Ta có \(5^x+5^{x+1}=750\)

\(\Rightarrow5^x+5^x.5=750\)

\(\Rightarrow5^x.\left(1+5\right)=750\)

\(\Rightarrow5^x.6=750\)

\(\Rightarrow5^x=125\)

\(\Rightarrow5^x=5^3\)

\(\Rightarrow x=3\)

\(5^x\) +  \(5^{x+1}\) =  \(750\)

=> \(5^x\) + \(5^x\) . \(5\)= \(750\)

=> \(5^x\). \(\left(1+5\right)\)=\(750\)

=> \(5^x\). \(6\)= \(750\)

=>        \(5^x\)= \(125\)

=>         \(5^x\) = \(5^3\)

=>         \(x=3\)

Bài 1:

a) \(4^{x+2}+4^x=68\)

\(\Rightarrow4^x\cdot\left(4^2+1\right)=68\)

\(\Rightarrow4^x\cdot17=68\)

\(\Rightarrow4^x=\dfrac{68}{17}\)

\(\Rightarrow4^x=4\)

\(\Rightarrow4^x=4^1\)

\(\Rightarrow x=1\)

b) \(5\cdot2^{x+4}-3\cdot2^x=308\)

\(\Rightarrow2^x\cdot\left(5\cdot2^4-3\right)=308\)

\(\Rightarrow2^x\cdot\left(5\cdot16-3\right)=308\)

\(\Rightarrow2^x\cdot77=308\)

\(\Rightarrow2^x=\dfrac{308}{77}\)

\(\Rightarrow2^x=4\)

\(\Rightarrow2^x=2^2\)

\(\Rightarrow x=2\)

c) \(4\cdot3^{x+1}+7\cdot3^x=513\)

\(\Rightarrow3^x\cdot\left(4\cdot3+7\right)=513\)

\(\Rightarrow3^x\cdot19=513\)

\(\Rightarrow3^x=\dfrac{513}{19}\)

\(\Rightarrow3^x=27\)

\(\Rightarrow3^x=3^3\)

\(\Rightarrow x=3\)

d) \(5^{x+4}-5^x=3120\)

\(\Rightarrow5^x\cdot\left(5^4-1\right)=3120\)

\(\Rightarrow5^x\cdot\left(625-1\right)=3120\)

\(\Rightarrow5^x\cdot624=3120\)

\(\Rightarrow5^x\cdot\dfrac{3120}{624}\)

\(\Rightarrow5^x=5\)

\(\Rightarrow5^x=5^1\)

\(\Rightarrow x=1\)

f) \(3\cdot4^{2x+1}-16^x=2816\)

\(\Rightarrow3\cdot4^{2x+1}-\left(4^2\right)^x=2816\)

\(\Rightarrow3\cdot4^{2x+1}-4^{2x}=2816\)

\(\Rightarrow4^{2x}\cdot\left(3\cdot4-1\right)=2816\)

\(\Rightarrow4^{2x}\cdot11=2816\)

\(\Rightarrow4^{2x}=\dfrac{2816}{11}\)

\(\Rightarrow4^{2x}=256\)

\(\Rightarrow\left(2^2\right)^{2x}=2^8\)

\(\Rightarrow2^{4x}=2^8\)

\(\Rightarrow4x=8\)

\(\Rightarrow x=2\)

Bài 2:

\(2^x+124=5^y\)

\(\Rightarrow5^y-2^x=124\)

\(\Rightarrow5^y-2^x=125-1\)

\(\Rightarrow\left\{{}\begin{matrix}5^y=125\\2^x=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}5^y=5^3\\2^x=2^0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y=3\\x=0\end{matrix}\right.\)

Vậy: .... 

a. 3^x+3^x+1+3^x+2=1053

=> 3^x.(1+3+3²)=1053

=> 3^x.(1+3+9)=1053

=> 3^x.13=1053

=> 3^x=1053:13

=> 3^x=81

=> 3^x=3⁴

=> x=4

b. 5^x.5¹⁹=5²⁰.5¹¹

=> 5^19+x=5²⁰⁺¹¹

=> 19+x=20+11

=> 19+x=31

=> x=31-19

=> x=12

\(3^x+3^{x+1}+3^{x+2}=1053\)

\(3^x\left(1+3+3^2\right)=1053\)

\(3^x\cdot13=1053\)

\(3^x=1053:13=81\)

\(3^x=3^4\)

=> x = 4 

A)x=4

B)x=12 100% đó

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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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 \(x=0\); \(x=1\); \(x=2\)

Bài 2

a) x=5

b)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.

5^x+5^x+1+5^x+2=?

5^x+5^x+1+5^x+2

=> (5^x+5^x+5^x)+1+2

= 3.5^x + 3

=> x vô nghiệm 

\(5^x+5^{x+1}+5^{x+2}\)

\(=5^x+5^x.5+5^x.5^2\)

\(=5^x\left(1+5+5^2\right)\)

\(=5^x.31\)