Tính nhanh
\(\frac{7}{5.9}+\frac{7}{9.11}+\frac{7}{11.13}+\frac{7}{13.15}+..+\frac{7}{2015.2017}\)
giup mk vs nha mk can gap lam
mk se tick cho ai lm dung
thank you, thank you.
K
Khách
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Những câu hỏi liên quan
T
20 tháng 6 2019
#)Giải :
Đặt \(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(A=\frac{1}{5}-\frac{1}{10}\)
\(A=\frac{1}{10}\)
T
4 tháng 4 2019
Ta có:
\(\frac{A}{5}=\frac{4}{35\cdot31}+\frac{6}{35\cdot41}+\frac{9}{50\cdot41}+\frac{7}{50\cdot57}=\frac{35-31}{35\cdot31}+\frac{41-35}{35\cdot41}+\frac{50-41}{50\cdot41}+\frac{57-50}{50\cdot57}\) ( Ps cuối là\(\frac{57-50}{50.57}\) nha).
\(=\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}\)
\(\Rightarrow\frac{A}{5}=5\cdot\left(\frac{1}{31}-\frac{1}{57}\right)\)
Tương tự:
\(\frac{B}{2}=\frac{7}{38.31}+\frac{5}{38.43}+\frac{3}{43.46}+\frac{11}{46.57}=\frac{38-31}{38.31}+\frac{41-38}{38.41}+\frac{46-43}{43.46}+\frac{57-46}{46.57}\)
\(=\frac{1}{31}-\frac{1}{38}+\frac{1}{38}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}+\frac{1}{46}-\frac{1}{57}\)
\(\Rightarrow\frac{B}{2}=2\cdot\left(\frac{1}{31}-\frac{1}{57}\right)\)
Từ đó, suy ra:
\(\frac{A}{B}=\frac{5}{2}\)
Vậy \(\frac{A}{B}=\frac{5}{2}\)
7 tháng 4 2017
\(A=81.\left[\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right].\frac{158158158}{711711711}\)
\(A=81.\left[\frac{12.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4.\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6.\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right].\frac{158}{711}\)
\(A=81.\left(\frac{12}{4}:\frac{5}{6}\right).\frac{2}{9}\)
\(A=81.3.\frac{6}{5}.\frac{2}{9}\)
\(A=\frac{324}{5}\)
Nhớ là: THANKS YOU VERY "MUCH" chứ không phải là THANKS YOU VERY "MATH"!!!
7 tháng 4 2017
\(A=81.\frac{158158158}{711711711}.\frac{12.\left(\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4.\left(\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5.\left(\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6.\left(\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\)
\(=81.\frac{158}{711}.\frac{12}{4}:\frac{5}{6}=\frac{1422}{79}.3.\frac{6}{5}=\frac{1422.3.6}{79.5}=\frac{25596}{395}\)
6 tháng 7 2017
Tính:
b) \(-2.\frac{-38}{21}.\frac{-7}{4}.-\frac{3}{8}\)
= \(\frac{\left(-2\right).\left(-38\right).\left(-7\right).\left(-3\right)}{21.4.8}\)
= \(\frac{-1.-19.1}{1.2.4}=\frac{19}{8}\)
// Học tốt!
NM
2
14 tháng 2 2017
\(\frac{x-7}{3}=\frac{-27}{7-x}\)
\(\Leftrightarrow\left(x-7\right)\left(7-x\right)=\left(-27\right).3\)
\(\Leftrightarrow7x-x^2-49+7x=-81\)
\(\Leftrightarrow-x^2+14x-49=-81\)
\(\Leftrightarrow-x^2+14x+32=0\)
\(\Leftrightarrow x^2-14x-32=0\)
\(\Leftrightarrow x^2+2x-16x-32=0\)
\(\Leftrightarrow x\left(x+2\right)-16\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-16\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-16=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=16\end{cases}}}\)
Đặt A là tên của biểu thức trên
2A = \(\frac{7.2}{5.9}+\frac{7.2}{9.11}+\frac{7.2}{11.13}+\frac{7.2}{13.15}+...+\frac{7.2}{2015.2017}\)
2A = \(7\left(\frac{2}{5.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{2015.2017}\right)\)
2A = \(7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)
2A = \(7\left(\frac{1}{5}-\frac{1}{2017}\right)\)
2A = \(7\cdot\frac{2012}{10085}\)
2A = \(\frac{14084}{10085}\)
A = \(\frac{14084}{10085}:2\)
A = \(\frac{7042}{10085}\)
\(\frac{7}{5.9}+\frac{7}{9.11}+\frac{7}{11.13}+\frac{7}{11.13}+...+\frac{7}{2015.2017}\)
\(=\frac{7}{5.9}+\frac{7}{2}.\left(\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}+...+\frac{2}{2015.2017}\right)\)
\(=\frac{7}{45}+\frac{7}{2}.\left(\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{2015}-\frac{1}{2017}\right)\)