tính 1/5*9 + 1/9*3 + 1/13*17 +.........+ 1/41*45
GIÚP MIK VỚI MIK ĐANG CẦN GẤP LÀM ƠN ,NHANH LÊN
K
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SC
5
31 tháng 7 2023
\(\left(x+\dfrac{1}{3}\right)\times\dfrac{9}{14}\times\dfrac{7}{3}-\dfrac{1}{3}=1:\dfrac{9}{5}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}-\dfrac{1}{3}=\dfrac{5}{9}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}=\dfrac{5}{9}+\dfrac{1}{3}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}=\dfrac{8}{9}\\ \Rightarrow x+\dfrac{1}{3}=\dfrac{8}{9}:\dfrac{3}{2}\\ \Rightarrow x+\dfrac{1}{3}=\dfrac{16}{27}\\ \Rightarrow x=\dfrac{16}{27}-\dfrac{1}{3}\\ \Rightarrow x=\dfrac{7}{27}\)
T
30 tháng 11 2018
b)\(\left(x-8\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x-8=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=2\end{cases}}\)
c) \(\left(x+1\right)+\left(x+2\right)+...+\left(x+10\right)=9x+200\)
\(\Leftrightarrow\left(x+x+...+x\right)+\left(1+2+...+10\right)=9x+200\) (10 số hạng x)
\(\Leftrightarrow10x+55=9x+200\Leftrightarrow x+55=200\)
\(\Leftrightarrow x=145\)
19 tháng 9 2021
\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)
\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)
\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)
25 tháng 6 2019
A = 1^2 + 3^2 + ... + 97^2 + 99^2
= 1.1 + 3.3 + ... + 97.97 + 99.99
> 1.2 + 2.3 + ... + 97.98 + 98.99
= 1.99 = 99
Suy ra A > 1
KT
21 tháng 7 2018
a) \(\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{23}\right)=\frac{31}{23}-\frac{7}{32}-\frac{8}{23}=1-\frac{7}{32}=\frac{25}{32}\)
b) \(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(=\frac{1}{3}-\left(\frac{79}{67}-\frac{12}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
\(=\frac{1}{3}-1+1=\frac{1}{3}\)
d) \(\frac{1}{7}.\frac{1}{3}+\frac{1}{7}.\frac{-1}{3}+\frac{17}{19}=\frac{1}{7}.\left(\frac{1}{3}-\frac{1}{3}\right)+\frac{17}{19}=\frac{17}{19}\)
e) \(\frac{3}{5}.\frac{7}{9}+\frac{7}{5}.\frac{2}{9}=\frac{7}{5}.\left(\frac{3}{9}+\frac{2}{9}\right)=\frac{7}{5}.\frac{5}{9}=\frac{7}{9}\)
\(\frac{1}{5\times9}+\frac{1}{9\times13}+\frac{1}{13\times17}+...+\frac{1}{41\times45}\)
= \(\frac{1}{4}\times\left(\frac{4}{5\times9}+\frac{4}{9\times13}+\frac{4}{13\times17}+...+\frac{4}{41\times45}\right)\)
= \(\frac{1}{4}\times\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\right)\)
= \(\frac{1}{4}\times\left(\frac{1}{5}-\frac{1}{45}\right)\)
= \(\frac{1}{4}\times\frac{8}{45}=\frac{2}{45}\)
\(A=\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+...+\frac{1}{41\cdot45}\)
\(4A=\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+...+\frac{4}{41\cdot45}\)
\(4A=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\)
\(4A=\frac{1}{5}-\frac{1}{45}\)
\(4A=\frac{8}{45}\)
\(A=\frac{2}{45}\)