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23 tháng 3 2020

Ta có thể thấy:

\(\frac{11}{29};\frac{9}{17};\frac{10}{19}< \frac{2}{3}\)

\(\Rightarrow\frac{11}{29}+\frac{9}{17}+\frac{10}{19}< 3\times\frac{2}{3}=2\)

Chúc bn hok tốt

30 tháng 4 2017

Ta có: Trong 3 phân số thì \(\frac{9}{17}\)là phân số lớn nhất

\(\Rightarrow\frac{9}{17}+\frac{9}{17}+\frac{9}{17}>\frac{11}{29}+\frac{9}{17}+\frac{10}{19}\)

\(\Rightarrow\frac{9}{17}\times3>A\)

\(\frac{9}{17}\times3=\frac{27}{17}< \frac{34}{17}=2\)

\(\Rightarrow2>\frac{9}{17}\times3>A\)

\(\Rightarrow A< 2\)

8 tháng 6 2020

\(\frac{4}{7}:\left(\frac{2}{5}.\frac{4}{7}\right)\)

\(=\frac{4}{7}:\frac{8}{35}\)

\(=\frac{4}{7}.\frac{35}{8}\)

\(=\frac{5}{2}\)

4/7:8/35

4/7×35/8

5/2

Ta có : 

\(\frac{-16}{32}=\frac{-16:16}{32:16}=\frac{-1}{2}\)

+)\(\frac{-1}{2}=\frac{x}{-10}\)

=> (-10) x (-1) = X x 2

=> 10 = X x 2

=> X = 10 : 2 

=> X = 5

+) \(\frac{-1}{2}=\frac{-7}{y}\)

=> (-1) x Y = (-7) x 2

=> -Y = -14

=> Y = 14

+)\(\frac{-1}{2}=\frac{z}{24}\)

=> (-1) x 24 = Z x 2

=> -24 = Z x 2

=> Z = -24 : 2

=> Z = -12

Kết luận : X = 5

                Y = 14

                Z = 12

7 tháng 5 2018

Bài 1 : 

Ta có :

\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)

Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)

Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)

Vậy \(A>B\)

Bài 2 :

Ta có :

\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)

\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)

\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)

\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)

Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên  \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)

Nên : \(M>4\)

Vậy \(M>4\)

Bài 3 : 

Ta có :

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)

Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)

\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)

\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)

\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)

\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)

\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)

\(\Rightarrow A< \frac{3}{4}\)

Vậy \(A< \frac{3}{4}\)

Bài 4 :

\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)

\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)

\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)

\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)

\(\Rightarrow A=\frac{1008}{2017}\)

Vậy \(A=\frac{1008}{2017}\)

\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)

\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)

\(1-\frac{1}{x+2}=\frac{2016}{2017}\)

\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)

\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)

\(\Rightarrow x+2=2017\)

\(\Rightarrow x=2017-2=2015\)

Vậy \(x=2015\)

23 tháng 4 2020

Bài giải

a) Ta có : \(\frac{4545+101}{6969-303}=\frac{45.101+101}{69.101-101.3}=\frac{101.\left(45+1\right)}{101.\left(69-3\right)}=\frac{101.46}{101.66}=\frac{23}{33}\)

b) Ta có : \(\frac{2929-101}{2.1919+404}=\frac{29.101-101}{2.19.101+4.101}=\frac{101.\left(29-1\right)}{101.\left(19.2+4\right)}=\frac{28}{42}=\frac{2}{3}\)

23 tháng 4 2020

a)\(\frac{4545+101}{6969-303}\)\(\frac{\left(4545:45\right)+101}{\left(6969:69\right)-303}\)\(\frac{101+101}{101-303}\)=\(\frac{202}{-202}\)=-1 

b)\(\frac{2929-101}{2.1919+404}\)\(\frac{2929-101}{3838+404}\)=\(\frac{\left(2929:29\right)-101}{\left(3838:38\right)+404}\)=\(\frac{101-101}{101+404}\)=\(\frac{0}{505}\)=0

học tốt 

\(=\frac{4}{7}\)

20 tháng 4 2020

\(\frac{1.2+1.4+3.6+4.8}{2.3+4.6+6.9+8.12}\)

=\(\frac{1.2}{2.3}\)+\(\frac{1.4}{4.6}\)+\(\frac{3.6}{6.9}\)+\(\frac{4.8}{8.12}\)

\(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{3}\)+\(\frac{1}{3}\)

\(\frac{2}{6}+\frac{1}{6}+\frac{2}{6}+\frac{2}{6}\)

=\(\frac{7}{6}\)

20 tháng 4 2020

Mình nghĩ đề bài phải là:

          \(\frac{1.2+2.4+3.6+4.8}{2.3+4.6+6.9+8.12}\)   *2.3 + 4.6 + 6.9 + 8.12 = 3.(1.2 + 2.4 + 3.6 + 4.8)*

\(=\)\(\frac{1\left(1.2+2.4+3.6+4.8\right)}{3\left(1.2+2.4+3.6+4.8\right)}\)

\(=\)\(\frac{1}{3}\)