Trả lời :

5⁶ : 5⁴ + 2³ . 2² - 225 : 15²

= 5^{6 - 4} + 2^{3 + 2} - 15² : 15²

= 5² + 2⁵ - 1

= 25 + 32 - 1

= 56 

5^6:5^4+2^3×2^2-15^2:15^2

5^2+2^5-1

25+32-1

56

\(5^6:5^4+2^3.2^2-225:15^2\)

\(=5^{6-4}+2^{3+2}-15^2:15^2\)

\(=5^2+2^5-1\)

\(=25+32-1\)

\(=56\)

\(5^{x+1}=125\)

\(5^{x+1}=5^3\)

\(x+1=3\)

\(x=2\)

b, \(5^{2x-3}-2.5^2=5^2.3\)

    \(5^{2x-3}=5^2.\left(3+2\right)\)

    \(5^{2x-3}=5^3\)

    \(2x-3=3\)

     \(x=3\)

c, \(2^x=32\)

    \(2^x=2^5\)

    \(x=5\)

Chúc em học tốt.

Trả lời:

( x - 5 )⁵ = 32

<=> ( x - 5 ) = 2⁵

=> x - 5 = 2

<=> x = 2 + 5 

<=> x = 7

Vậy x = 7

Sửa lại:

\(\left(x-5\right)^5=32\)

\(\left(x-5\right)^5=2^5\)

\(x-5=2\)

\(x=7\)

\(\left(x-5\right)^5=32^5\)

\(x-5=32\)

\(x=32+5\)

\(x=37\)

\(2^x+2^{x+2}=32.\left(2^2+1\right)\)

\(\Rightarrow2^x+2^{x+2}=32.5\)

\(\Rightarrow2^x+2^{x+2}=160\)

\(\Rightarrow2^x\left(1+4\right)=160\)

\(\Rightarrow2^x.5=160\)

\(\Rightarrow2^x=160:5=32\)

\(\Rightarrow2^x=2^5\)

\(\Rightarrow x=5\)

Vậy x = 5

a: \(16^x< 32^4\)

=>\(2^{4x}< 2^{20}\)

=>4x<20

=>\(x< 5\)

=>0<=x<5

=>\(x\in\left\{0;1;2;3;4\right\}\)

b: \(9< 3^x< 81\)

=>\(3^2< 3^x< 3^4\)

=>2<x<4

=>x=3

c: \(25< 5^x< 125\)

=>\(5^2< 5^x< 5^3\)

=>2<x<3

mà x là số tự nhiên

nên \(x\in\varnothing\)

cô ơi cho con hỏi 

11 mũ 5 chia cho 11 mũ n trừ 4 bằng 11 mũ 5 

giúp tính bài giải cho con nhé

\(\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{29}\cdot9^{10}-7\cdot2^{29}\cdot27^6}\)

\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot2^{27}\cdot3^{20}}{5\cdot2^{29}\cdot3^{20}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{2^{29}\cdot3^{18}\left(5\cdot2-3^2\right)}{2^{29}\cdot3^{18}\left(5\cdot3^2-7\right)}\)

\(=\dfrac{10-9}{5\cdot9-7}=\dfrac{1}{38}\)

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: .... 

\(23+3x=5^6:5^3\)                         \(9^{x-1}=9\)                            \(x^4=16\)                                              \(2^x:2^5=1\)

\(23+3x=5^2\)                                => x-1= 0                                     \(x^4=2^4\)                                               \(2^x=1.2^5\)

\(23+3x=25\)                                x= 0+1                                           => x= 2                                                       \(2^x=2^5\)

           3x= 25-23                                       x= 1                                                                                                                => x= 5

          3x= 2

           x= \(\frac{2}{3}\)

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