Bài 2: a: <

        b: >

        c :=

Thông cảm nhé bài 1 mình ko biết

Bài 2 : So sánh: 

10⁷⁵⁰⁰ = 10^12.625 = ( 10¹²)⁶²⁵

11⁵⁰⁰⁰ = 11^8.625 = ( 11⁸)⁶²⁵ 

  Dễ thấy 10¹² > 11⁸ nên 10⁷⁵⁰⁰ > 11⁵⁰⁰⁰ 

Chúc bạn học tốt!!!

c) 100² và 97 . 103

100² = 10000

97 . 103 = 9991

Vì 10000 > 9991 

Nên 100² > 97 . 103

ta có: \(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^9}=1\)

mà \(1+3+3^2+...+3^9>1+3+3^2+...+3^8\)

\(\Rightarrow B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}>1\)

\(\Rightarrow A< B\)

Câu hỏi của nguyen van nam - Toán lớp 6 - Học toán với OnlineMath

kieu nay la ko tinh ra ket qua hay so sanh

A=1+C; voi C=5^9/(1+...5^8)=1/(1/5^9+1/5^8+...+1/5)

B=1+D;voi D=3^9/(1+..3^8)=1/(1/3^9+1/3^8+...+1/3)

C=1/E; voi E=(1/5^9+1/5^8+...+1/5)

D=1/f; voi F=(1/3^9+1/3^8+...+1/3)

=> F-E=(1/3-1/5)+...+(1/3^9-1/5^9) >0=> F>E

=> C>D=> A>B

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K mk nhé

a)\(3\frac{4}{10}>2\frac{9}{10}\)

b)\(3\frac{4}{10}< 3\frac{9}{10}\)

c)\(5\frac{1}{10}>2\frac{9}{10}\)

d)\(3\frac{4}{10}=3\frac{2}{5}\)

Hok tốt!~

a) 3 9/10 > 2 9/10

b) 3 4/10 < 3 9/10

c) 5 1/10 > 2 9/10

d) 3 4/10 = 3 2/5

a) \(3\cdot24^{10}=3\cdot6^{10}\cdot4^{10}=3\cdot3^{10}\cdot2^{10}\cdot2^{20}\)

\(=3^{11}\cdot2^{30}\)

\(4^{30}=2^{30}\cdot2^{30}=2^{30}\cdot4^{15}\)

Ta có \(4^{15}>3^{15}>3^{11}\) nên \(4^{15}>3^{11}\)

Khi đó \(4^{15}\cdot2^{30}>3^{11}\cdot2^{30}\) hay \(4^{30}>3\cdot24^{10}\)

b) \(\dfrac{3}{1^2\cdot2^2}+\dfrac{5}{2^2\cdot3^2}+...+\dfrac{19}{9^2\cdot10^2}\)

\(=\dfrac{3}{1\cdot4}+\dfrac{5}{4\cdot9}+...+\dfrac{19}{81\cdot100}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{81}-\dfrac{1}{100}\)

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

Vậy dãy trên nhỏ hơn 1

a/

\(4^{30}=\left(2^2\right)^{30}=2^{60}=2^{30}.2^{30}=\left(2^2\right)^{15}.2^{30}=4^{15}.2^{30}\)

\(3.24^{10}=3.3^{10}.\left(2^3\right)^{10}=3^{11}.2^{30}< 3^{15}.2^{30}\)

\(\Rightarrow4^{30}=4^{15}.2^{30}>3^{15}.2^{30}>3^{11}.2^{30}=3.24^{10}\)

b/

\(=\dfrac{2^2-1^2}{1^2.2^2}+\dfrac{3^2-2^2}{2^2.3^2}+\dfrac{4^2-3^2}{3^2.4^2}+...+\dfrac{10^2-9^2}{9^2.10^2}=\)

\(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{9^2}-\dfrac{1}{10^2}=\)

\(=1-\dfrac{1}{10^2}< 1\)

a) 4³⁰ = (2²)³⁰ = 2⁶⁰ = 2³⁰.2³⁰ = 1073741824.2³⁰

3.24¹⁰ = 3.(3.2³)¹⁰ = 3.3¹⁰.2³⁰ = 3¹¹.2³⁰ = 177147.2³⁰

Do 1073741824 > 177147

⇒ 1073741824.2³⁰ > 177147.2³⁰

Vậy 4³⁰ > 3.24¹⁰

b) 3/(1².2²) + 5/(2².3²) + ... + 19/(9².10²)

= 1/1² - 1/2² + 1/2² - 1/3² + ... + 1/9² - 1/10²

= 1 - 1/100

= 99/100

Mà 99/100 < 1

⇒ 3/(1².2²) + 5/(2².3²) + 7/(3².4²) + ... + 19/(9².10²) < 1

ta thấy: A= 10^m+2/10^m-1>10^m/10^m-1

mà B=10^m/10^m-3<10^m/10^m-1         (m thuộc N * ) ;

=> A<B

tương tự (phân số trung gian)

đang cần gấp ạ 

ÉT Ô ÉT

2/-9<8/-9

-2/5<-3/4

\(a.\dfrac{2}{-9}và\dfrac{8}{-9}\)

Ta có: \(\dfrac{-2}{9}và\dfrac{-8}{9}\)

vì -2 > -8 

-> \(\dfrac{2}{-9}>\dfrac{8}{-9}\)

\(b.\dfrac{-2}{5}và\dfrac{-3}{4}\)

Ta có: \(\dfrac{-2}{5}=\dfrac{-8}{20};\dfrac{-3}{4}=\dfrac{-15}{20}\)

Vì -8 > -15

=> \(\dfrac{-2}{5}>\dfrac{-3}{4}\)