|3x - 1| - 1/2 = 0,75
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1 tháng 12 2019
\(1\frac{1}{3}x-\left(\frac{3}{2}\right)^2=-0,75\)
\(\frac{4}{3}x-\frac{9}{4}=-\frac{3}{4}\)
\(\frac{4}{3}x=-\frac{3}{4}+\frac{9}{4}\)
\(\frac{4}{3}x=\frac{3}{2}\)
\(x=\frac{3}{2}:\frac{4}{3}\)
\(x=\frac{3}{2}.\frac{3}{4}\)
\(x=\frac{9}{8}\)
Vậy \(x=\frac{9}{8}\)
TT
4
PT
Phạm Trần Hoàng Anh
CTVHS
19 tháng 6 2020
ok ko sao
2/3x : (-0.75)=1
2/3x=-0.75
2/3x=-3/4
x=-3/4:2/3
x=-9/8
Y
17 tháng 8 2023
\(\dfrac{1}{2}-\left|2-3x\right|=\sqrt{\dfrac{19}{16}}-\sqrt{\left(-0,75\right)^2}\\ \Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt{19}}{4}-\dfrac{3}{4}\\ \Rightarrow\left|2-3x\right|=\dfrac{1}{2}-\dfrac{\sqrt{19}-3}{4}\)
\(\Rightarrow\left|2-3x\right|=\dfrac{5-\sqrt{19}}{4}\)
\(TH_1:x\le\dfrac{2}{3}\\ 2-3x=\dfrac{5-\sqrt{19}}{4}\\ \Rightarrow3x=\dfrac{3+\sqrt{19}}{4}\\ \Rightarrow x=\dfrac{3+\sqrt{19}}{12}\left(tm\right)\)
\(TH_2:x>\dfrac{2}{3}\\ 3x-2=\dfrac{5-\sqrt{19}}{4}\\ \Rightarrow3x=\dfrac{13-\sqrt{19}}{4}\\ \Rightarrow x=\dfrac{13-\sqrt{19}}{12}\left(tm\right)\)
Vậy \(x\in\left\{\dfrac{3+\sqrt{19}}{12};\dfrac{13-\sqrt{19}}{12}\right\}\)
17 tháng 8 2023
\(\dfrac{1}{2}-\left|2-3x\right|=\sqrt[]{\dfrac{19}{16}}-\sqrt[]{\left(-0,75\right)^2}\)
\(\Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt[]{19}}{4}-0,75\)
\(\Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt[]{19}}{4}-\dfrac{3}{4}\)
\(\Rightarrow\left|2-3x\right|=\dfrac{1}{2}-\dfrac{\sqrt[]{19}}{4}+\dfrac{3}{4}\)
\(\Rightarrow\left|2-3x\right|=\dfrac{5-\sqrt[]{19}}{4}\)
\(\Rightarrow\left[{}\begin{matrix}2-3x=\dfrac{5-\sqrt[]{19}}{4}\\2-3x=\dfrac{-5+\sqrt[]{19}}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=2-\dfrac{5-\sqrt[]{19}}{4}\\3x=2-\dfrac{\sqrt[]{19}-5}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=\dfrac{3+\sqrt[]{19}}{4}\\3x=\dfrac{13-\sqrt[]{19}}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt[]{19}}{12}\\x=\dfrac{13-\sqrt[]{19}}{12}\end{matrix}\right.\)
LC
0
5 tháng 2 2022
a: \(\Leftrightarrow x\cdot\dfrac{3}{5}=\dfrac{-1}{7}+\dfrac{1}{2}=\dfrac{1}{2}-\dfrac{1}{7}=\dfrac{5}{14}\)
\(\Leftrightarrow x=\dfrac{5}{14}:\dfrac{3}{5}=\dfrac{25}{42}\)
b: =>|3x-1|=2
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2\\3x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
c: \(\Leftrightarrow7\left(37-x\right)=3\left(x-13\right)\)
=>259-7x=3x-39
=>-10x=-298
hay x=29,8
d: =>x=3/4+2/3=9/12+8/12=17/12
18 tháng 6 2020
a, \(\left(2x-1\right)^3=8\)
\(\Leftrightarrow\left(2x-1\right)^3=2^3\)
\(\Leftrightarrow2x-1=2\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
b, \(\left(x-1\right)^x+2=\left(x-1\right)^x+4\)
\(\Leftrightarrow\left(x-1\right)^x+2-\left(x-1\right)^x-4=0\)
\(\Leftrightarrow-2\ne0\)=> vô nghiệm
c, \(3^x-1+5.3^{3x}-1=162\)
Đề sai.
24 tháng 10 2021
Bài 2:
a: \(\left(3x-3\right)^2-\left(5x-3\right)^2=0\)
\(\Leftrightarrow\left(3x-3-5x+3\right)\left(3x-3+5x-3\right)=0\)
\(\Leftrightarrow-2x\left(8x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{4}\end{matrix}\right.\)
\(\left|3x-1\right|-\dfrac{1}{2}=0,75=\dfrac{3}{4}\)
\(\left|3x-1\right|=\dfrac{3}{4}+\dfrac{1}{2}=\dfrac{5}{4}\)
⇔ \(\left[{}\begin{matrix}3x-1=\dfrac{5}{4}\\3x-1=-\dfrac{5}{4}\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}3x=\dfrac{9}{4}\\3x=-\dfrac{1}{4}\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{1}{12}\end{matrix}\right.\)
|3x-1| -1/2= 0,75
=>|3x-1|= 0,75 -1/2
=>|3x-1|= -1/2
=> 3x-1= -1/2 hoặc 3x-1= 1/2
=> 3x= 1/2 hoặc 3x= 3/2
=>x=1/6 hoặc x=1/2