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|5\(x\) - 4| = |\(x+2\)|
\(\left[{}\begin{matrix}5x-4=x+2\\5x-4=-x-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}4x=6\\6x=2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
vậy \(x\in\) { \(\dfrac{1}{3};\dfrac{3}{2}\)}
|2\(x\) - 3| - |3\(x\) + 2| = 0
|2\(x\) - 3| = | 3\(x\) + 2|
\(\left[{}\begin{matrix}2x-3=3x+2\\2x-3=-3x-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{5}\end{matrix}\right.\)
vậy \(x\in\){ -5; \(\dfrac{1}{5}\)}
+ Nếu \(x\ge5\) thì |x-1|+|x-3|+|x-5| = x-1+x-3+x-5 = 3x-9 = 5x-5
=> -9+5 = 5x-3x
=> -4 = 2x
=>x=-2 (loại vì x<5)
+ Nếu \(3\le x<5\) thì |x-1|+|x-3|+|x-5| = x-1 + x-3 - x+5 = x+1 = 5x-5
=> 1+5 = 5x-x
=> 6 = 4x
=> x= 6/4 = 1,5 (loại vì x<3)
+ Nếu \(1\le x<3\) thì |x-1|+|x-3|+|x-5| = x-1 - x+3 - x+5 = 7-x = 5x-5
=> 7+5 = 5x+x
=> 12 = 6x
=> x=2 (thõa mãn)
+ Nếu x<1 thì |x-1|+|x-3|+|x-5| = 1-x+3-x+5-x = 9 - 3x = 5x-5
=> 9+5 = 5x+3x
=> 14 = 8x
=> x=14/8 = 1,75 (loại vì x<1)
Vậy x=2 thõa mãn yêu cầu đề bài
=>(2x+3).(10x+2)=(5x+2).(4x+5)
=>(2x.10x)+(2x.2)+(3.10x)+(3.2)=(5x.4x)+(5x.5)+(2.4x)+(2.5)
=>20x2+4x+30x+6=20x2+25x+8x+10
=>20x2-20x2+4x-8x+30x-25x=10-6
=>0+4x-8x+30x-25x=4
=>-4x+30x-25x=4
=>26x-25x=4
=>x=4
B)=>(3x-1).(5x-34)=(40-5x).(25-3x)
=>15x2-102x-5x+34=1000-120x-125x+15x2
=>15x2-107x+34=1000-245x+15x2
=>15x2-15x2-107x+245x=1000-34
=>0-107x+245x=966
=>138x=966
=>x=7
A,=>(2x+3).(10x+2)=(5x+2).(4x+5)
=>(2x.10x)+(2x.2)+(3.10x)+(3.2)=(5x.4x)+(5x.5)+(2.4x)+(2.5)
=>20x2+4x+30x+6=20x2+25x+8x+10
=>20x2-20x2+4x-8x+30x-25x=10-6
=>0+4x-8x+30x-25x=4
=>-4x+30x-25x=4
=>26x-25x=4
=>x=4
Ta có : \(x^2:\frac{3}{5}=\frac{-3^2}{5}:5x\)
=> \(x^2.5x=\frac{-3^2}{5}.\frac{3}{5}\)
=> \(x^3.5=\frac{-3^3}{5^2}\)
=> \(x^3=\frac{-3^3}{5^3}\)
=> \(x^3=\left(-\frac{3}{5}\right)^3\)
=> \(x=-\frac{3}{5}\)
\(x^2\div\frac{3}{5}=\frac{-3^2}{5}\div5x\)
\(\Leftrightarrow x^2\times5x=\frac{-3^2}{5}\times\frac{3}{5}\)
\(\Leftrightarrow5x^3=-\frac{27}{25}\)
\(\Leftrightarrow x^3=-\frac{27}{125}\)
\(\Leftrightarrow x^3=\left(-\frac{3}{5}\right)^3\)
\(\Leftrightarrow x=-\frac{3}{5}\)
Ta có : \(\left|-5x\right|=\frac{5}{3}\)
\(\Leftrightarrow5\left|x\right|=\frac{5}{3}\)
\(\Leftrightarrow\left|x\right|=\frac{1}{3}\Leftrightarrow x=\pm\frac{1}{3}\)