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a) \(\left|2x-3\right|-\dfrac{5}{2}=\dfrac{1}{3}\)
\(\left|2x-3\right|=\dfrac{1}{3}+\dfrac{5}{2}=\dfrac{2}{6}+\dfrac{15}{6}\)
\(\left|2x-3\right|=\dfrac{17}{6}\)
\(+)2x-3=\dfrac{17}{6}\Rightarrow2x=\dfrac{35}{6}\Rightarrow x=\dfrac{35}{12}\)
\(+)2x-3=\dfrac{-17}{6}\Rightarrow2x=\dfrac{1}{6}\Rightarrow x=\dfrac{1}{12}\)
vậy...
\(\left|x-1\right|+3x=1\\ \Rightarrow\left|x-1\right|=1-3x\\ \Rightarrow\left\{{}\begin{matrix}x-1=1-3x\\x-1=-1+3x\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4x=2\\-2x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)
Dấu ngoặc vuông nhé
thánh bấm nhầm
a.
để x \(\in Z\) thì \(2x-3\inƯ_{\left(7\right)}\left\{1;-1;7;-7\right\}\)
| 2x-3 | 1 | -1 | 7 | -7 |
| x | 2 | 1 | 5 | -2 |
Vậy x ={2;1;5;-2}
câu b mik ko pt lm đâu nhé sorry
c.
\(\frac{3x+2}{x+1}=\frac{3x+3-1}{x+1}=3-\frac{1}{x+1}\)
để x \(\in Z\) thì \(x+1\inƯ_{\left(1\right)}\left\{1;-1\right\}\)
| x+1 | 1 | -1 | ||
| x | 0 | -2 |
câu d giống câu a rồi nhé
a) \(\left|2x-5\right|=x+1\)
+) Xét \(x\ge\frac{5}{2}\) có:
\(2x-5=x+1\)
\(\Rightarrow x=6\) ( t/m )
+) Xét \(x< \frac{5}{2}\) có:
\(-\left(2x-5\right)=x+1\)
\(\Rightarrow-2x+5=x+1\)
\(\Rightarrow-3x=-4\)
\(\Rightarrow x=\frac{4}{3}\) ( t/m )
Vậy \(x\in\left\{6;\frac{4}{3}\right\}\)
b) \(\left|3x-2\right|-1=x\)
+) Xét \(x\ge\frac{2}{3}\) có:
\(3x-2-1=x\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\frac{3}{2}\) ( t/m )
+) Xét \(x< \frac{2}{3}\) có:
\(2-3x-1=x\)
\(\Rightarrow1=4x\)
\(\Rightarrow x=\frac{1}{4}\) ( t/m )
Vậy \(x\in\left\{\frac{3}{2};\frac{1}{4}\right\}\)
Các phần c, d tương tự
a: \(\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\\left(3x+8+2x+4\right)\left(3x+8-2x-4\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\\left(5x+12\right)\left(x+4\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
b: \(\Leftrightarrow\left|4x+2\right|=x+15\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-15\\\left(4x+2+x+15\right)\left(4x+2-x-15\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-15\\\left(5x+17\right)\left(3x-13\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{17}{5};\dfrac{13}{3}\right\}\)
c: =>3x+7>=0
hay x>=-7/3
d: =>|2x-5|=-2x+5
=>2x-5<=0
hay x<=5/2
b)\(\left|21x-5\right|=\left|3x-7\right|\)
\(\Leftrightarrow\begin{cases}21x-5=3x-7\\21x-5=7-3x\end{cases}\)
\(\Leftrightarrow\begin{cases}9x=-1\\24x=12\end{cases}\)
\(\Leftrightarrow\begin{cases}x=-\frac{1}{9}\\x=\frac{1}{2}\end{cases}\)
a)\(\left|2x-7\right|=3\)
\(\Rightarrow2x-7=\pm3\)
Nếu \(2x-7=3\)
\(\Rightarrow2x=10\)
\(\Rightarrow x=5\)
Nếu \(2x-7=-3\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
đầu bài trên tớ làm luôn nhá !!!
a, / 3x+1/= 5-3
/ 3x+1/= 2
3x+1=2
x+1 = 2:3
x+1 = 2 phần 3
x= 2/3 -1
x= -1/3