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18 tháng 7 2019

\(\left(\frac{13}{21}+\frac{17}{20}-\text{[}\frac{13}{20}-\frac{28}{21}\text{]}\right):\frac{176}{95}\)

\(\left(\frac{13}{21}+\frac{17}{20}-\frac{13}{20}+\frac{28}{21}\right).\frac{95}{176}\)

\(\frac{131}{60}.\frac{95}{176}=\frac{2489}{2112}\)

sửa đề :3

1 tháng 9 2019

a, \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)

\(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{13}{90}\)

\(\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)

\(\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)

\(\frac{1}{x+1}=\frac{18}{90}-\frac{13}{90}\)

\(\frac{1}{x+1}=\frac{1}{18}\)

⇒ x + 1 = 18

⇒ x = 17

Vậy x = 17

b, \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{49.3}{148}\)

\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{147}{148}\)

\(1-\frac{1}{x+3}=\frac{147}{148}\)

\(\frac{1}{x+3}=1-\frac{147}{148}\)

\(\frac{1}{x+3}=\frac{1}{148}\)

⇒ x + 3 = 148

⇒ x = 145

Vậy x = 145

27 tháng 8 2015

45/26          

4 tháng 10 2021

yutyugubhujyikiu

20 tháng 9 2020

A) \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}\)

\(=\frac{\left(x+10\right)-\left(x+3\right)}{\left(x+3\right)\left(x+10\right)}+\frac{\left(x+21\right)-\left(x+10\right)}{\left(x+10\right)\left(x+21\right)}+\frac{\left(x+34\right)-\left(x+21\right)}{\left(x+21\right)\left(x+34\right)}\)

\(=\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}\)

\(=\frac{1}{x+3}-\frac{1}{x+34}\)

\(=\frac{\left(x+34\right)-\left(x+3\right)}{\left(x+3\right)\left(x+34\right)}\)\(=\frac{x}{\left(x+3\right)\left(x+34\right)}\)

\(\Rightarrow\left(x+34\right)-\left(x+3\right)=x\)

\(\Rightarrow x=31\)

Vậy, x = 31 

20 tháng 9 2020

Bạn áp dụng: \(\frac{k}{x\cdot\left(x+k\right)}=\frac{1}{x}-\frac{1}{x+k}\) với    \(x,k\inℝ;x\ne0;x\ne-k\)

Chứng minh: \(\frac{1}{x}-\frac{1}{x+k}=\frac{x+k}{x\left(x+k\right)}-\frac{x}{x\left(x+k\right)}=\frac{x+k-x}{x\left(x+k\right)}=\frac{k}{x\left(x+k\right)}\)

24 tháng 9 2016

a ) \(\left(-\frac{40}{51}.0,32.\frac{17}{20}\right):\frac{64}{75}\)

\(=\left(-\frac{40}{51}.\frac{8}{25}.\frac{17}{20}\right):\frac{64}{75}\)

\(=\left(\frac{-40.8.17}{51.25.20}\right):\frac{64}{75}\)

\(=\left(\frac{-16}{75}\right).\frac{75}{64}\)

\(=\frac{-1}{1}.\frac{1}{4}=-\frac{1}{4}\)

 

 

 

24 tháng 9 2016

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