a) 16¹⁹ và 8²⁵ 

Ta có :

16¹⁹ = ( 2⁴ )¹⁹ = 2⁷⁶

8²⁵ = ( 2³ )²⁵ = 2⁷⁵

Vì 2⁷⁶ > 2⁷⁵ Nên 16¹⁹ > 8²⁵

b) 5³⁶ và 11²⁴

Ta có :

5³⁶ = ( 5³ )¹² = 125¹²

11²⁴ = ( 11² )¹² = 121¹²

Vì 125¹² > 121¹² Nên 5³⁶ > 11²⁴

1.

\(M=3^0+3^1+......+3^{50}.\)

\(\Rightarrow3M=3+3^2+.......+3^{51}\)

\(\Rightarrow3M-M=\left(3+3^2+.......+3^{51}\right)-\left(3^0+3+.....+3^{50}\right)\)

\(\Rightarrow2M=3^{51}-1\)

\(\Rightarrow M=\frac{3^{51}-1}{2}\)

2.

\(a,\)Ta có : \(16^{19}=\left(2^4\right)^{19}=2^{76}\)

                     \(8^{25}=\left(2^3\right)^5=2^{75}\)

Vì \(2^{76}>2^{75}\Rightarrow16^{19}>8^{25}\)

\(b,\)Ta có : \(5^{36}=\left(5^3\right)^{12}=125^{12}\)

                      \(11^{24}=\left(11^2\right)^{12}=121^{12}\)

Vì \(125^{12}>121^{12}\Rightarrow5^{36}>11^{24}\)

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? crisdevergamemer

A = 1 + 2 +2² +...+ 2⁹⁹

2A = 2 + 2² + 2³ +....+2¹⁰⁰

2A - A =2 + 2² + 2³ +....+2¹⁰⁰ -1 + 2 +2² +...+ 2⁹⁹

A = 2¹⁰⁰ -1

B = 4⁵⁰ +1

B = 2¹⁰⁰ + 1

=> B > A

A=1+2+2²+...+2⁹⁹

A= 1 + 2¹⁰⁰ -1

A = 2¹⁰⁰

B=4⁵⁰ + 1 = 2¹⁰⁰ + 1

\(\Rightarrow\)A<B

cái A mk suy ra dc nho quy nạp toán học

\(A=1+2+2^2+...+2^{99}\)

\(A=1+2^{100}-1\)

\(A=2^{100}\)

\(B=4^{50}+1=2^{100}+1\)

\(\Rightarrow B>A\)

\(S=\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+\frac{7}{3^2\cdot4^2}+...+\frac{99}{49^2\cdot50^2}\)

\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+.....+\frac{1}{49^2}-\frac{1}{50^2}\)

\(=1-\frac{1}{50^2}=\frac{2499}{2500}\)

\(T=\frac{1}{\left(2-1\right)\left(2+1\right)}+\frac{1}{\left(3-1\right)\left(3+1\right)}+...+\frac{1}{\left(50-1\right)\left(50+1\right)}\)

\(=\frac{1}{1\cdot3}+\frac{1}{2\cdot4}+\frac{1}{3\cdot5}+...+\frac{1}{49\cdot51}\)

\(=\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{1}{2}\cdot\left(1+\frac{1}{2}-\frac{1}{51}\right)=\frac{151}{204}\)

Vì \(\frac{2499}{2500}>\frac{151}{204}\)nên S>T

JOKER_Võ Văn Quốc, T = \(\frac{1}{2}.\left(1-\frac{1}{51}+\frac{1}{2}-\frac{1}{50}\right)\)mới đúng
Sẽ dễ hơn nếu bạn chia ra 2 vế \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)và \(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{48+50}\)

Ta có :

\(\frac{1}{50}>\frac{1}{100}\)

\(\frac{1}{51}>\frac{1}{100}\)

............

\(\frac{1}{98}>\frac{1}{100}\)

\(\frac{1}{99}>\frac{1}{100}\)

\(\Rightarrow\frac{1}{50}+\frac{1}{51}+....+\frac{1}{98}+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+....+\frac{1}{100}=\frac{50.1}{100}=\frac{1}{2}\)

\(\Rightarrow M>\frac{1}{2}\)

Ta có: \(\frac{1}{50}>\frac{1}{51}>....>\frac{1}{99}\)

\(\Rightarrow M>\frac{1}{99}+\frac{1}{99}+...+\frac{1}{99}=\frac{50}{99}>\frac{50}{100}=\frac{1}{2}\)

 Vậy M > 1/2

Ta có :

M = \(\frac{1}{50}+\frac{1}{51}+...+\frac{1}{98}+\frac{1}{99}\)

M > \(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\)( 50 số hạng )

\(=\frac{1}{100}.50=\frac{1}{2}\)

Vậy M > \(\frac{1}{2}\)

Ta có : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{99.100}.\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)\(< 1\)

Vậy : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{99.100}< 1\)

Đặt :

\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\)

   \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

  \(=1-\frac{1}{100}\)

 \(=\frac{99}{100}\)

Vậy  \(A=\frac{99}{100}\)

Vì \(\frac{99}{100}< 1\)nên \(A< 1\)

Học tốt #

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}< 1\)

1/3+1/3²+...+1/3⁹⁹<1/2