Tìm x I3x-5I=Ix+2I
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(\Leftrightarrow\left\{{}\begin{matrix}3x-2>-4\\3x-2< 4\end{matrix}\right.\Leftrightarrow-\dfrac{2}{3}< x< 2\)
c: \(\Leftrightarrow\left[{}\begin{matrix}3x-1>5\\3x-1< -5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -\dfrac{4}{3}\end{matrix}\right.\)
d: \(\Leftrightarrow\left[{}\begin{matrix}3x+1>x-2\\3x+1< -x+2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x>-3\\4x< 1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-\dfrac{3}{2}\\x< \dfrac{1}{4}\end{matrix}\right.\)
Xét x<-2=>x+2<0=>Ix+2I=-x-2
=>3x+6<3.(-2)+6=>3x+6<0=>I3x+6I=-3x-6
=>I3x+6I-Ix+2I=7
=>-3x-6+x+2=7
=>(-3x+x)-(6-2)=7
=>-2x-4=7
=>-2x=7+4
=>-2x=11
=>x=11:(-2)
=>x=-11/2
Xét x>_-2=>x+2>_0=>Ix+2I=x+2
=>3x+6>_3.(-2)+6=>3x+6>_0=>I3x+6I=3x+6
=>I3x+6I-Ix+2I=7
=>3x+6+x+2=7
=>(3x+x)+(6+2)=7
=>4x+8=7
=>4x=7-8
=>4x=-1
=>x=-1/4
Vậy x=-11/2,-1/4
Vì :
|x + 2| ≥ 0
|x + 5| ≥ 0
|x + 9| ≥ 0
|x + 11| ≥ 0
=> |x + 2| + |x + 5| + |x + 9| + |x + 11| ≥ 0
Hay 5x ≥ 0 => x ≥ 0
=> |x + 2| + |x + 5| + |x + 9| + |x + 11| = x + 2 + x + 5 + x + 9 + x + 11
= 4x + 27 = 5x
=> x = 27
Vậy x = 27
HAPPY NEW YEAR !!!
Ta có :
\(\left|x+2\right|\ge0\)
\(\left|x+5\right|\ge0\)
\(\left|x+9\right|\ge0\)
\(\left|x+2\right|\ge0\)
a) \(\left|2x\right|=3-x\)
\(\Rightarrow\orbr{\begin{cases}2x=3-x\\2x=x-3\end{cases}}\Rightarrow\orbr{\begin{cases}2x+x=3\\2x-x=-3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=3\\x=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
b) \(\left|x-1\right|=2x-1\)
\(\Rightarrow\orbr{\begin{cases}x-1=2x-1\\x-1=1-2x\end{cases}}\Rightarrow\orbr{\begin{cases}x-2x=-1+1\\x+2x=1+1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-x=0\\3x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{2}{3}\end{cases}}\)
a) \(\left|\left|x-1\right|-1\right|=2\Rightarrow\orbr{\begin{cases}\left|x-1\right|-1=2\\\left|x-1\right|-1=-2\end{cases}}\Rightarrow\orbr{\begin{cases}\left|x-1\right|=3\\\left|x-1\right|=-1\left(l\right)\end{cases}}\)
TH1: x - 1 = 3
x = 4
TH2: x - 1 = - 3
x = - 2
b) Tương tự câu a.
c) \(\left|\left|2x-3\right|-x+1\right|=42-8\)
\(\left|\left|2x-3\right|-x+1\right|=34\)
TH1: \(\left|2x-3\right|-x+1=34\)
\(\left|2x-3\right|-x=33\)
Với \(x\ge\frac{3}{2}\), ta có \(2x-3-x=33\Rightarrow x=36\) (tm)
Với \(x< \frac{3}{2}\), ta có \(3-2x-x+1=34\Rightarrow-3x=30\Rightarrow x=-10\left(tm\right)\)
TH2: \(\left|2x-3\right|-x+1=-34\)
\(\left|2x-3\right|-x=-35\)
Với \(x\ge\frac{3}{2}\), ta có \(2x-3-x=-35\Rightarrow x=-32\) (l)
Với \(x< \frac{3}{2}\), ta có \(3-2x-x+1=-34\Rightarrow-3x=38\Rightarrow x=\frac{38}{3}\left(l\right)\)
d) Tương tự câu c.
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 ...
/x+1/>= 0
/x+3/>=0
=>/x+1/+/x+3/>=0
=>3x>=0
=> x>=0
=> /x+1/=x+1 ;/x+3/=x+3=> x+1+x+3=3x=>2x+4=3x =>x=4