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A. 2.\(|3x+1|\)=\(\frac{3}{4}\)-\(\frac{5}{8}\)
2.\(|3x+1|\)=1/8
\(|3x+1|\)=1/8:2
\(|3x+1|\)=1/16
TH1 : 3x+1=1/16
3x=1/16-1
3x=-15/16
x=-15/16:3
x=-5/16
a,\(\frac{3}{4}-2.\left|3x+1\right|=\frac{5}{8}\)
\(\Rightarrow2.\left|3x+1\right|=\frac{3}{4}-\frac{5}{8}=\frac{6}{8}-\frac{5}{8}=\frac{1}{8}\)
\(\Rightarrow\left|3x+1\right|=\frac{1}{8}.\frac{1}{2}=\frac{1}{16}\)
\(\Rightarrow\orbr{\begin{cases}3x+1=\frac{1}{16}\\3x+1=\frac{-1}{16}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}3x=\frac{1}{16}-1=\frac{-15}{16}\\3x=\frac{-1}{16}-1=\frac{-17}{16}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-15}{16}.\frac{1}{3}=\frac{-5}{16}\\x=\frac{-17}{16}.\frac{1}{3}=\frac{-17}{48}\end{cases}}\)
Vậy....
b,\(\left|3x+2\right|-\left|x-3\right|=\frac{7}{2}\left(1\right)\)
Ta có bảng xét dấu
x | \(\frac{-2}{3}\) 3 |
3x+2 | - 0 + | + |
x-3 | - | - 0 + |
Nếu x<\(\frac{-2}{3}\) thì \(\left|3x+2\right|-\left|x-3\right|\) \(=-3x-2-3+x\)
\(=-2x-5\)
Từ (1) \(\Rightarrow-2x-5=\frac{7}{2}\)
\(\Rightarrow-2x=\frac{7}{2}+5=\frac{17}{2}\)
\(\Rightarrow x=\frac{17}{2}\cdot\frac{-1}{2}=\frac{-17}{4}\)(thỏa mãn x<\(\frac{-2}{3}\)
Nếu \(\frac{-2}{3}\le x\le3\)thì \(\left|3x+2\right|-\left|x-3\right|=3x+2-\left(3-x\right)\)
\(=3x+2-3+x\)
\(=2x-1\)
Từ (1)\(\Rightarrow\)\(2x-1=\frac{7}{2}\)
\(\Rightarrow2x=\frac{9}{2}\)
\(\Rightarrow x=\frac{9}{4}\)(thỏa mãn......
Còn trưonwfg hợp cuối bạn tự làm nốt nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
\(|\dfrac{4}{3}x-\dfrac{3}{4}|=\left|-\dfrac{1}{3}\right|.\left|x\right|\Leftrightarrow|\dfrac{4}{3}x-\dfrac{3}{4}|=\dfrac{1}{3}.\left|x\right|\left(1\right)\)
Tìm nghiệm \(\dfrac{4}{3}x-\dfrac{3}{4}=0\Leftrightarrow\dfrac{4}{3}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.\dfrac{3}{4}\Leftrightarrow x=\dfrac{9}{16}\)
\(x=0\)
Lập bảng xét dấu :
\(x\) \(0\) \(\dfrac{9}{16}\)
\(\left|\dfrac{4}{3}x-\dfrac{3}{4}\right|\) \(-\) \(0\) \(-\) \(0\) \(+\)
\(\left|x\right|\) \(-\) \(0\) \(+\) \(0\) \(+\)
TH1 : \(x< 0\)
\(\left(1\right)\Leftrightarrow-\dfrac{4}{3}x+\dfrac{3}{4}=\dfrac{1}{3}.\left(-x\right)\)
\(\Leftrightarrow-\dfrac{4}{3}x+\dfrac{3}{4}=-\dfrac{1}{3}.x\)
\(\Leftrightarrow\dfrac{4}{3}x-\dfrac{1}{3}x=\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{3}{4}\) (loại vì không thỏa \(x< 0\))
TH2 : \(0\le x\le\dfrac{9}{16}\)
\(\left(1\right)\Leftrightarrow-\dfrac{4}{3}x+\dfrac{3}{4}=\dfrac{1}{3}x\)
\(\Leftrightarrow\dfrac{4}{3}x+\dfrac{1}{3}x=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.\dfrac{3}{5}\Leftrightarrow x=\dfrac{9}{20}\) (thỏa điều kiện \(0\le x\le\dfrac{9}{16}\))
TH3 : \(x>\dfrac{9}{16}\)
\(\left(1\right)\Leftrightarrow\dfrac{4}{3}x-\dfrac{3}{4}=\dfrac{1}{3}x\)
\(\Leftrightarrow\dfrac{4}{3}x-\dfrac{1}{3}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}\) (thỏa điều kiện \(x>\dfrac{9}{16}\))
Vậy \(x\in\left\{\dfrac{9}{20};\dfrac{3}{4}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x+1\right)\left(x^2-x\right)+x\left(5+x-2x^2\right)=3x+7\)
\(2x^3-2x^2+x^2-x+5x+x^2-2x^3=3x+7\)
\(5x-x=3x+7\)
\(4x-3x=7\)
\(x=7\)
(2x+1)(x^2-x)+x(-2x^2+x+5)=3x+7
=>2x^3-2x^2+x^2-x-2x^3+x^2+5x=3x+7
=>-x^2-x+x^2+5x=3x+7
=>4x=3x+7
=>x=7
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có : |2x - 5| = x + 1
\(\Leftrightarrow\orbr{\begin{cases}2x-5=-x-1\\2x-5=x+1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x+x=-1+5\\2x-x=1+5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=4\\x=6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=6\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(|2x-7|=12\Leftrightarrow\orbr{\begin{cases}2x-7=12\\2x-7=-12\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=19\\2x=-5\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{19}{2}\\x=-\frac{5}{2}\end{cases}}}\)
\(|4x+3|=|3x-1|\Leftrightarrow\orbr{\begin{cases}4x+3=3x-1\\4x+3=1-3x\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\7x=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-4\\x=-\frac{2}{7}\end{cases}}}\)
\(|3x+5|=2x+9\left(ĐKXĐ:2x+9\ge0\Leftrightarrow x\ge-\frac{9}{2}\right)\)
\(\Leftrightarrow\orbr{\begin{cases}3x+5=2x+9\\3x+5=-2x-9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\5x=-14\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\left(tm\right)\\x=-\frac{14}{5}\end{cases}}}\left(tm\right)\)
Tự KL cho mỗi phần
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(3x-\dfrac{2}{5}=0=>3x=\dfrac{2}{5}=>x=\dfrac{2}{15}\)
b)\(\left(x-3\right)\left(2x+8\right)=0=>\left[{}\begin{matrix}x-3=0\\2x=-8\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)
c)\(3x^2-x-4=0=>3x^2+3x-4x-4=0=>\left(3x-4\right)\left(x+1\right)=0\)
\(=>\left[{}\begin{matrix}3x=4\\x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(2\(x\) + 3)2 + (3\(x\) - 2)4 =0
Vì:
(2\(x\) + 3)2 ≥ 0
(3\(x\) - 2)4 ≥ 0
Nên :
(2\(x\) + 3)2 + (3\(x\) - 2)4 = 0
⇔ \(\left\{{}\begin{matrix}2x+3=0\\3x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(x\) \(\in\) \(\varnothing\)
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=2x^3-2x^2-5x-10-2x^2+4x+x^2(2x-3)-x(x+1)-3x+2
=2x^3-4x^2-4x-8+2x^3-6x^2-x^2+x
=4x^3-11x^2-3x-8
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,2\left|3x-1\right|+1=5\)
\(\Leftrightarrow2\left|3x-1\right|=4\)
\(\Leftrightarrow\left|3x-1\right|=2\Leftrightarrow\left[{}\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
2.l3x-1l+1=5
=>2.l3x-1l=4
=>l3x-1l=2
TH1:3x-1=2 =>x=1
TH2:3x-1=-2 =>x=-1/3
`Answer:`
\(\left|1-3x\right|=\left|2x+5\right|\)
\(\Leftrightarrow\left|-3x+1\right|=\left|2x+5\right|\)
\(\Leftrightarrow\orbr{\begin{cases}-3x+1=2x+5\\-3x+1=-\left(2x+5\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-3x+1-2x=2x+5-2x\\-3x+1=-2x-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-5x+1=5\\-3x+1+2x=-2x-5+2x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-5x+1-1=5-1\\-x+1=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-5x=4\\-x=-6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{4}{5}\\x=6\end{cases}}\)