\(A=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-...+\frac{1}{2^{2014}}-\frac{1}{2^{2016}}\)

\(\Rightarrow2^2A=1-\frac{1}{2^2}+\frac{1}{2^4}-\frac{1}{2^6}+\frac{1}{2^8}-...+\frac{1}{2^{2012}}-\frac{1}{2^{2014}}\)

\(\Rightarrow2^2A+A=1+\left(\frac{1}{2^2}-\frac{1}{2^2}\right)+\left(\frac{1}{2^4}-\frac{1}{2^4}\right)+...+\left(\frac{1}{2^{2014}}-\frac{1}{2^{2014}}\right)-\frac{1}{2^{2016}}\)

\(\Rightarrow5A=1-\frac{1}{2^{2016}}< 1\Rightarrow A< \frac{1}{5}=0,2\)

đây là toán lớp 2 hả?

đây là toán lớp mấy thế

A= \(2^0+2^1+2^2+...+2^{50}\)

\(\Rightarrow\)2A =2(\(2^0+2^1+2^2+...+2^{50}\))

\(\Rightarrow\)2A= \(2+2^2+2^3+2^4+...+2^{51}\)

\(\Rightarrow\)2A-A= (\(2+2^2+2^3+2^4+...+2^{51}\))-(\(2+2^2+2^3+2^4+...+2^{50}\))

\(\Rightarrow\)A= \(2^{51}-1\)

a, S = 2 + 2² + 2³ + 2⁴ + ... + 2⁹⁹ + 2¹⁰⁰. 2S = 2² + 2³ + 2⁴ + 2⁵ + ... + 2¹⁰⁰ + 2¹⁰¹ => 2S - S = S = (2² + 2³ + 2⁴ + 2⁵ + ... + 2¹⁰⁰ + 2¹⁰¹) - (2 + 2² + 2³ + 2⁴ + ... + 2⁹⁹ + 2¹⁰⁰) = 2¹⁰¹ - 2. Vậy S = 2¹⁰¹ - 2. b, S = 2 + 2² + 2³ + 2⁴ + ... + 2⁹⁹ + 2¹⁰⁰ = (2 + 2²) + (2³ + 2⁴) + ... + (2⁹⁹ + 2¹⁰⁰) = 2.(1 + 2) + 2³.(1 + 2) + ... + 2⁹⁹.(1 + 2) = (1 + 2).(2 + 2³ + ... + 2⁹⁹) = 3.(2 + 2³ + ... + 2⁹⁹) => S ⋮ 3. Vậy S ⋮ 3 (đpcm)

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.