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a, |x+1| +2x \(=\)-2
\(\Leftrightarrow\) |x+1|\(=\)-2-2x (*)
TH1: Nếu x+1\(\ge0\)\(\Leftrightarrow x\ge-1\)thì \(\left|x+1\right|=x+1\)
Thay vào (*) ta có:
\(x+1=-2-2x\)
\(\Leftrightarrow x+2x=-2-1\)
\(\Leftrightarrow3x=-3\)
\(\Leftrightarrow x=-1\)(TMĐK)
TH2: Nếu \(x+1< 0\Leftrightarrow x< -1\Rightarrow\left|x+1\right|=-\left(x+1\right)\)
Thay vào (*), ta có:
\(-x-1=\)\(-2-2x\)
\(\Leftrightarrow-x+2x=-2+1\)
\(\Leftrightarrow x=-1\)(kTMĐK)
Vậy S\(=\){-1}
b, \(x^2-6x+9=1\)
\(\Leftrightarrow x^2-2.3.x+3^2\)
\(\Leftrightarrow\left(x-3\right)^2=1\)
\(\Leftrightarrow\left(x-3\right)^2-1^2=0\)
\(\Leftrightarrow\left(x-3-1\right)\left(x-3+1\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
Vậy S\(=\){4;2}
a) |x + 1| + 2x = 2
\(\Rightarrow\) |3x + 1| = 2
\(\Rightarrow\) |3x| = 1
\(\Rightarrow\) |x| = 1 : 3 = \(\dfrac{1}{3}\)
\(\Rightarrow\) x = \(\dfrac{1}{3}\) hoặc \(-\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow\left|x+1\right|=-2x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =-1\\\left(-2x-2\right)^2-\left(x+1\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =-1\\3\left(x+1\right)^2=0\end{matrix}\right.\Leftrightarrow x=-1\)
b: \(\Leftrightarrow\left(x-3\right)^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow\left|x+1\right|=-2x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(-2x-2\right)^2-\left(x+1\right)^2=0\\-2x-2>=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3\left(x+1\right)^2=0\\x< =-1\end{matrix}\right.\Leftrightarrow x=-1\)
b: \(\Leftrightarrow\left(x-3\right)^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\Leftrightarrow x\in\left\{4;2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) |6x-3|=15
=> 6x-3 = 15 hoặc 6x-3 = -15
=> x=3 hoặc x=-2
2) x+xy+y=9
<=> x(y+1) +y=9
<=> x(y+1) +(y+1) = 10
<=> (x+1)(y+1)=10= -2.-5 =-5.-2 = -1.-10 = -10.1 = 2.5=5.2=1.10=10.1
Từ đây có thể tìm đc x và y nhé!
CHÚC BẠN HỌC TỐT!
Bài 1:
\(\left|6x-3\right|=15\Rightarrow\orbr{\begin{cases}6x-3=-15\\6x-3=15\end{cases}\Rightarrow}\orbr{\begin{cases}6x=-12\\6x=18\end{cases}\Rightarrow}\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
Bài 2:
\(x+xy+y=9\Leftrightarrow x+xy+y+1=10\Leftrightarrow x\left(1+y\right)+\left(y+1\right)=10\)
\(\Leftrightarrow\left(x+1\right)\left(y+1\right)=10\)
Ta có bảng sau:
x+1 | -10 | -5 | -2 | -1 | 1 | 2 | 5 | 10 |
y+1 | -1 | -2 | -5 | -10 | 10 | 5 | 2 | 1 |
x | -11 | -6 | -3 | -2 | 0 | 1 | 4 | 9 |
y | -2 | -3 | -6 | -11 | 9 | 4 | 1 | 0 |
Vậy có 8 cặp số nguyên thỏa mãn là ........
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
a) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(x=\dfrac{-13}{12}\)
b) \(2.\left(x-\dfrac{1}{3}\right)=\left(\dfrac{1}{3}\right)^2+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{1}{9}+\dfrac{5}{9}\)
\(2.\left(x-\dfrac{1}{3}\right)=\dfrac{2}{3}\)
\(x-\dfrac{1}{3}=\dfrac{2}{3}:2\)
\(x-\dfrac{1}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{1}{3}\)
\(x=\dfrac{2}{3}\)
c) \(\left|2x-\dfrac{3}{4}\right|-\dfrac{3}{8}=\dfrac{1}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{8}+\dfrac{3}{8}\)
\(\left|2x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{1}{2}\\2x-\dfrac{3}{4}=\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{8}\\x=\dfrac{1}{8}\end{matrix}\right.\)
d) \(\dfrac{2}{3}x+\dfrac{1}{6}x=3\dfrac{5}{8}\)
\(x.\left(\dfrac{2}{3}+\dfrac{1}{6}\right)=\dfrac{29}{8}\)
\(x.\dfrac{5}{6}=\dfrac{29}{8}\)
\(x=\dfrac{29}{8}:\dfrac{5}{6}\)
\(x=\dfrac{87}{20}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Quy đồng với mẫu số chung là 30
\(\frac{15x}{30}+\frac{30x+10}{30}+\frac{-36x-54}{30}=\frac{5}{30}\)
=> 9x - 44 = 5
=> x = \(\frac{49}{9}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left|x+1\right|+2x=-2\)
\(\Leftrightarrow\left|x+1\right|+2\left(x+1\right)=0\)
\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
b) \(x^2-6x+9=1\)
\(\Leftrightarrow x^2-6x+8=0\)
\(\Leftrightarrow x^2-2x-4x+8=0\)
\(\Leftrightarrow x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\x-4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=4\end{array}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có: 3 x = 3 2 nên x = 2
b, Ta có: 5 x = 5 3 nên x = 3
c, Ta có: 3 x + 1 = 3 2 nên x +1 = 2, do đó x = 1
d, Ta có: 6 x - 1 = 6 2 nên x - 1 = 2, đo đó x = 3
e) Ta có: 3 2 x + 1 = 3 3 nên 2x +1 = 3, do đó x = 1
f) Ta có: x 50 = x nên x 50 - x = 0 , do đó x x 49 - 1 = 0 = 0
Vì thế x = 0 hoặc x = 1