Tìm x , biết
5 - ( 1 - 3x ) = 2x +4
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`(-2 1/3 ) .( - (-6)/4 )`
`= - (2xx3+1)/3 . 6/4`
`= - 7/3 . 6/4`
`= -42/12`
`= -7/2`
\(\left(-2\dfrac{1}{3}\right).\left(-\dfrac{-6}{4}\right)\)
\(=\left(-\dfrac{7}{3}\right).\dfrac{3}{2}=-\dfrac{7}{2}\)
\(\left|3x+5\right|=x+1\)
TH1: \(3x+5=x+1\left(x\ge-\dfrac{5}{3}\right)\)
\(\Rightarrow3x-x=1-5\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\left(ktm\right)\)
TH2: \(3x-5=-\left(x+1\right)\left(x< -\dfrac{5}{3}\right)\)
\(\Rightarrow3x-5=-x-1\)
\(\Rightarrow3x+x=-1+5\)
\(\Rightarrow4x=4\)
\(\Rightarrow x=1\)
Vậy không có x thõa mãn
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\(\left|2x-3\right|=2x-3\)
\(\Rightarrow2x-3=2x-3\left(x\ge\dfrac{3}{2}\right)\)
\(\Rightarrow0=0\) (luôn đúng)
Nên mọi x đề thỏa mãn khi \(x\ge\dfrac{3}{2}\)
Vậy: ...
|3x + 5| = x + 1
TH1: x ≥log ) -5/3
(1) ⇒ 3x + 5 = x + 1
3x - x = 1 - 5
2x = -4
x = -2 (loại)
*) TH2: x < -5/3
(1) ⇒ 3x + 5 = -x - 1
3x + x = -1 - 5
4x = -6
x = -3/2 (loại)
Vậy không tìm được x thỏa mãn yêu cầu
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|2x - 3| = 2x - 3 (2)
*) TH1: x 3/2
(2) ⇒ 2x - 3 = 2x - 3
0x = 0 (luôn đúng với mọi x ≥ 3/2)
*) TH2: x < 3/2
(2) ⇒ 2x - 3 = 3 - 2x
2x + 2x = 3 + 3
4x = 6
x = 3/2 (loại)
Vậy x ≥ 3/2
\(|-2x+1,5|=\dfrac{1}{4}\Rightarrow-2x+1,5=\pm\dfrac{1}{4}\)
\(-2x+1,5=\dfrac{1}{4}\Rightarrow-2x=1,5-0,25\Rightarrow-2x=1,25\Rightarrow x=1,25:\left(-2\right)\Rightarrow x=...\)
\(-2x+1,5=-\dfrac{1}{4}\Rightarrow-2x=-0,25-1,5\Rightarrow-2x=1,75\Rightarrow x=1,75:\left(-2\right)\Rightarrow x=...\)
\(\dfrac{3}{2}-|1.\dfrac{1}{4}+3x|=\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{3}{2}-\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{5}{4}\)
\(\Rightarrow1.\dfrac{1}{4}+3x=\pm\dfrac{5}{4}\)
\(1.\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow3x=\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=1\Rightarrow x=3\)
\(1.\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow3x=-\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=-\dfrac{3}{2}x=...\)
a) \(\left|4x-1\right|-\left|3x-\dfrac{1}{2}\right|=0\\ \Leftrightarrow\left|4x-1\right|=\left|3x-\dfrac{1}{2}\right|\\ \Leftrightarrow\left[{}\begin{matrix}4x-1=3x-\dfrac{1}{2}\\4x-1=\dfrac{1}{2}-3x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}4x-3x=1-\dfrac{1}{2}\\4x+3x=\dfrac{1}{2}+1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\7x=\dfrac{3}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{14}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{3}{14}\right\}\) là nghiệm của pt.
b) \(\left|x-1\right|-2x=\dfrac{1}{2}\\ \Leftrightarrow\left|x-1\right|=2x+\dfrac{1}{2}\left(ĐK:x\ge\dfrac{-1}{4}\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x+\dfrac{1}{2}\\x-1=-2x-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-2x=1+\dfrac{1}{2}\\x+2x=1-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-x=\dfrac{3}{2}\\3x=\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\left(ktmđk\right)\\x=\dfrac{1}{6}\left(tmđk\right)\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{6}\) là nghiệm của pt.
Lời giải:
a.
$|4x-1|-|3x-\frac{1}{2}|=0$
$\Leftrightarrow |4x-1|=|3x-\frac{1}{2}$
\(\Leftrightarrow \left[\begin{matrix} 4x-1=3x-\frac{1}{2}\\ 4x-1=\frac{1}{2}-3x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=\frac{3}{14}\end{matrix}\right.\)
b. Nếu $x\geq 1$ thì:
$|x-1|-2x=\frac{1}{2}$
$\Leftrightarrow x-1-2x=\frac{1}{2}$
$\Leftrightarrow -x-1=\frac{1}{2}$
$\Leftrightarrow x=\frac{-3}{2}$ (vô lý vì $x\geq 1$)
Nếu $x< 1$ thì:
$1-x-2x=\frac{1}{2}$
$\Leftrightarrow x=\frac{1}{6}$ (tm)
\(\frac{1}{x+2}+\frac{5}{x-2}=\frac{3x-12}{x^2-4}\)
ĐKXĐ \(x\ne\pm2\)
\(\frac{1}{x+2}+\frac{5}{x-2}=\frac{3x-12}{x^2-4}\)
\(\Leftrightarrow\frac{x-2+5\left(x+2\right)}{x^2-4}=\frac{3x-12}{x^2-4}\)
\(\Leftrightarrow x-2+5x+10=3x-12\)
\(\Leftrightarrow6x+8=3x-12\)
\(\Leftrightarrow3x=20\Leftrightarrow x=\frac{20}{3}\left(tm\right)\)
5-(1-3x)=2x+4
5-1+3x=2x+4
5-1-4 =2x-3x
0 = x (2-3)
0 = x.(-1)
(-1) \(\ne\) 0 =>x=0
\(5-\left(1-3\chi\right)=2\chi+4\)
\(\Rightarrow5-1+3\chi=2\chi+4\)
\(\Rightarrow4-4=2\chi-3\chi\)
\(\Rightarrow0\ne-\chi\Leftrightarrow\chi\in\varnothing\)
SUN