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Ta có : \(x-\frac{3}{1.5}-\frac{3}{5.9}-\frac{3}{9.13}-\frac{3}{13.17}-\frac{3}{17.21}=\frac{2}{7}\)
=> \(x-\left(\frac{3}{1.5}+\frac{3}{5.9}+\frac{3}{9.13}+\frac{3}{13.17}+\frac{3}{17.21}\right)=\frac{2}{7}\)
=> \(x-\frac{3}{4}\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+\frac{4}{17.21}\right)=\frac{2}{7}\)
=> \(x-\frac{3}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{3}-\frac{1}{17}+\frac{1}{17}-\frac{1}{21}\right)=\frac{2}{7}\)
=> \(x-\frac{3}{4}\left(1-\frac{1}{21}\right)=\frac{2}{7}\)
=> \(x-\frac{3}{4}.\frac{20}{21}=\frac{2}{7}\)
=> \(x-\frac{5}{7}=\frac{2}{7}\)
=> x = 1
Vậy x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
a) -x+\(\frac{2}{5}\)=\(\frac{3}{10}\)
-x=\(\frac{3}{10}\)\(-\)\(\frac{2}{5}\)
\(-1.x\)=\(\frac{3}{10}-\frac{4}{10}\)
\(-1.x=\frac{-1}{10}\)
\(x=\frac{-1}{10}:-1\)
\(x=\frac{1}{10}\)
b)\(-x+\frac{3}{10}=1\)
\(-x=1-\frac{3}{10}\)
\(-x=\frac{10}{10}-\frac{3}{10}\)
\(-x=\frac{7}{10}\)
\(=>x=\frac{-7}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-15=0\\\left(2x-15\right)^2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=15\\2x-15=1;-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{15}{2}\\2x=16;14\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{15}{2}\\x=8;7\end{cases}}\)
\(\Rightarrow\left(2x-15\right)^5=\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-15=0\\\left(2x-15\right)^2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=15\\2x-15=1;-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=7,5\\x=8;7\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1 + 3 + 5 + .....+ x = 25
1 + 3 + 5 + 7 + x = 25
16 + x =25
x = 25 - 16
x = 9
nhớ k cho mk nha , mk trả lời nhanh nhất !(> - ^)
Vì 1 + 3 + 5 + ..... + x = 25
=>\(\frac{\text{ (x - 1) . [(x - 1) : 2 + 1]}}{2}\) = 25
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2009}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2003}{2009}\)
\(\Rightarrow2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2003}{2009}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2003}{2009}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2003}{2009}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{2009}\div2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4018}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4018}\)
\(\Rightarrow\frac{1}{x+1}=\frac{3}{2009}\)
\(\Rightarrow\frac{3}{3\left(x+1\right)}=\frac{3}{2009}\)
\(\Rightarrow3\left(x+1\right)=2009\)
\(\Rightarrow3x+3=2009\)
\(\Rightarrow3x=2006\)
\(\Rightarrow x=\frac{2006}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Mik có nhầm lẫn nhé
l2x - 1l = 16
suy ra 2x-1 = 16 và -16
nếu 2x-1 = 16 nếu 2x-1 = -16
2x = 16+1 2x = (-16) +1
2x=17 2x = -15
x=17:2 x = (-15) :2
x = 8,5 x = -7,5
Vậy x = 8,5 và -7,5
Vì ko có giá trị tuyệt đối nào có giá trị bằng 0
suy ra : x ko có giá trị