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1) Ta có: \(\frac{-4}{7}-\frac{11}{19}+\frac{13}{19}\cdot\frac{-3}{7}+\frac{2}{19}:\frac{-7}{4}\)

\(=\frac{-4}{7}-\frac{11}{19}-\frac{39}{133}-\frac{8}{133}\)

\(=\frac{-76}{133}-\frac{77}{133}-\frac{39}{133}-\frac{8}{133}\)

\(=\frac{-200}{133}\)

2) Ta có: \(\left(\frac{-4}{9}+\frac{3}{5}\right):\frac{1}{\frac{1}{5}}+\left(\frac{1}{5}-\frac{5}{9}\right):\frac{1}{\frac{1}{5}}\)

\(=\left(\frac{-4}{9}+\frac{3}{5}\right)\cdot\frac{1}{5}+\left(\frac{1}{5}-\frac{5}{9}\right)\cdot\frac{1}{5}\)

\(=\frac{1}{5}\left(\frac{-4}{9}+\frac{3}{5}+\frac{1}{5}-\frac{5}{9}\right)\)

\(=\frac{1}{5}\left(-1+\frac{4}{5}\right)\)

\(=\frac{1}{5}\cdot\frac{-1}{5}=\frac{-1}{25}\)

3) Ta có: \(\frac{4}{5}-\left(-\frac{2}{7}\right)-\frac{7}{10}\)

\(=\frac{4}{5}+\frac{2}{7}-\frac{7}{10}\)

\(=\frac{56}{70}+\frac{20}{70}-\frac{49}{70}\)

\(=\frac{27}{70}\)

4) Ta có: \(\frac{2}{7}-\left(-\frac{13}{15}+\frac{4}{9}\right)-\left(\frac{5}{9}-\frac{2}{15}\right)\)

\(=\frac{2}{7}+\frac{13}{15}-\frac{4}{9}-\frac{5}{9}+\frac{2}{15}\)

\(=\frac{2}{7}+1-1=\frac{2}{7}\)

Viết đầu bài khó hiểu quá :((

        \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2+3^2}+...+\dfrac{19}{9^2-10^2}\)

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

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

\(\frac{3}{1^2\cdot2^2}+\frac{5}{2^2\cdot3^2}+...+\frac{19}{9^2\cdot10^2}\)\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{9^2}-\frac{1}{10^2}=1-\frac{1}{10^2}=\frac{99}{100}\)<1

a)\(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{23.27}=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{23}-\frac{1}{27}=\frac{1}{3}-\frac{1}{27}=\frac{8}{27}\)

b)\(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)

c)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{11.13}+\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{9.10}=\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{13}\right)+2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=\frac{1}{3}-\frac{1}{13}+2\left(1-\frac{1}{10}\right)=\frac{10}{39}+\frac{9}{5}=\frac{401}{195}\)

ai giúp mình với rồi mình tink cho nha cảm ơn các bạn nhiều 

\(A=\dfrac{3}{1^2+2^2}+\dfrac{5}{2^2+3^2}+...+\dfrac{19}{9^2+10^2}\) (sửa \(1^22^2\) thành \(1^2+2^2\))

Ta có : \(\left(1+2\right)^2=1^2+2^2+2.1.2\Rightarrow1^2+2^2< \left(1+2\right)^2\)

\(\Rightarrow1^2+2^2< 3^2=3.3\)

\(\Rightarrow\dfrac{3}{1^2+2^2}< \dfrac{1}{3}< 1\)

Tương tự \(\dfrac{5}{2^2+3^2}< \dfrac{1}{5}< 1\)

\(.....\)

\(\dfrac{9}{9^2+10^2}< \dfrac{1}{19}< 1\)

\(\Rightarrow A=\dfrac{3}{1^2+2^2}+\dfrac{5}{2^2+3^2}+...+\dfrac{19}{9^2+10^2}< 1.9=9< 1\)

\(\Rightarrow dpcm\)

10 < 10 + 3

11 + 2=2 + 11

9 < 10 + 9

10 = 10 + 0

17 – 4 > 14 - 3

18 – 4 >12

15 > 15 – 1

17 + 1<17 + 2

12+ 5 > 16

16 =19 - 3

15 – 4 =10 + 1

19 – 3 >11

sorry nhầm dấu = thành + và cm nó bé hơn 1

sory em chịu em mới lp 4