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Bài 2
a. \(-1\frac{2}{3}-|2x-1|:\frac{3}{5}=-2\)
\(|2x-1|:\frac{3}{5}=\frac{5}{3}-2\)
\(|2x-1|:\frac{3}{5}=-\frac{1}{3}\)
\(|2x-1|=-\frac{1}{5}\)
Vì giá trị tuyệt đối luôn \(\ge0\)với mọi x
mà \(-\frac{1}{5}< 0\)
=> \(x\in\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
.........................
= \(\frac{1}{2}\). ( \(\frac{2}{1.3}\) + \(\frac{2}{3.5}\) + \(\frac{2}{5.7}\) ... + \(\frac{2}{x.\left(x+2\right)}\) )
= \(\frac{1}{2}\) . ( 1 - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{5}\) + \(\frac{1}{5}\) - \(\frac{1}{7}\) + ... + \(\frac{1}{x}\)- \(\frac{1}{x+2}\) )
= ................
Bạn tự làm tiếp nhé ! Chúc bạn học tốt :)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có
\(\frac{x+1\text{0}}{8}+\frac{x+11}{7}=-2\)
\(\Leftrightarrow\frac{7\left(x+1\text{0}\right)}{7\cdot8}+\frac{8\left(x+11\right)}{8\cdot7}=-\frac{112}{56}\)
\(\Leftrightarrow\frac{7x+7\text{0}+8x+88}{56}=-\frac{112}{56}\)
\(\Leftrightarrow15x+158=-112\)
\(\Leftrightarrow15x=-27\text{0}\)
\(\Leftrightarrow x=-18\)
Vậy ...........
chúc bn học tốt
\(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)
\(\Leftrightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2-\left(\frac{1}{4}\right)^2=0\)
\(\Leftrightarrow\left(\frac{1}{x}-\frac{2}{3}+\frac{1}{4}\right)\left(\frac{1}{x}-\frac{2}{3}-\frac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\frac{1}{x}-\frac{5}{12}\right)\left(\frac{1}{x}-\frac{11}{12}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}-\frac{5}{12}=0\\\frac{1}{x}-\frac{11}{12}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}=\frac{5}{12}\\\frac{1}{x}=\frac{11}{12}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{12}{11}\\x=\frac{12}{5}\end{cases}}\)
Vậy....
\(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)
\(\Rightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2=\frac{1}{16}\)
\(\Rightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2=\left(\frac{1}{4}\right)^2\)
\(\Rightarrow\frac{1}{x}-\frac{2}{3}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{x}=\frac{11}{12}\)
\(\Rightarrow x=\frac{11}{12}\)