\(\dfrac{1}{3^{400}}\)và \(\dfrac{1}{4^{300}}\)
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21 tháng 4 2017

Ta có :

\(\dfrac{1}{3^{400}}=\dfrac{1}{\left(3^4\right)^{100}}=\dfrac{1}{81^{100}}\)

\(\dfrac{1}{4^{300}}=\dfrac{1}{\left(4^3\right)^{100}}=\dfrac{1}{64^{100}}\)

\(81^{100}>64^{100}\)

\(\Rightarrow\dfrac{1}{81^{100}}< \dfrac{1}{61^{100}}\)

Vậy ...

21 tháng 4 2017

phân số cùng tử số mà phân số nó to hơn thì phân số nó bé hơn

=>1/3^400>1/4^300

17 tháng 4 2017

ời giải:

Giải bài 158 trang 64 SGK Toán 6 Tập 2 | Giải toán lớp 6

17 tháng 4 2017

a) \(\dfrac{-1}{-4}\)=\(\dfrac{1}{4}>0\)

\(\dfrac{3}{-4}< 0\)

\(\Rightarrow\dfrac{1}{4}>\dfrac{3}{-4}hay\dfrac{-1}{-4}>\dfrac{3}{-4}\)

b) Ta có:

\(\dfrac{15}{17}=1-\dfrac{2}{17}\\ \)

\(\dfrac{25}{27}=1-\dfrac{2}{27}\\ \\ \)

\(\dfrac{2}{17}>\dfrac{2}{27}\left(17< 27\right)\)

\(\Rightarrow1-\dfrac{2}{17}< 1-\dfrac{2}{27}\)hay \(\dfrac{15}{17}< \dfrac{25}{27}\)

13 tháng 3 2018

a,A<B

b,A,<B

c,A<B

13 tháng 3 2018

a, \(A-B=\frac{3}{8^3}+\frac{7}{8^4}-\frac{7}{8^3}-\frac{3}{8^4}==\left(\frac{7}{8^4}-\frac{3}{8^4}\right)-\left(\frac{7}{8^3}-\frac{3}{8^3}\right)=\frac{4}{8^4}-\frac{4}{8^3}< 0\)

Vậy A < B

b, \(A=\frac{10^7+5}{10^7-8}=\frac{10^7-8+13}{10^7-8}=1+\frac{13}{10^7-8}\)

\(B=\frac{10^8+6}{10^8-7}=\frac{10^8-7+13}{10^8-7}=1+\frac{13}{10^8-7}\)

Vì \(10^7-8< 10^8-7\Rightarrow\frac{1}{10^7-8}>\frac{1}{10^8-7}\Rightarrow\frac{13}{10^7-8}>\frac{13}{10^8-7}\Rightarrow A>B\)

c,Áp dụng nếu \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{a+n}\) có:

 \(B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)

Vậy A < B

13 tháng 10 2017

\(3^{400}=\left(3^4\right)^{100}=81^{100}\)

\(4^{300}=\left(4^3\right)^{100}=64^{100}\)

Vì \(81^{100}>64^{100}\Rightarrow3^{400}>4^{300}\)

13 tháng 10 2017

\(3^{400}\)và \(4^{300}\)

Ta có : \(3^{400}=\left(3^4\right)^{100}=81^{100}\)

\(4^{300}=\left(4^3\right)^{100}=64^{100}\)

Vì \(81^{100}>64^{100}\)nên \(3^{400}>4^{300}\)

6 tháng 9 2017

Đặt \(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{300}\)

\(=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{600}\)

\(=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{24.25}\)

\(=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{24.25}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{24}-\dfrac{1}{25}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{25}\right)\)

\(=2\cdot\dfrac{23}{50}=\dfrac{23}{25}\)

Vậy A = \(\dfrac{23}{25}\).

6 tháng 9 2017

It's very easy :)))))

Call expression is A, we have:

\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{300}.\)

\(\Rightarrow2A=2\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+...+\dfrac{1}{300}\right).\)

\(\Rightarrow2A=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{600}.\)

\(\Rightarrow2A=\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+...+\dfrac{2}{24.25}.\)

\(\Rightarrow2A=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\right).\)

\(\Rightarrow2A=2\left(\dfrac{1}{2}-\dfrac{1}{25}\right).\)

\(\Rightarrow2A=2\left(\dfrac{25}{50}-\dfrac{2}{50}\right).\)

\(\Rightarrow2A=2.\dfrac{23}{50}=\dfrac{23}{25}.\)

Vậy.....

a: 51/56=1-5/56

61/66=1-5/66

mà -5/56<-5/66

nên 51/56<61/66

b: 41/43<1<172/165

c: \(\dfrac{101}{506}>0>-\dfrac{707}{3534}\)

12 tháng 3 2017

Bài 2:

Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};....;\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=2-\dfrac{1}{100}< 2\)

Vậy A < 2

Bài 3:

D = \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right)....\left(1-\dfrac{1}{2015}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}......\dfrac{2014}{2015}\)

\(=\dfrac{1.2......2014}{2.3......2015}=\dfrac{1}{2015}\)

Bài 4:

A = \(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}......\dfrac{899}{900}\)

\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}........\dfrac{29.31}{30.30}\)

\(=\dfrac{1.2.3......29}{2.3.4.......30}.\dfrac{3.4.5......31}{2.3.4.....30}\)

\(=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)

23 tháng 3 2017

Ta có:

\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{299}+\dfrac{1}{300}>\dfrac{1}{300}.200=\dfrac{200}{300}=\dfrac{2}{3}\)

\(\Rightarrow\) biểu thức trên lớn hơn \(\dfrac{2}{3}\).

18 tháng 4 2016

\(\frac{1}{3^{400}}=\frac{1}{\left(3^4\right)^{100}}=\frac{1}{81^{100}}\)

\(\frac{1}{4^{300}}=\frac{1}{\left(4^3\right)^{100}}=\frac{1}{64^{100}}\)

vì \(\frac{1}{81^{100}}<\frac{1}{64^{100}}\)nên \(\frac{1}{3^{400}}<\frac{1}{4^{300}}\)

\(\frac{1}{3^{400}}>\frac{1}{4^{300}}\)

chắc thế!! 47568

27 tháng 3 2017

A=\(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-3}\)

A=\(\left(\dfrac{2}{7}+\dfrac{11}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{3}+\dfrac{5}{-3}\right)+\dfrac{-3}{8}\)

A=\(2+\dfrac{-4}{3}+\dfrac{-3}{8}\)

A=\(\dfrac{7}{24}\)

B=\(\dfrac{3}{17}+\dfrac{-5}{13}+\dfrac{-18}{35}+\dfrac{14}{17}+\dfrac{17}{-35}+\dfrac{-8}{13}\)

B=\(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-18}{35}+\dfrac{17}{-35}\right)+\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)\)

B=\(\dfrac{17}{17}+\dfrac{-35}{35}+\dfrac{-13}{13}\)

B=\(1+\left(-1\right)+\left(-1\right)=-1\)

C=\(\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)

C=\(\dfrac{-3}{17}+\dfrac{2}{3}+\dfrac{3}{17}=\left(\dfrac{-3}{17}+\dfrac{3}{17}\right)+\dfrac{2}{3}\)

C=0+\(\dfrac{2}{3}=\dfrac{2}{3}\)

D=\(\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)

D=\(\dfrac{-1}{6}+\dfrac{5}{-12}+\dfrac{7}{12}\)

D=\(\dfrac{-2}{12}+\dfrac{-5}{12}+\dfrac{7}{12}=\left(\dfrac{-2}{12}+\dfrac{-5}{12}\right)+\dfrac{7}{12}\)

D=\(\dfrac{-7}{12}+\dfrac{7}{12}=0\)