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3-/x-5/=-7+/-8/
3-/x-5/=-7+8
3-/x-5/=1
/x-5/=3-1
/x-5/=2
=>x-5=2 hoặc x-5=-2
x=5+2 x=5+(-2)
x=7 x=3
vậy x\(\in\){3;7}
\(1.\)\(3-\left|X-5\right|=-7+\left|-8\right|\)
\(3-\left|X-5\right|=-7+8\)
\(3-\left|X-5\right|=1\)
\(\left|X-5\right|=3-1\)
\(\left|X-5\right|=2\)
\(\Rightarrow X-5=2\)\(ho\text{ặ}c\)\(X-5=\left(-2\right)\)
\(N\text{ếu}\)\(X-5=2\) \(N\text{ếu}\)\(X-5=\left(-2\right)\)
\(X=5+2\) \(X=\left(-2\right)+5\)
\(X=7\) \(X=3\)
\(\frac{7}{4}.x-\frac{3}{2}=\frac{1}{2}x-3\)
\(\Rightarrow\frac{7}{4}x-\frac{1}{2}x=\frac{3}{2}-3\)
\(\Rightarrow\frac{5}{4}x=-\frac{3}{2}\)
\(\Rightarrow x=-\frac{3}{2}.\frac{4}{5}=-\frac{6}{5}\)
7x/4-3/2=x/2-3
=> 7x/4-6/4=x/2-6/2
=>7x-6/4=x-6/2
=>(7x-6)*2=(x-6)*4
=>14x-12=4x-24
=>10x=-24+12
=>10x=-12=>x=-6/5
Vậy x=-6/5
1, => x + 12 = 0 => x = -12
x - 3 = 0 => x = 3
=> x \(\in\) { -12; 3 }
1; (\(x\) + 12)(\(x\) - 3) = 0
\(\left[{}\begin{matrix}x+12=0\\x-3=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-12\\x=3\end{matrix}\right.\)
Vậy \(x\) \(\in\) { -12; 3}
1.
2|x-6|+7x-2=|x-6|+7x
2|x-6| - |x-6|=7x-(7x-2)
|x-6| = 2
=>x-6 = +2
*x-6=2 *x-6 = -2
x =2+6 x = (-2)+6
x =8 x = 4
2.
|x-5|-7(x+4)=5-7x
|x-5|-7x-28 =5-7x
|x-5|-28 =5-7x+7x
|x-5|-28 = 5
|x-5| = 5+28
|x-5| = 33
=>x-5 = +33
*x-5=33 *x-5=-33
x =38 x = -28
3.
3|x+4|-2(x-1)=7-2x
3|x+4|-2x+2 =7-2x
3|x+4|-2 =7-2x+2x
3|x+4|-2 =7
3|x+4| =7+2
3|x+4| = 9
|x+4| =9:3
|x+4| = 3
=>x+4 = +3
*x+4=3 *x+4=-3
x =-1 x = -7
\(\dfrac{3}{7}\cdot x+\dfrac{1}{4}=\dfrac{1}{2}\\ \dfrac{3}{7}\cdot x=\dfrac{1}{2}-\dfrac{1}{4}\\ \dfrac{3}{7}\cdot x=\dfrac{1}{4}\\ x=\dfrac{1}{4}:\dfrac{3}{7}\\ x=\dfrac{1}{4}\cdot\dfrac{7}{3}\\ x=\dfrac{7}{12}\)
Vậy: ...
\(\dfrac{3}{7}\) \(\times\) \(x\) + \(\dfrac{1}{4}\) = \(\dfrac{1}{2}\)
\(\dfrac{3}{7}\) \(\times\) \(x\) = \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\)
\(\dfrac{3}{7}\) \(\times\) \(x\) = \(\dfrac{1}{4}\)
\(x\) = \(\dfrac{1}{4}\) : \(\dfrac{3}{7}\)
\(x\) = \(\dfrac{7}{12}\)
Vậy \(x=\dfrac{7}{12}\)