I5x-2I\(\le\)13
I2x-3I\(\ge\)7
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\(\left|5x-4\right|=\left|x+2\right|\)
\(\Rightarrow\orbr{\begin{cases}5x-4=x+2\\5x-4+x+2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}5x-x=2+4\\6x-2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}4x=6\\6x=2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{3}\end{cases}}\)
b tương tự
\(\left|5x-4\right|=\left|x+2\right|\)
\(\Rightarrow\orbr{\begin{cases}5x-4=x+2\\5x-4=-\left(x+2\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}5x-x=4+2\\5x-4=-x-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}4x=6\\5x+x=4-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{6}{4}\\6x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{2}{6}=\frac{1}{3}\end{cases}}\)
\(\text{b) }\left|2+3x\right|=\left|4x-3\right|\)
\(\Rightarrow\orbr{\begin{cases}2+3x=4x-3\\2+3x=-\left(4x-3\right)\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-4x+3x=-2-3\\2+3x=-4x+3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-1x=-5\\4x+3x=-2+3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\7x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{7}\end{cases}}\)
a) \(|5x-3|-x=\text{}6\)
\(\Rightarrow|5x-3|=6+x\left(1\right)\)
Vì \(\Rightarrow|5x-3|\ge0\)
\(\Rightarrow6+x\ge0\)
\(\Rightarrow x\ge-6\)
(1) xảy ra\(\Leftrightarrow\orbr{\begin{cases}5x-3=6+x\\5x-3=-6-x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x-x=6+3\\5x+x=-6+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x=9\\6x=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=-2\end{cases}}\)
Vậy ...
\(\left|5x-3\right|\ge7\)
\(\Rightarrow5x-3\ge\pm7\)
TH1 :
\(5x-3\ge7\)
\(\Rightarrow x\ge2\)
TH2 :
\(5x-3\ge-7\)
\(\Rightarrow x\ge-\frac{4}{5}\)
Vậy ................
Tìm x,y biết
a) 2I2x-3I=\(\frac{1}{2}\)
b)7,5-3I5-2xI=-4,5
c)I3x-4I+I3y+5I=0
d)3,7+I4,3-xI=0
e)4-I5x-2I=1
a) \(2\left|2x-3\right|=\frac{1}{2}\)
\(\left|2x-3\right|=\frac{1}{2}:2\)
\(\left|2x-3\right|=\frac{1}{4}\)
\(\orbr{\begin{cases}2x-3=\frac{1}{4}\\2x-3=-\frac{1}{4}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}2x=\frac{13}{4}\\2x=\frac{11}{4}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{13}{8}\\x=\frac{11}{8}\end{cases}}\)
b)\(7,5-3\left|5-2x\right|=-4,5\)
\(3\left|5-2x\right|=12\)
\(\left|5-2x\right|=4\)
\(\orbr{\begin{cases}5-2x=4\\5-2x=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=1\\2x=9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{9}{2}\end{cases}}}\)
|5x-3| -x =7
|5x-3| = 7 +x
Th1: 5x-3 = 7+x
4x =10
x= 10/4
Th2: 5x -3 = -(7+x)
5x -3 = -7 -x
6x = -4
x= -2/3
Vậy............
/5x-3/-x =7
/5x-3/ = 7+x
=>\(\orbr{\begin{cases}5x-3=7+x\\5x-3=-\left(7+x\right)\end{cases}}\)=>\(\orbr{\begin{cases}5x-x=3+7\\5x-3=-7-x\end{cases}}\)=>\(\orbr{\begin{cases}4x=10\\5x+x=3-7\end{cases}}\)=>\(\orbr{\begin{cases}x=2,5\\6x=-4\end{cases}}\)=>\(\orbr{\begin{cases}x=2,5\\x=-\frac{2}{3}\end{cases}}\)
|5x-3|-x=7
=>|5x-3|=7+x
=>5x-3=-(7+x) hoặc 7+x
Nếu 5x-3=-(7+x)
=>5x-3=-x-7
=>6x=-4
=>x=\(-\frac{2}{3}\)
Nếu 5x-3=7+x
=>4x=10
=>x=\(\frac{5}{2}\)
I5x-3I-x=7
=> |5x-3|=7+x
=> TH1 :
5x-3=7+x
=> 5x-x=7+3
=> 4x = 10
=> x=2.5
TH2 :
5x-3=-(x+7)
=> 5x=-(x+7)+3
=> 5x=-x+((-7)+3)
=> 5x=-x+(-4)
=> 5x+x=-4
=> 6x=-4
=> x = -2/3
I5x - 2I\(\le13\Rightarrow-13\le5x-2\le13\Rightarrow-11\le5x\le15\Rightarrow-2,2\le x\le3\)
I2x - 3I\(\ge7\Rightarrow2x-3\le-7\)hoặc\(2x-3\ge7\Rightarrow2x\le-4\)hoặc\(2x\ge10\Rightarrow x\le-2\)hoặc\(x\ge5\)