5=35\)

\(\Rightarrow\) \(\left(4x+28\right).3+55=175\)

\(\Rightarrow\) \(\left(4x+28\right).3=120\)

\(\Rightarrow\) \(4x+28=40\)

\(\Rightarrow\) \(4x=12\)

\(\Rightarrow\) \(x=3\)

1. a) 4.415.8.25.125

= (4.25). (8.125).415

= 100.1000.415

= 100000.415

= 41500000

b) 2.31.12+4.42.6+8.27.3

= (2.31.12)+(4.42.6)+(8.27.3)

= (2.12).31+(4.6).42+(8.3).27

= 24.31+24.42+24.27

= 24 (31+42+27)

= 24.100

= 2400

Bài 6 :

a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)

b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)

c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)

d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)

Bài 7 :

a) \(3^x+3^{x+2}=9^{17}+27^{12}\)

\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)

\(\Rightarrow10.3^x=3^{34}+3^{36}\)

\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)

\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)

b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)

\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)

\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)

c) Bài C bạn xem lại đề

d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)

\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)

\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)

\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)

\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)

\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)

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