So sánh
A = \(\dfrac{3^{10}+1}{3^9+1}\) và B = \(\dfrac{3^9+1}{3^8+1}\)
So sánh hai phân số:
a) \(\dfrac{1}{5}\) và \(\dfrac{3}{5}\) b) \(\dfrac{9}{10}\) và \(\dfrac{3}{10}\) c) \(\dfrac{7}{12}\) và \(\dfrac{11}{12}\) d) \(\dfrac{7}{8}\) và \(\dfrac{5}{8}\)
e) \(\dfrac{17}{100}\) và \(\dfrac{23}{100}\) g) \(\dfrac{4}{10}\) và \(\dfrac{1}{10}\) h) \(\dfrac{100}{100}\) và \(\dfrac{49}{100}\) k) \(\dfrac{15}{15}\) và \(\dfrac{2}{15}\)
a) \(< \)
b) \(>\)
c) \(< \)
d) \(>\)
e) \(< \)
g) \(>\)
h) \(>\)
k) \(>\)
So sánh:
a) \(\dfrac{-9}{4}\) và \(\dfrac{1}{3}\).
b) \(\dfrac{-8}{3}\) và \(\dfrac{4}{-7}\).
c) \(\dfrac{9}{-5}\) và \(\dfrac{7}{-10}\).
em trả lời ccaua này hi vọng thầy còn nhớ em
a) -9/4<`1/3
a) \(\dfrac{-9}{4}< 0\)
\(0< \dfrac{1}{3}\)
Do đó: \(\dfrac{-9}{4}< \dfrac{1}{3}\)
So sánh các phân số:
a,\(\dfrac{1}{5}và\dfrac{3}{5}\) b\(\dfrac{8}{21}và\dfrac{8}{9}\) c,\(\dfrac{3}{5}và1\) d,\(\dfrac{7}{5}và1\)
`a)1<3`
`=>1/5<3/5`
`b)21>9`
`=>8/21<8/9`
`c)3/5<5/5=1`
`d)7/5>5/5=1`
a)3/5>1/5 b)8/21<8/9 c)3/5<1 d)7/5>1
so sánh các hỗn số sau:
\(7\dfrac{4}{5}\) và \(9\dfrac{1}{2}\)
\(7\dfrac{1}{6}\) và \(3\dfrac{4}{5}\)
\(9\dfrac{9}{1}\) và \(5\dfrac{8}{6}\)
\(7\dfrac{4}{5}và9\dfrac{1}{2}\\ Tacó:7< 9\\ \Rightarrow7\dfrac{4}{5}< 9\dfrac{1}{2}\\ 7\dfrac{1}{6}và3\dfrac{4}{5}\\ Tacó:7>3\\ \Rightarrow7\dfrac{1}{6}>3\dfrac{4}{5}\)
Câu cuối không phải hỗn số
A= \(\left(\dfrac{1}{2}-1\right)\)\(\left(\dfrac{1}{3}-1\right)\).........\(\left(\dfrac{1}{10}-1\right)\). So sánh A với \(\dfrac{-1}{9}\)
B= \(\left(\dfrac{1}{4}-1\right)\)\(\left(\dfrac{1}{9}-1\right)\)...........\(\left(\dfrac{1}{100}-1\right)\). So sánh B với \(\dfrac{-11}{21}\)
a: \(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\)
\(=-\dfrac{1}{10}\)
9<10
=>1/9>1/10
=>\(-\dfrac{1}{9}< -\dfrac{1}{10}\)
=>\(A>-\dfrac{1}{9}\)
b: \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{10}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{11}{10}\)
\(=\dfrac{-1}{10}\cdot\dfrac{11}{2}=\dfrac{-11}{20}\)
20<21
=>\(\dfrac{11}{20}>\dfrac{11}{21}\)
=>\(-\dfrac{11}{20}< -\dfrac{11}{21}\)
=>\(B< -\dfrac{11}{21}\)
1) So sánh :
a) \(3^{2^3}\) và (32)3 b) (-8)9 và (-32)5 c) 221 và 314
2) Cho \(\dfrac{a}{b}=\dfrac{c}{d}.\) Chứng minh rằng :
a)\(\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\) b) \(\dfrac{ab}{cd}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
Mk săpp thi rồi nên hơi nhiều bài mong mn giúp mk !!!
\(1,\\ a,3^{2^3}=3^8>3^6=\left(3^2\right)^3\\ b,\left(-8\right)^9=\left(-2\right)^{27}< \left(-2\right)^{25}=\left(-32\right)^5\\ c,2^{21}=8^7< 9^7=3^{14}\\ 2,\)
\(a,\) Áp dụng tcdtsbn:
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)
\(b,\) Sửa: \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow a=bk;c=dk\)
\(\Leftrightarrow\dfrac{ab}{cd}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2};\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2}{d^2}\\ \LeftrightarrowĐpcm\)
Cho A = \(\dfrac{n^9+1}{n^{10}+1}\) và B = \(\dfrac{n^8+1}{n^9+1}\) trong đó n\(\in\)N; n>1. Hãy so sánh nghịch đảo của A và B rồi so sánh A với B
A = \(\dfrac{n^9+1}{n^{10}+1}\)
\(\dfrac{1}{A}\) = \(\dfrac{n^{10}+1}{n^9+1}\) = n - \(\dfrac{n-1}{n^9+1}\)
B = \(\dfrac{n^8+1}{n^9+1}\)
\(\dfrac{1}{B}\) = \(\dfrac{n^9+1}{n^8+1}\) = n - \(\dfrac{n-1}{n^8+1}\)
Vì n > 1 ⇒ n - 1> 0
\(\dfrac{n-1}{n^9+1}\) < \(\dfrac{n-1}{n^8+1}\)
⇒ n - \(\dfrac{n-1}{n^9+1}\) > n - \(\dfrac{n-1}{n^8+1}\)⇒ \(\dfrac{1}{A}>\dfrac{1}{B}\)
⇒ A < B
So sánh:
a) 430 và 3.2410
b) \(\dfrac{3}{1^2.2^2}\) + \(\dfrac{5}{2^2.3^2}\) + \(\dfrac{7}{3^2.4^2}\) +...+\(\dfrac{19}{9^2.10^2}\) và 1
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
\(a,\dfrac{1}{3}và\dfrac{2}{5}\) \(b,\dfrac{3}{7}và\dfrac{8}{9}\) \(c,\dfrac{3}{5}và\dfrac{7}{10}\) \(d,\dfrac{1}{3}và\dfrac{5}{6}\)
a) \(\dfrac{1}{3}=\dfrac{1.5}{3.5}=\dfrac{5}{15}\)
\(\dfrac{2}{5}=\dfrac{2.3}{5.3}=\dfrac{6}{15}\)
Vậy quy đồng mẫu số hai phân số 1/3 và 2/5 ta được 5/15 và 6/15
b)\(\dfrac{3}{7}=\dfrac{3.9}{7.9}=\dfrac{27}{63}\)
\(\dfrac{8}{9}=\dfrac{8.7}{9.7}=\dfrac{56}{63}\)
Vậy...
c)\(\dfrac{3}{5}=\dfrac{3.2}{5.2}=\dfrac{6}{10}\) giữ nguyên phân số 7/10
d)\(\dfrac{1}{3}=\dfrac{1.2}{3.2}=\dfrac{2}{6}\) giữ nguyên phân số 5/6