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![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(\frac{3}{4}x+3-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}x+3\)
\(\frac{9}{12}x-\frac{8}{12}x-\frac{2}{12}x=4+3-3-4+1\)
\(\left(\frac{9}{12}-\frac{8}{12}-\frac{2}{12}\right)x=1\)
\(-\frac{1}{12}x=1\)
\(x=-12\)
b)
\(\frac{x+1}{3}=\frac{x-2}{4}\)
\(\Rightarrow4\left(x+1\right)=3\left(x-2\right)\)
\(4x+4=3x-6\)
\(4x-3x=-6-4\)
\(x=-10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi d là ƯCLN của 12n+1 và 30n+2
=> 12n+1 chia hết cho d. 30n+2 chia hết cho d
=> (12n+1) - (30n+2) chia hết cho d
=.> 5(12n+1) - 2(30n+2) chia hết cho d
=> 1 chia hết cho d
Ta có d C Ư(1) = [-1;1]
Vây phân số \(\frac{12n+1}{30n+2}\)là phân số tối giản
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3\left(2x-\frac{5}{4}\right)=\left(3-1\frac{1}{2}\right)\left(x-\frac{1}{2}\right)\)
\(\Leftrightarrow6x-\frac{15}{4}=\frac{3}{2}x+\frac{1}{12}\)
\(\Leftrightarrow\frac{9}{2}x+\frac{3}{4}=\frac{15}{4}\)
\(\Leftrightarrow\frac{9}{2}x=3\)
\(\Leftrightarrow x=\frac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(x-\left|1\frac{1}{6}\right|=\frac{5}{21}\)
\(\Rightarrow x-\frac{5}{21}=\left|1\frac{1}{6}\right|\)
\(\Rightarrow x-\frac{5}{21}=\frac{7}{6}\)
\(\Rightarrow x=\frac{7}{6}+\frac{5}{21}=\frac{49}{42}+\frac{10}{42}=\frac{59}{42}\)
2) \(x+\left|-1\frac{2}{3}\right|=\left|-\frac{3}{4}\right|\)
\(\Rightarrow x+\left|-1\frac{2}{3}\right|=\frac{3}{4}\)
\(\Rightarrow x-\frac{3}{4}=-\left|-1\frac{2}{3}\right|\)
\(\Rightarrow x-\frac{3}{4}=-1\frac{2}{3}\)
\(\Rightarrow x-\frac{3}{4}=-\frac{5}{3}\)
\(\Rightarrow x=-\frac{5}{3}+\frac{3}{4}=-\frac{11}{12}\)
3) \(\left|x-\frac{1}{3}\right|=\frac{5}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{1}{3}=\frac{5}{2}\\x-\frac{1}{3}=-\frac{5}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{5}{2}+\frac{1}{3}=\frac{17}{6}\\x=-\frac{5}{2}+\frac{1}{3}=-\frac{13}{6}\end{matrix}\right.\)
4) \(\left|x+\frac{2}{3}\right|=0\)
\(\Rightarrow x+\frac{2}{3}=0\)
\(\Rightarrow x=0-\frac{2}{3}=-\frac{2}{3}\)
5) \(\left|x+2\right|=\frac{1}{3}-\frac{1}{5}\)
\(\Rightarrow\left|x+2\right|=\frac{2}{15}\)
\(\Rightarrow\left[{}\begin{matrix}x+2=\frac{2}{15}\\x+2=-\frac{2}{15}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{2}{15}-2=-\frac{28}{15}\\x=-\frac{2}{15}-2=-\frac{32}{15}\end{matrix}\right.\)
6) \(\left|x-4\right|=\frac{1}{5}-\left(\frac{1}{2}-\frac{5}{4}\right)\)
\(\Rightarrow\left|x-4\right|=\frac{19}{20}\)
\(\Rightarrow\left[{}\begin{matrix}x-4=\frac{19}{20}\\x-4=-\frac{19}{20}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{19}{20}+4=\frac{99}{20}\\x=-\frac{19}{20}+4=\frac{61}{20}\end{matrix}\right.\)
7) \(\left|x-\frac{5}{4}\right|=-\frac{1}{3}\)
Vì \(\left|x-\frac{5}{4}\right|\ge0\)
=> Không có giá trị x thỏa mãn với điều kiện trên
\(a;\)\(\left(x+\frac{1}{3}\right)^2=x+\frac{1}{3}\)
\(\Leftrightarrow\left(x+\frac{1}{3}\right)^2-\left(x+\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left(x+\frac{1}{3}\right)\left(x+\frac{1}{3}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{3}=0\\x+\frac{1}{3}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{3}\\x=\frac{2}{3}\end{cases}}}\)
b)\(\left(x-\frac{1}{4}\right)^3=\left(x-\frac{1}{4}\right)^2\)
\(\Leftrightarrow\left(x-\frac{1}{4}\right)^3-\left(x-\frac{1}{4}\right)^2=0\)
\(\Leftrightarrow\left(x-\frac{1}{4}\right)^2\left(x-\frac{1}{4}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-\frac{1}{4}\right)^2=0\\\left(x-\frac{1}{4}-1\right)=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{4}=0\\x-\frac{5}{4}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=\frac{5}{4}\end{cases}}}\)
\(\left(x+\frac{1}{3}\right)^2-\left(x+\frac{1}{3}\right)\)=0
\(\left(x+\frac{1}{3}\right)\left(x+\frac{1}{3}-1\right)=0\)
\(\left(x+\frac{1}{3}\right)\left(x-\frac{2}{3}\right)=0\)
x=-1/3 hoặc x=2/3
b tương tự