tìm x thuộc Z
I x + 1I - I x - 3 I = 0
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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.
\(\left|3x-1\right|=\left|2x+5\right|\)
\(\Rightarrow\orbr{\begin{cases}3x-1=2x+5\\3x-1+2x+5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x-2x=5+1\\5x+4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=6\\x=-\frac{4}{5}\end{cases}}\)
Ta có: \(\hept{\begin{cases}\left(x-1\right)^2\ge0\\\left|3y-1\right|\ge0\\\left|z+2\right|\ge0\end{cases}}\Rightarrow\left(x-1\right)^2+\left|3y-1\right|+\left|z+2\right|\ge0\)
Dấu "="\(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\\left|3y-1\right|=0\\\left|z+2\right|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\3y-1=0\\x+2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\y=\frac{1}{3}\\z=-2\end{cases}}\)
Vậy x = 1, \(y=\frac{1}{3}\),z = -2
a, !x-1!=0
\(\Rightarrow x-1=0\)
\(\Rightarrow x=0+1\)
\(\Rightarrow x=1\)
Vậy x=1
b,-11.!3x-1!=-22
\(\Rightarrow!3x-1!=-22:\left(-11\right)\)
\(\Rightarrow!3x-1!=2\)
\(\Rightarrow\orbr{\begin{cases}3x-1=2\\3x-1=-2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=2+1\\3x=-2+1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=3\\3x=-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3:3\\x=-1:3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-1}{3}\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=1\\x=-\frac{1}{3}\end{cases}}\)
c, !x!<2
\(\Rightarrow x\in\left\{-1;0;1\right\}\)
Vậy \(x\in\left\{-1;0;1\right\}\)
Hok tốt nhé!!
A )
| x - 1 | = 0
x = 0 + 1
x = 1
B )
-11 . | 3x - 1 | = -22
| 3x - 1 | = -22 : ( -11 )
| 3x - 1 | = 2
3x = 2 + 1
3x = 3
x = 3 : 3
x = 1
Ta có:\(\hept{\begin{cases}\left|x-2\right|\ge0\\\left|1,5-y\right|\ge0\\\left|3-z\right|\ge0\end{cases}\Rightarrow\left|x-2\right|+\left|1,5-y\right|+\left|3-z\right|\ge0}\)
Để \(\left|x-2\right|+\left|1,5-y\right|+\left|3-z\right|=0\) thì \(\hept{\begin{cases}\left|x-2\right|=0\\\left|1,5-y\right|=0\\\left|3-z\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=1,5\\z=3\end{cases}}}\)
Vì |x-2| ; |1,5-y| ; |3-z| đều >= 0 nên VT >= 0
=> VT= 0 <=> x-2=0;1,5-y=0;3-z=0
<=> x=2;y=1,5;z=3
a) |x+1|+|x+2+|x+3|=4x
<=> x+1+x+2+x+3=4x
<=> 3x+6=4x
<=> 6=4x-3x
<=> x=6
\(\left|x+\frac{1}{2}\right|+\left|y-\frac{3}{4}\right|+\left|z+1\right|=0\)
\(\Rightarrow\hept{\begin{cases}\left|x+\frac{1}{2}\right|=0\\\left|y-\frac{3}{4}\right|=0\\\left|z+1\right|=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=0-\frac{1}{2}\\y=0+\frac{3}{4}\\z=0-1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{-1}{2}\\y=\frac{3}{4}\\z=-1\end{cases}}\)