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26 tháng 6 2018

Bài 1 và Bài 2 dễ, bn có thể tự làm được!

Bài 3:

a) ta có: 1020 = (102)10 = 10010

=> 10010>910

=> 1020>910

b) ta có: (-5)30 = 530 =( 53)10 = 12510 ( vì là lũy thừa bậc chẵn)

(-3)50 = 350 = (35)10= 24310

=> 12510 < 24310

=> (-5)30 < (-3)50

c) ta có: 648 = (26)8= 248

1612 = ( 24)12 = 248

=> 648 = 1612

d) ta có: \(\left(\frac{1}{16}\right)^{10}=\left(\frac{1}{2^4}\right)^{10}=\frac{1}{2^{40}}\)

\(\left(\frac{1}{2}\right)^{50}=\frac{1}{2^{50}}\)

\(\Rightarrow\frac{1}{2^{40}}>\frac{1}{2^{50}}\)

\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)

26 tháng 6 2018

3.a) Ta có: 910=(32)10=320

Mà 1020<320

Nên 1020<910

c)Ta có:648 =(82)8=816

1612=(23)12=836

vì 816<836

Nên 648<162

              

28 tháng 7 2015

A = \(-\frac{1.3}{2.2}.-\frac{2.4}{3.3}.\cdot\cdot\cdot-\frac{2013.2015}{2014.2014}=-\frac{\left(1.2.3...2013\right).\left(3.4.5....2015\right)}{\left(2.3....2014\right).\left(2.3....2014\right)}=-\frac{2.2015}{2014}=-\frac{4030}{2014}<-\frac{1}{2}\)

VẬy A < B 

30 tháng 7 2015

A<B

mình đúng nha

5 tháng 11 2015

x<y nhé bạn :)

 

...
Đọc tiếp

\(\left(\frac{-5}{12}+\frac{7}{4}-\frac{3}{8}\right)-\left[4\frac{1}{2}-7\frac{1}{3}\right]-\left(\frac{1}{4}-\frac{5}{2}\right)\)

\(\left[2\frac{1}{4}-5\frac{3}{2}\right]-\left(\frac{3}{10}-1\right)-5\frac{1}{2}+\left(\frac{1}{3}-\frac{5}{6}\right)\)

\(\frac{4}{7}-\left(3\frac{2}{5}-1\frac{1}{2}\right)-\frac{5}{21}+\left[3\frac{1}{2}-4\frac{2}{3}\right]\)

\(\frac{1}{8}-1\frac{3}{4}+\left(\frac{7}{8}-3\frac{7}{2}+\frac{3}{4}\right)-\left[\frac{7}{4}-\frac{5}{8}\right]\)

\(\left(\frac{3}{5}-2\frac{1}{10}+\frac{11}{20}\right)-\left[\frac{-3}{4}+1\frac{7}{2}\right]\)

\(\left[-2\frac{1}{5}-2\frac{2}{3}\right]-\left(\frac{1}{15}-5\frac{1}{2}\right)+\left[\frac{-1}{6}+\frac{1}{3}\right]\)

\(1\frac{1}{8}-\left(\frac{1}{15}-\frac{1}{2}+\frac{-1}{6}\right)+\left[\frac{5}{4}+\frac{3}{2}\right]\)

\(\frac{5}{6}-\left(1\frac{1}{3}-1\frac{1}{2}\right)+\left[\frac{5}{12}-\frac{3}{4}-\frac{1}{6}\right]\)

\(1\frac{1}{4}-\left(\frac{7}{12}-\frac{2}{3}-1\frac{3}{8}\right)+\left[\frac{5}{24}-2\frac{1}{2}\right]-\frac{1}{6}-\left[\frac{-3}{4}\right]\)

\(-2\frac{1}{5}+2\frac{3}{10}-\left(\frac{6}{20}-\left[\frac{2}{8}-1\frac{1}{2}\right]\right)+\left[\frac{7}{20}-1\frac{1}{4}\right]\)

\(-\left[1\frac{2}{3}-3\frac{1}{2}+\frac{1}{4}\right]+\left(\frac{2}{6}-\frac{5}{12}\right)-\left(\frac{1}{3}-\left[\frac{1}{4}-\frac{1}{3}\right]\right)\)

\(-\frac{4}{5}-\left(1\frac{1}{10}-\frac{7}{10}\right)+\left[\frac{3}{4}-1\frac{1}{5}\right]+1\frac{1}{2}\)

\(\frac{3}{21}-\frac{5}{14}+\left[1\frac{1}{3}-5\frac{1}{2}+\frac{5}{14}\right]-\left(\frac{1}{6}-\frac{3}{7}+\frac{1}{3}\right)\)

\(-1\frac{2}{5}+\left[1\frac{3}{10}-\frac{7}{20}-1\frac{1}{4}\right]-\left(\frac{1}{5}-\left[\frac{3}{4}-1\frac{1}{2}\right]\right)\)

\(2\frac{1}{3}-\left(\frac{1}{2}-2\frac{1}{6}+\frac{3}{4}\right)+\left[\frac{5}{12}-1\frac{1}{3}\right]-\frac{7}{8}+3\frac{1}{2}\)

\(2\frac{1}{4}-1\frac{3}{5}-\left(\frac{9}{20}-\frac{7}{10}\right)+\left[1\frac{3}{5}-2\frac{1}{2}\right]+\frac{3}{4}\)

\(\left[\frac{8}{3}-5\frac{1}{4}+\frac{1}{6}\right]-\frac{7}{4}+\frac{-5}{12}-\left(1-1\frac{1}{2}+\frac{1}{3}\right)\)

\(\left(\frac{1}{4}-\left[1\frac{1}{4}-\frac{7}{10}\right]+\frac{1}{2}\right)-2\frac{1}{5}-1\frac{3}{10}+\left[1-\frac{1}{2}\right]\)

TRÌNH BÀY GIÚP MÌNH NHA 

0
22 tháng 12 2015

a,>

b,=

c,>

Chắc đấy! Tick nhé!

Bài 1:

a) Ta có: \(25\cdot\left(\frac{-1}{5}\right)^3+\frac{1}{5}-2\cdot\left(\frac{-1}{2}\right)^2-\frac{1}{2}\)

\(=25\cdot\frac{-1}{125}+\frac{1}{5}-2\cdot\frac{1}{4}-\frac{1}{2}\)

\(=-\frac{1}{5}+\frac{1}{5}-\frac{1}{2}-\frac{1}{2}\)

\(=\frac{-2}{2}=-1\)

b) Ta có: \(35\frac{1}{6}:\left(\frac{-4}{5}\right)-46\frac{1}{6}:\left(\frac{-4}{5}\right)\)

\(=\frac{211}{6}\cdot\frac{-5}{4}-\frac{277}{6}\cdot\frac{-5}{4}\)

\(=\frac{-5}{4}\cdot\left(\frac{211}{6}-\frac{277}{6}\right)\)

\(=\frac{-5}{4}\cdot\left(-11\right)=\frac{55}{4}\)

c) Ta có: \(\left(\frac{-3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)

\(=\frac{-7}{20}\cdot\frac{7}{3}+\frac{7}{20}\cdot\frac{7}{3}\)

\(=\frac{7}{3}\cdot\left(-\frac{7}{20}+\frac{7}{20}\right)=\frac{7}{3}\cdot0=0\)

d) Ta có: \(\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{18}\right)+\frac{7}{8}\cdot\left(\frac{1}{36}-\frac{5}{12}\right)\)

\(=\frac{7}{8}\cdot6+\frac{7}{8}\cdot\frac{-7}{18}\)

\(=\frac{7}{8}\cdot\left(6+\frac{-7}{18}\right)\)

\(=\frac{7}{8}\cdot\frac{101}{18}=\frac{707}{144}\)

e) Ta có: \(\frac{1}{6}+\frac{5}{6}\cdot\frac{3}{2}-\frac{3}{2}+1\)

\(=\frac{1}{6}+\frac{15}{12}-\frac{3}{2}+1\)

\(=\frac{2}{12}+\frac{15}{12}-\frac{18}{12}+\frac{12}{12}\)

\(=\frac{11}{12}\)

f) Ta có: \(\left(-0,75-\frac{1}{4}\right):\left(-5\right)+\frac{1}{15}-\left(-\frac{1}{5}\right):\left(-3\right)\)

\(=\left(-1\right):\left(-5\right)+\frac{1}{15}-\frac{1}{15}\)

\(=\frac{1}{5}\)