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!!!

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)

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 \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)

\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)

\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)

\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)

\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)

Bài 3:

a)\(2009-\left|x-2009\right|=x\)

\(\Rightarrow\left|x-2009\right|=2009-x\)

\(\Rightarrow\left|x-2009\right|=-\left(x-2009\right)\)

Vì GTTĐ của số âm bằng số đối của nó

\(\Rightarrow x-2009\le0\)

\(\Rightarrow x\le2009\)

Vậy với mọi \(x\le2009\) đều thỏa mãn

b)\(\left|3x+2\right|=\left|5x-3\right|\)

\(\Rightarrow3x+2=5x-3\) hoặc \(3x+2=3-5x\)

\(\Rightarrow2x=5\) hoặc \(8x=1\)

\(\Rightarrow x=\frac{5}{2}\) hoặc \(x=\frac{1}{8}\)

Đúng rùi bn

BN mún hỏi j vậy, đây k phải câu hỏi, mà có thì phải là toán lớp 6

\(A=3-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}\)

\(A=3-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\)

\(A=3-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)\)

\(A=3-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)

\(A=3-\left(1-\frac{1}{8}\right)\)

\(A=3-\frac{5}{8}\)

\(A=\frac{19}{8}\)

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) 

    \(\left(\frac{59}{9}-\frac{117}{8}+x\right):\frac{109}{9}=\frac{1}{2}\)

    \(\frac{59}{9}-\frac{117}{8}+x=\frac{1}{2}\cdot\frac{109}{9}\)

    \(\frac{59}{9}-\frac{117}{8}+x=\frac{109}{18}\)

   \(x=\frac{109}{18}-\frac{59}{9}+\frac{117}{8}\)

\(x=\frac{113}{8}\)

( \(\left(y+\frac{1}{3}\right)+\left(y+\frac{2}{9}\right)+\left(y+\frac{1}{27}\right)+\left(y+\frac{1}{81}\right)=\frac{56}{81}\)

   \(y+\frac{1}{3}+y+\frac{2}{9}+y+\frac{1}{27}+y+\frac{1}{81}=\frac{56}{81}\)

\(4y+\frac{1}{3}+\frac{2}{9}+\frac{1}{27}+\frac{1}{81}=\frac{56}{81}\)

\(4y+\frac{49}{81}=\frac{56}{81}\)

\(4y=\frac{7}{81}\)

y      =  7/81:4

y       = 7/324

\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{1}{2}+\frac{1}{6}+....+\frac{1}{32}\)

\(\left(\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right):x=\frac{1}{1\times2}+\frac{1}{2\times3}+.....+\frac{1}{11\times12}\)

\(\frac{15}{16}:x=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{12}\)

\(\frac{15}{16}:x=1-\frac{1}{12}\)

\(\frac{15}{16}:x=\frac{11}{12}\)

\(x=\frac{15}{16}:\frac{11}{12}\)

\(x=\frac{45}{44}\)

Tính \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)

2 x A = \(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)

2 x A - A = A =  \(1-\frac{1}{16}=\frac{15}{16}\)

Tính \(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{11\times12}\)

 \(B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{11}-\frac{1}{12}=\frac{1}{1}-\frac{1}{12}=\frac{11}{12}\)

Ta có: \(\frac{15}{16}:x=\frac{11}{12}\Rightarrow x=\frac{15}{16}:\frac{11}{12}=\frac{15}{16}\times\frac{12}{11}=\frac{45}{44}\)

Vậy...

kết quả là \(\frac{45}{44}\)