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1) \(|5x-3|=|7-x|\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=7-x\\5x-3=x-7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}6x=10\\4x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)
Vậy...
2) \(2.|3x-1|-3x=7\)
\(\Leftrightarrow2.|3x-1|=7+3x\)
\(\Leftrightarrow\orbr{\begin{cases}2.\left(3x-1\right)=7+3x\\2.\left(3x-1\right)=-7-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}6x-2=7+3x\\6x-2=-7-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=9\\9x=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{9}\end{cases}}\)
Vậy...
a: \(\Leftrightarrow\left|\dfrac{5}{3}x\right|=\dfrac{1}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{5}{3}=\dfrac{1}{6}\\x\cdot\dfrac{5}{3}=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}:\dfrac{5}{3}=\dfrac{3}{30}=\dfrac{1}{10}\\x=-\dfrac{1}{10}\end{matrix}\right.\)
b: \(\Leftrightarrow\left|\dfrac{3}{4}x-\dfrac{3}{4}\right|=\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x-1\right|=\dfrac{3}{2}:\dfrac{3}{4}=2\)
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
c: \(\Leftrightarrow\left|x+\dfrac{3}{5}\right|=\left|x-\dfrac{7}{3}\right|\)
\(\Leftrightarrow x+\dfrac{3}{5}=\dfrac{7}{3}-x\)
=>2x=44/15
hay x=22/15
1 +1 = 3, 3 voi 3 la 4, 4 voi 1 la ba, 3 ngon tay that deu
a. | x - 1/7 | + 3/7 = 0
<=> | x - 1/7 | = - 3/7
Mà \(\left|x-\frac{1}{7}\right|\ge0\forall x\)
=> Không có x tm đề bài
b. | x + 1/4 | - 3/4 = 5%
<=> | x + 1/4 | = 4/5
<=> \(\orbr{\begin{cases}x+\frac{1}{4}=\frac{4}{5}\\x+\frac{1}{4}=-\frac{4}{5}\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{11}{20}\\x=-\frac{21}{20}\end{cases}}\)
c. | - x + 2/5 | + 1/2 = 3,5
<=> | - x + 2/5 | = 3
<=> \(\orbr{\begin{cases}-x+\frac{2}{5}=3\\-x+\frac{2}{5}=-3\end{cases}}\)<=>\(\orbr{\begin{cases}x=-\frac{13}{5}\\x=\frac{17}{5}\end{cases}}\)
\(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
<=> \(\frac{x-2}{7}.\frac{x+3}{5}.\frac{x+4}{3}=0\)
<=> \(\frac{x-2}{7}=0\)hoặc \(\frac{x+3}{5}=0\); \(\frac{x+4}{3}=0\)
Nếu \(\frac{x-2}{7}=0\)<=> \(x-2=0\)<=> \(x=2\)
Nếu \(\frac{x+3}{5}=0\)<=> \(x+3=0\) <=> \(x=3\)
Nếu \(\frac{x+4}{3}=0\)<=> \(x+4=0\)<=> \(x=4\)
Vây x= 2 hoặc 3; 4
a)Ta có: |x-5/4|-|x+2/3|=0
=>|x-5/4|=|x+2/3|
*Xét x>_5/4=>x-5/4>_0=>|x-5/4|=x-5/4
=>x+2/3>0=>|x+2/3|=x+2/3
=>|x-5/4|=|x+2/3|
=>x-5/4=x+2/3
=>x-x=2/3+5/4
=>0=23/12
=>Vô lí
*Xét -2/3<_x<5/4=>x-5/4<0=>|x-5/4|=5/4-x
=>x+2/3>_0=>|x+2/3|=x+2/3
=>|x-5/4|=|x+2/3|
=>5/4-x=x+2/3
=>5/4-2/3=x+x
=>7/12=2x
=>x=7/24
*Xét x<-2/3=>x-5/4<0=>|x-5/4|=5/4-x
=>x+2/3<0=>|x+2/3|=-x-2/3
=>|x-5/4|=|x+2/3|
=>5/4-x=-x-2/3
=>x-x=5/4+2/3
=>0=23/12
=>Vô lí
Vậy x=7/24
\(\left(x+\frac{1}{4}\right).\left(x-\frac{3}{7}\right)=0\)
=> \(x+\frac{1}{4}=0\) hoặc \(x-\frac{3}{7}=0\)
<=> \(x=\frac{-1}{4}\) \(x=\frac{3}{7}\)
vậy x = \(\frac{-1}{4}\) hoặc x = \(\frac{3}{7}\)
\(\left(x+\frac{1}{4}\right)\left(x-\frac{3}{7}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{4}=0\\x-\frac{3}{7}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=\frac{3}{7}\end{cases}}\)