x=1

vì 1²⁵=1(1 mũ bao nhiêu cũng bằng 1)

\(1^x=1\)

\(0^x=0\)

VÌ 1 mũ bao nhiêu cx =1

0mux bao nhiêu cx =0

\(x^{25}=x\)

\(\Rightarrow x^{25}-x=0\)

\(\Rightarrow x\left(x^{24}-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x^{24}-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}}\)

Vậy\(x=0;\pm1\)

x¹⁰ = 1^x

=> x¹⁰ = 1 (vì 1 mũ mấy cũng bằng 1)

=> x¹⁰ = 1¹⁰

=> x = 1.

\(x^{10}=1^x\)

\(\Rightarrow x^{10}=1\)

\(\Rightarrow x^{10}=1^{10}\)

\(\Rightarrow x=1\)

Bài 1:

a) \(8^5\cdot8^2=8^7\)

b) \(9^3\cdot3^2=\left(3^2\right)^3\cdot3^2=3^6\cdot3^2=3^8\)

c) \(2^7\cdot5^7=10^7\)

d) \(27^6:3^3=\left(3^3\right)^6:3^3=3^{18}:3^3=3^{15}\)

Bài 2:

a) \(x^6:x^3=125\)

\(\Rightarrow x^3=125\)

\(\Rightarrow x=5\)

b) \(x^{20}=x\)

\(\Rightarrow x^{20}-x=0\)

\(\Rightarrow x\left(x^{19}-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{19}-1=0\Rightarrow x=1\end{matrix}\right.\)

c) \(3^x\cdot3=243\)

\(\Rightarrow3^x=81\)

\(\Rightarrow x=4\)

d) \(2x-138=2^3\cdot3^2\)

\(\Rightarrow2x-138=72\)

\(\Rightarrow2x=200\)

\(\Rightarrow x=100\)

Giải:

Bài 1:

a) \(8^5.8^2=8^{5+2}=8^7\)

b) \(9^3.3^2=3^6.3^2=3^{6+2}=3^8\)

c) \(2^7.5^7=\left(2.5\right)^7=10^7\)

d) \(27^6:3^3=3^{18}:3^3=3^{18-3}=3^{15}\)

Bài 2:

a) \(x^6:x^3=x^{6-3}=x^3=125\)

\(\Leftrightarrow x=5\)

b) \(x^{20}=x\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=1\end{matrix}\right.\)

c) \(3^x.3=243\)

\(\Leftrightarrow3^{x+1}=243\)

\(\Leftrightarrow3^{x+1}=3^5\)

\(\Leftrightarrow x+1=5\Leftrightarrow x=4\)

d) \(2.x-138=2^3.3^2\)

\(\Leftrightarrow2.x-138=8.9\)

\(\Leftrightarrow2.x-138=72\)

\(\Leftrightarrow2.x=72+138\)

\(\Leftrightarrow2.x=210\Leftrightarrow x=105\)

Chúc bạn học tốt!

Bài 1 : Viết kết quả sau dưới dạng một lũy thừa :

a) 8⁵ . 8² = 8⁷

b) 9³ . 3² = 9³ . 9 = 9⁴

c) 2⁷ . 5⁷ = 10⁷

d) 27⁶ : 3³ = 27⁶ : 27 = 27⁵

Bài 2 : Tìm x biết :

a) x⁶ : x³ = 125

\(\Rightarrow\) x³ = 5³

\(\Rightarrow\) x = 5.

b) x²⁰ = x

- Nếu x = 0 thì 0²⁰ = 0 (chọn)

- Nếu x = 1 thì 1²⁰ = 1 (chọn)

- Nếu x \(\ge\) 2 thì x²⁰ = x (vô lí)

Vậy x \(\in\) {0; 1}.

c) 3^x . 3 = 243

\(\Rightarrow\) 3^x . 3 = 3⁵

3^x = 3⁵ : 3

3^x = 3⁴

\(\Rightarrow\) x = 4.

d) 2x - 138 = 2³ . 3²

2x - 138 = 8 . 9

2x - 138 = 72

2x = 72 + 138

2x = 210

x = 210 : 2

x = 105.

Bài 3 : So sánh :

c) 4⁴ < 64⁷

Bài 4 : Cho n! =1.2.3.4...........n . Tính 5! - 4!.

5! - 4! = 120 - 24 = 96.

2^161>2^160, 2^160=2^4*40=16^40

16^40>13^40

=> 2^161>13^40

Lần sau viết cái đề rõ rõ ra nhs!!!

a) \(A=2+2^2+2^3+................+2^{100}\)

\(\Rightarrow2A=2^2+2^3+2^4+................+2^{100}+2^{101}\)

\(\Rightarrow2A-A=\left(2^2+2^3+..............+2^{100}+2^{101}\right)-\left(2+2^2+............+2^{100}\right)\)

\(\Rightarrow A=2^{101}-2\)

b) \(B=1+3+3^2+..................+3^{2009}\)

\(\Rightarrow3B=3+3^2+3^3+..................+3^{2009}+3^{2010}\)

\(\Rightarrow3B-B=\left(3+3^2+...............+3^{2010}\right)-\left(1+3+3^2+.............+3^{2009}\right)\)

\(\Rightarrow2B=3^{2010}-1\)

\(\Rightarrow B=\dfrac{3^{2010}-1}{2}\)

c) \(C=4+4^2+4^3+................+4^n\)

\(\Rightarrow4C=4^2+4^3+.................+4^n+4^{n+1}\)

\(\Rightarrow4C-C=\left(4^2+4^3+.............+4^n+4^{n+1}\right)-\left(4+4^2+............+4^n\right)\)

\(\Rightarrow3C=4^{n+1}-4\)

\(\Rightarrow C=\dfrac{4^{n+1}-4}{3}\)

\(A=2+2^2+2^3+...+2^{100}\)

\(2A=2^2+2^3+2^4+...+2^{101}\)

\(\Rightarrow2A-A=2^{101}-2\)

\(\Rightarrow A=2^{101}-2\)

Vậy \(A=2^{101}-2\).

\(B=1+3^2+3^3+...+3^{2009}\)

\(3B=3+3^3+3^4+...+3^{2010}\)

\(\Rightarrow3B-B=3^{2010}-7\)

\(\Rightarrow2B=3^{2010}-7\)

\(\Rightarrow B=\dfrac{3^{2010}-7}{2}\)

Vậy \(B=\dfrac{3^{2010}-7}{2}\).

\(C=4+4^2+4^3+...+4^n\)

\(4C=4^2+4^3+4^4+...+4^{n+1}\)

\(\Rightarrow4C-C=4^{n+1}-4\)

\(\Rightarrow3C=4^{n+1}-4\)

\(\Rightarrow C=\dfrac{4^{n+1}-4}{3}\)

Vậy \(C=\dfrac{4^{n+1}-4}{3}\).