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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 9 2024
Bài 1:
- \(\dfrac{11}{2}x\) + 1 = \(\dfrac{1}{3}x-\dfrac{1}{4}\)
- \(\dfrac{11}{2}\)\(x\) - \(\dfrac{1}{3}\)\(x\) = - \(\dfrac{1}{4}\) - 1
-(\(\dfrac{33}{6}\) + \(\dfrac{2}{6}\))\(x\) = - \(\dfrac{5}{4}\)
- \(\dfrac{35}{6}\)\(x\) = - \(\dfrac{5}{4}\)
\(x=-\dfrac{5}{4}\) : (- \(\dfrac{35}{6}\))
\(x\) = \(\dfrac{3}{14}\)
Vậy \(x=\dfrac{3}{14}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 9 2024
Bài 2: 2\(x\) - \(\dfrac{2}{3}\) - 7\(x\) = \(\dfrac{3}{2}\) - 1
2\(x\) - 7\(x\) = \(\dfrac{3}{2}\) - 1 + \(\dfrac{2}{3}\)
- 5\(x\) = \(\dfrac{9}{6}\) - \(\dfrac{6}{6}\) + \(\dfrac{4}{6}\)
- 5\(x\) = \(\dfrac{7}{6}\)
\(x\) = \(\dfrac{7}{6}\) : (- 5)
\(x\) = - \(\dfrac{7}{30}\)
Vậy \(x=-\dfrac{7}{30}\)
2 tháng 4 2020
1: Ta có: \(2x+x\left(x-5\right)=3x^2-x\)
\(\Leftrightarrow2x+x^2-5x-3x^2+x=0\)
\(\Leftrightarrow-2x^2-2x=0\)
\(\Leftrightarrow-2x\left(x+1\right)=0\)
Vì -2≠0
nên \(\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy: x∈{0;-1}
2) Ta có: \(15-5\left(1-2x\right)=12-x\)
\(\Leftrightarrow15-5+10x-12+x=0\)
\(\Leftrightarrow11x-2=0\)
\(\Leftrightarrow11x=2\)
hay \(x=\frac{2}{11}\)
Vậy: \(x=\frac{2}{11}\)
3) Ta có: \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
\(\Leftrightarrow\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}-5=0\)
\(\Leftrightarrow\frac{-13}{3}-\frac{4}{3}x=0\)
\(\Leftrightarrow\frac{4}{3}x=\frac{-13}{3}\)
hay \(x=\frac{-13}{3}:\frac{4}{3}=\frac{-13}{4}\)
Vậy: \(x=\frac{-13}{4}\)
4) Ta có: \(\left|x-\frac{4}{5}\right|=\frac{3}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{5}\\x-\frac{4}{5}=\frac{-3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{5}\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{5};\frac{7}{5}\right\}\)
BL
2 tháng 4 2020
1. \(2x+x\left(x-5\right)=3x^2-x\)
\(\Leftrightarrow2x+x^2-5x=3x^2-x\)
\(\Leftrightarrow\left(2x-5x+x\right)+\left(x^2-3x^2\right)=0\)
\(\Leftrightarrow-2x-2x^2=0\)
\(\Leftrightarrow-2x\left(1+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\1+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
2. \(15-5\left(1-2x\right)=12-x\)
\(\Leftrightarrow15-5+10x=12-x\)
\(\Leftrightarrow\left(15-5-12\right)+\left(10x+x\right)=0\)
\(\Leftrightarrow-2+11x=0\)
\(\Leftrightarrow11x=2\Leftrightarrow x=\frac{2}{11}\)
3. \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
\(\Leftrightarrow\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
\(\Leftrightarrow\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}-5\right)-\left(\frac{1}{3}x+x\right)=0\)
\(\Leftrightarrow-\frac{13}{3}-\frac{4}{3}x=0\)
\(\Leftrightarrow-\frac{4}{3}x=\frac{13}{3}\Leftrightarrow x=-\frac{13}{4}\)
4. \(\left|x-\frac{4}{5}\right|=\frac{3}{5}\)
\(\Rightarrow x-\frac{4}{5}=-\frac{3}{5}\) hoặc \(x-\frac{4}{5}=\frac{3}{5}\)
\(TH1:x-\frac{4}{5}=-\frac{3}{5}\Rightarrow x=\frac{1}{5}\)
\(TH2:x-\frac{4}{5}=\frac{3}{5}\Rightarrow x=\frac{7}{5}\)
VM
0
KH
22 tháng 9 2018
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
21 tháng 7 2017
\(2x+\frac{1}{2}=\frac{-5}{3}\)
\(2x=\frac{-5}{3}-\frac{1}{2}\)
\(2x=\frac{-10}{6}-\frac{3}{6}\)
\(2x=\frac{-13}{6}\)
\(x=\frac{-13}{6}:2\)
\(x=\frac{-13}{12}\)
29 tháng 6 2019
\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x=-\frac{13}{4}\)
\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)
29 tháng 6 2019
\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)
\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)
\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)
\(\Leftrightarrow x=-\frac{6}{11}\)
d,e,f Tương tự
Ta có:
\(\left|x+\frac{1}{5}\right|=\left|2x-\frac{2}{3}\right|\)
\(\Leftrightarrow\left|x+\frac{1}{5}\right|=\left|\frac{2x\cdot3-2}{3}\right|\)
\(\Leftrightarrow\left|x+\frac{1}{5}\right|=\left|\frac{6x-2}{3}\right|\)
\(\Leftrightarrow\left|\frac{5x+1}{5}\right|=\left|\frac{6x-2}{3}\right|\)
\(\Leftrightarrow\left|5x+1\right|=\frac{10\left|3x+1\right|}{3}\)
\(\Leftrightarrow\sqrt{\left(x+\frac{1}{5}\right)^2}=\sqrt{\left(2x+\frac{2}{3}\right)^2}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{7}{15}\\x=-\frac{13}{45}\end{cases}}\)
\(\left|x+\dfrac{1}{5}\right|=\left|2x-\dfrac{2}{3}\right|\)
Trường hợp 1: \(x+\dfrac{1}{5}=2x-\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{1}{5}+\dfrac{2}{3}=\dfrac{13}{15}\)
Trường hợp 2: \(x+\dfrac{1}{5}=\dfrac{2}{3}-2x\)
\(\Rightarrow3x=\dfrac{2}{3}-\dfrac{1}{5}=\dfrac{7}{15}\Rightarrow x=\dfrac{7}{45}\)