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a)\(\left|3-x\right|-2=x+2\)
\(TH1:3-x-2=x+2\)
\(\Leftrightarrow1-x=x+2\)
\(\Leftrightarrow1-x-x-2=0\)
\(\Leftrightarrow-1-2x=0\)
\(\Leftrightarrow x=-\frac{1}{2}\)
\(TH2:-\left(3-x\right)-2=x+2\)
\(\Leftrightarrow x-3-2=x+2\)
\(\Leftrightarrow x-5=x+2\)
\(\Leftrightarrow xkoTM\)
b)\(\left|2x+1\right|+x=2x-5\)
\(TH1:2x+1+x=2x-5\)
\(\Leftrightarrow3x+1-2x+5=0\)
\(\Leftrightarrow x+6=0\)
\(\Leftrightarrow x=-6\)
\(TH2:-2x-1+x=2x-5\)
\(\Leftrightarrow-x-1-2x+5=0\)
\(\Leftrightarrow4-3x=0\)
\(\Leftrightarrow x=\frac{4}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
| 3-x | -2=x+2
=) | 3-x | =x+2+2=x+4
=) 3-x = x+4 hoặc = -(x+4)=-x-4
TH1 : 3-x=x+4
=) 3-4=x+x
=) -1 = 2x
=) \(x=\frac{-1}{2}\)
TH2 : 3-x=-x-4
=) 3+4=-x+x
=) 7=0 (Loại vì \(7\ne0\))
=) \(x\in\)rỗng
Vậy \(x=\frac{-1}{2}\)
b) | 2x+1 | + x = 2x-5
=) | 2x+1 | = 2x-5-x
=) 2x+1 = 2x-5-x hoặc = -(2x-5-x) = -2x+5+x
TH1 : 2x+1 = 2x-5-x
=) 2x-2x=-5-x-1
=) 0 = -5-1-x = -6-x
=) x = (-6)-0=-6
TH2 : 2x+1 = -2x+5+x
=) 2x+2x = 5+x-1
=) 4x = 4+x
=) 4x-x = 4
=) 3x = 4
=) \(x=\frac{4}{3}\)
Vậy \(x=\left\{-6,\frac{4}{3}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1: (x-1)+|(y-1)2|=0. vì (y-1)2 >=0 với mọ y. nên phương trình đúng khi x-1=0 và y-1=0 nên x=1 và y=1
![](https://rs.olm.vn/images/avt/0.png?1311)
/x-3/ -(-3)=4
=>/x-3/ +3=4
=>/x-3/=1
=>x-3=1 hoặc x-3=-1
=>x=4 hoặc x=2
Vậy x=4 hoặc x=2
Các câu khác làm tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2 :
a, \(\left|x-\frac{5}{3}\right|< \frac{1}{3}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{5}{3}< \frac{1}{3}\\x-\frac{5}{3}< -\frac{1}{3}\end{cases}\Leftrightarrow\orbr{\begin{cases}x< 2\\x< \frac{4}{3}\end{cases}}}\)
b, \(\frac{2}{5}< \left|x-\frac{7}{5}\right|< \frac{3}{5}\)
\(\orbr{\begin{cases}\frac{2}{5}< x-\frac{7}{5}< \frac{3}{5}\\\frac{2}{5}< -x+\frac{7}{5}< \frac{3}{5}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{9}{5}< x< 2\\1>x>\frac{4}{5}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1)|5-2x|=|x+4|\)
\(\Leftrightarrow\orbr{\begin{cases}5-2x=x+4\\5-2x=-x-4\end{cases}\Leftrightarrow\orbr{\begin{cases}-2x-x=4-5\\-2x+x=-4-5\end{cases}\Leftrightarrow}\orbr{\begin{cases}-3x=-1\\-x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=9\end{cases}}}\)
Vậy \(x=\frac{1}{3};x=9\)
\(2)|x-1|=|2x+5|\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=2x+5\\x-1=-2x-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x-2x=5+1\\x+2x=-5+1\end{cases}\Leftrightarrow}\orbr{\begin{cases}-x=4\\3x=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-4\\x=-\frac{4}{3}\end{cases}}}\)
Vậy \(x=-4;x=-\frac{4}{3}\)
\(3)|x+1|+|x+2|+|x+3|=0\left(1\right)\)
Ta có: \(|x+1|\ge0\forall x;|x+2|\ge0\forall x;|x+3|\ge0\forall x\)
\(\Leftrightarrow|x+1|+|x+2|+|x+3|\ge0\forall x\)
\(\left(1\right)\Leftrightarrow|x+1|+|x+2|+|x+3|=0\)
\(\Leftrightarrow\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=0\)
\(\Leftrightarrow x+1+x+2+x+3=0\)
\(\Leftrightarrow\left(x+x+x\right)+\left(1+2+3\right)=0\)
\(\Leftrightarrow3x+6=0\)
\(\Leftrightarrow3x=-6\)
\(\Leftrightarrow x=-6:3\)
\(\Leftrightarrow x=-2\)
Vậy x=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 2:
a: Để A là số nguyên thì \(x-1\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{2;0;4;-2\right\}\)
b: Để -4/2x-1 là số nguyên thì \(2x-1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{1;0;\dfrac{3}{2};-\dfrac{1}{2};\dfrac{5}{2};-\dfrac{3}{2}\right\}\)
c: Để 3x+7/x-1là số nguyên thì \(3x-3+10⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)
hay \(x\in\left\{2;0;3;-1;6;-4;11;-9\right\}\)
d: Để 4x-1/x-3 là số nguyên thì \(4x-12+11⋮x-3\)
\(\Leftrightarrow x-3\in\left\{1;-1;11;-11\right\}\)
hay \(x\in\left\{4;2;14;-8\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) 2. I2x-3l = 1/2
|2x-3| =1/2:2
|2x-3| =1/4
=>2x-3 =1/4 hoặc 2x-3 =-1/4
2x =1/4+3 2x =-1/4+3
2x =13/4 2x =11/4
x =13/4:2 x =11/4:2
x =13/8 x =11/8
vậy x=13/8 hoặc 11/8
tich dung cho minh nhe
Ta có : |3 - x| - 2 = x + 2
<=> |3 - x| = x + 4
Ta có : |3 - x| \(\ge0\)
Nên x \(\le3\)
3 - x = x + 4
=> 3 - 4 = x + x
=> 2x = -1
=> x = -1/2(Chọn)