Question

[Linh Linh]

Giúp mik vs

Bài 1:

a) so sánh 3²⁰⁰ và 2³⁰⁰

b)71⁵⁰ và 37⁷⁵

c)\(\dfrac{201201}{202202}\)và \(\dfrac{201201201}{202202202}\)

Bài 2:

a) Cho A=\(\dfrac{1}{1^2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{50^2}\) chứng minh rằng A<2

b)Cho B=2¹+2²+2³+...+2³⁰ chứng minh rằng B chia hết cho 21

Answer 1

[Khôngg Tồnn Tạii]

Bài 1:

a) Ta có:

\(3^{200}=\left(3^2\right)^{100}=9^{100}\)

\(2^{300}=\left(2^3\right)^{100}=8^{100}\)

Vì \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)

b) Ta có:

\(71^{50}=\left(71^2\right)^{25}=5041^{25}\)

\(37^{75}=\left(37^3\right)^{25}=50653^{25}\)

Vì \(5041^{25}< 50653^{25}\Rightarrow71^{50}< 37^{75}\)

c) Ta có:

\(\frac{201201}{202202}=\frac{201.1001}{202.1001}=\frac{201}{202}\)

\(\frac{201201201}{202202202}=\frac{201.1001001}{202.1001001}=\frac{201}{202}\)

\(\Rightarrow\frac{201201}{202202}=\frac{201201201}{202202202}\)

Answer 2

[Khôngg Tồnn Tạii]

Bài 2:

a) \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}\)

Ta có: \(\frac{1}{1^2}=1;\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};....;\frac{1}{50^2}< \frac{1}{49.50}\)

\(\Rightarrow\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)

\(\Rightarrow A< 1+1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)

\(\Rightarrow A< 1+1-\frac{1}{50}\)

\(\Rightarrow A< 2-\frac{1}{50}< 2\)

b) \(B=2^1+2^2+2^3+...+2^{30}\) (Có 30 số hạng)

\(\Rightarrow B=\left(2^1+2^2+...+2^5+2^6\right)+\left(2^7+2^8+2^9+...+2^{12}\right)+...+\left(2^{25}+2^{26}+...+2^{29}+2^{30}\right)\)

(có \(30:6=5\) nhóm)

\(\Rightarrow B=1\left(2^1+2^2+...+2^6\right)+2^6\left(2^1+2^2+...+2^6\right)+.....+2^{24}\left(2^1+2^2+...+2^6\right)\)

\(\Rightarrow B=1.126+2^6.126+2^{12}.126+...+2^{24}.126\)

\(\Rightarrow B=126.\left(1+2^6+2^{12}+...+2^{24}\right)\)

\(\Rightarrow B=21.6.\left(1+2^6+2^{12}+...+2^{24}\right)⋮21\)

\(\Rightarrow B⋮21\)