Đặt t = 5x, ta có (1)⇔ 1/5.t2 + 5t = 250 ⇔ t2 + 25t - 1250 = 0

⇔ t = 25 hoặc t = -50(loại)

⇔ 5x ⇔ x = 2.

a: \(4x^3+12=120\)

=>\(4x^3=108\)

=>\(x^3=27=3^3\)

=>x=3

b: \(\left(x-4\right)^2=64\)

=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)

c: (x+1)^3-2=5^2

=>\(\left(x+1\right)^3=25+2=27\)

=>x+1=3

=>x=2

d: 136-(x+5)^2=100

=>(x+5)^2=36

=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)

e: \(4^x=16\)

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

=>x=2

f: \(7^x\cdot3-147=0\)

=>\(3\cdot7^x=147\)

=>\(7^x=49\)

=>x=2

g: \(2^{x+3}-15=17\)

=>\(2^{x+3}=32\)

=>x+3=5

=>x=2

h: \(5^{2x-4}\cdot4=10^2\)

=>\(5^{2x-4}=\dfrac{100}{4}=25\)

=>2x-4=2

=>2x=6

=>x=3

i: (32-4x)(7-x)=0

=>(4x-32)(x-7)=0

=>4(x-8)*(x-7)=0

=>(x-8)(x-7)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)

k: (8-x)(10-2x)=0

=>(x-8)(x-5)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)

m: \(3^x+3^{x+1}=108\)

=>\(3^x+3^x\cdot3=108\)

=>\(4\cdot3^x=108\)

=>\(3^x=27\)

=>x=3

n: \(5^{x+2}+5^{x+1}=750\)

=>\(5^x\cdot25+5^x\cdot5=750\)

=>\(5^x\cdot30=750\)

=>\(5^x=25\)

=>x=2

Ta có: \(100< 5^{2x-1}\le5^6\)

\(\Leftrightarrow5^2< 5^{2x-1}\le5^6\)

\(\Leftrightarrow2x-1\in\left\{3;5\right\}\)

\(\Leftrightarrow2x\in\left\{4;6\right\}\)

hay \(x\in\left\{2;3\right\}\)

Ta có: 100 < 52x – 1 < 56

=> 52 < 100 < 52x-1 < 56

=> 2 < 2x – 1 < 6

=> 2 + 1 < 2x < 6 + 1

=> 3 < 2x < 7

Vì x ∈ N nên suy ra: x ∈ {2; 3} là thỏa mãn.

Ta có 100=5².4

\(\Rightarrow5^3\le5^{2x-1}< 5^6\)

\(\Rightarrow3\le2x-1< 6\)

\(\Rightarrow4\le2x< 7\)

\(\Rightarrow2\le x< 3,5\)

Mà \(x\) là số tự nhiên

\(\Rightarrow x=2\) hoặc \(x=3\)

x ϵ {2;3}

Tham Khảo]

(5^2x.5^x+2) : 25 = 125²

5^2x.5^x+2 = 125².25 = 5⁶.5²

5^2x+x+2 = 5⁸

5^3x+2 = 5⁸

=> 3x + 2 = 8

=> 3x = 6

=> x = 2

Ta có: \(\left(5^{2x}\cdot5^{x+2}\right):25=125^2\)

\(\Leftrightarrow5^{3x+2}=5^6\cdot5^2=5^8\)

hay x=2

\(5^{2x+1}\cdot5^{2x+2}-5^x\cdot5^{3x+2}=100\)

\(\Leftrightarrow5^{4x+3}-5^{4x+2}=100\)

\(\Leftrightarrow625^x\left(5^3-5^2\right)=100\)

\(\Leftrightarrow625^x=1\)

hay x=0

\(\Rightarrow5^{2x+1+2x+2}-5^{x+3x+2}=100\\ \Rightarrow5^{4x+3}-5^{4x+2}=100\\ \Rightarrow5^{4x+2}\left(5-1\right)=100\\ \Rightarrow5^{4x+2}=25=5^2\\ \Rightarrow4x+2=2\Rightarrow x=0\)

\(5^{2x+1}.5^{2x+2}-5^x.5^{3x+2}=100\)

\(5^{4x+3}-5^{4x+2}=100\)

\(5^{4x+2}\left(5-1\right)=100\)

\(5^{4x+2}=25\)

4x+2=2

4x=0

x=0

a). 5^x = 125

=>5^x = 5³

=> x = 3

b) 3^2x = 81

=> 3^2x = 3⁴

=> 2x = 4

=> x = 2

c). 5^2x-3 – 2.5² = 5² .3

=>5^2x: 5³ = 5² .3 + 2.5²

=>5^2x: 5³ = 5² .5

=>
5^2x = 5² .5.5³

=>5^2x = 5⁶

=> 2x = 6

=> x=3

a) 5x = 125

x = 125 : 5

x = 25

b) 32x = 81

x = 81 : 32

x = 81/ 32

c) 52x - 3 - 2.52 = 52.3

52x - 3 - 104 = 156

x - 3 - 104 = 156 : 52

x - 3 - 104 = 3

x - 3 = 3 + 104

x - 3 = 107

x = 107 + 3

x = 110

Ko chắc lắm , sai thì Sorry nhé .

\(a,\\ 5x=125\\ < =>x=125:5\\ < =>x=25\)

\(b,\\ 32x=81\\ < =>x=81:32\\ < =>x=\frac{81}{32}\)

\(c,\\ 52x-3-2.52=52.3\\ < =>52x-3-104=156\\ < =>52x-107=156\\ < =>52x=156+107\\ < =>52x=263\\ < =>x=x=\frac{263}{52}\)

Em check lại đề nha

100 sao bé hơn 5 được

làm gì có số nào lớn hơn 5 mà bé hơn 100