Tìm y:\(\left[y-\frac{1}{2}\right]x\left[\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right]=\frac{1}{3}\)
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18 tháng 8 2016
\(\left(y-\frac{1}{2}\right):\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{1}{3}\)
\(\Leftrightarrow\left(y-\frac{1}{2}\right):\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow\left(y-\frac{1}{3}\right):\left(1-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow\left(y-\frac{1}{2}\right):\frac{9}{10}=\frac{1}{3}\)
\(\Leftrightarrow\left(y-\frac{1}{2}\right)=\frac{3}{10}\)
\(\Leftrightarrow y=\frac{4}{5}\)
10 tháng 11 2016
Bài 1:
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{101}\right|=101x\)
Ta thấy:
\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{101}\right)=101x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{101}\right)=0\)
\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\frac{10}{11}=0\)
\(\Rightarrow10x=-\frac{10}{11}\Rightarrow x=-\frac{1}{11}\)(loại,vì x\(\ge\)0)
10 tháng 11 2016
Bài 2:
Ta thấy: \(\begin{cases}\left(2x+1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)
\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\)
Mà \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)
23 tháng 6 2015
1, \(\frac{1}{2}-\left(6\frac{5}{9}+x-\frac{117}{8}\right):\left(12\frac{1}{9}\right)=0\)
\(\left(\frac{6.9+5}{9}+x-\frac{117}{8}\right):\frac{12.9+1}{9}=\frac{1}{2}\)
( . là nhân nha)
\(\left(\frac{59}{9}-\frac{117}{8}+x\right):\frac{109}{9}=\frac{1}{2}\)
\(\frac{59}{9}-\frac{117}{8}+x=\frac{1}{2}\cdot\frac{109}{9}\)
\(\frac{59}{9}-\frac{117}{8}+x=\frac{109}{18}\)
\(x=\frac{109}{18}-\frac{59}{9}+\frac{117}{8}\)
\(x=\frac{113}{8}\)
23 tháng 6 2015
( \(\left(y+\frac{1}{3}\right)+\left(y+\frac{2}{9}\right)+\left(y+\frac{1}{27}\right)+\left(y+\frac{1}{81}\right)=\frac{56}{81}\)
\(y+\frac{1}{3}+y+\frac{2}{9}+y+\frac{1}{27}+y+\frac{1}{81}=\frac{56}{81}\)
\(4y+\frac{1}{3}+\frac{2}{9}+\frac{1}{27}+\frac{1}{81}=\frac{56}{81}\)
\(4y+\frac{49}{81}=\frac{56}{81}\)
\(4y=\frac{7}{81}\)
y = 7/81:4
y = 7/324
Đặt \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{90}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{9.10}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{10}\)
\(A=1-\frac{1}{10}\)
\(A=\frac{9}{10}\)
\(=>\left[y-\frac{1}{2}\right]x\frac{9}{10}=\frac{1}{3}\)
\(y-\frac{1}{2}=\frac{1}{3}:\frac{9}{10}\)
\(y-\frac{1}{2}=\frac{10}{27}\)
\(=>y=\frac{10}{27}+\frac{1}{2}\)
\(y=\frac{20+27}{54}=\frac{47}{54}\)
Vậy \(y=\frac{47}{54}\)
Ủng hộ mk nha!!!