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`a)|2x-15|=13`
`**2x-15=13`
`<=>2x=28`
`<=>x=14.`
`**2x-15=-13`
`<=>2x=-2`
`<=>x=-1.`
`b)|7x+3|=66`
`**7x+3=66`
`<=>7x=63`
`<=>x9`
`**7x+3=-66`
`<=>7x=-69`
`<=>x=-69/7`
`c)|5x-2|=0`
`<=>5x-2=0`
`<=>5x=2`
`<=>x=2/5`
\(a,\Leftrightarrow\left[{}\begin{matrix}2x-5=13\\2x-5=-13\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)
Vậy ...
\(b,\Leftrightarrow\left[{}\begin{matrix}7x+3=66\\7x+3=-66\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-\dfrac{69}{7}\end{matrix}\right.\)
Vậy ...
\(c,\Leftrightarrow5x-2=0\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
a) I2x-5I=13
<=> 2x-5=13 hoặc 2x-5=-13
<=> 2x=18 hoặc 2x=-8
<=> x=9 hoặc x=-4
b) I7x+3I=66
<=> 7x+3=66 hoặc 7x+3=-66
<=> 7x=63 hoặc 7x=-69
<=> x=9 hoặc x=\(\frac{-69}{7}\)
a) \(\left|2x-5\right|=13\)
\(\Leftrightarrow2x-5=\pm13\)
\(\Leftrightarrow\orbr{\begin{cases}2x-5=13\\2x-5=-13\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=18\\2x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=9\\2x=-4\end{cases}}}\)
Vậy \(x\in\left\{9;-4\right\}\)
b) \(\left|7x+3\right|=66\)
\(\Leftrightarrow7x+3=\pm66\)
\(\Leftrightarrow\orbr{\begin{cases}7x+3=66\\7x+3=-66\end{cases}\Leftrightarrow\orbr{\begin{cases}7x=63\\7x=-69\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=9\\x=\frac{-69}{7}\end{cases}}}\)
Vậy \(x\in\left\{9;\frac{-69}{7}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left|2x-5\right|=13\)
\(\Rightarrow\orbr{\begin{cases}2x-5=13\\2x-5=-13\end{cases}}\Rightarrow\orbr{\begin{cases}2x=18\\2x=-8\end{cases}}\Rightarrow\orbr{\begin{cases}x=9\\x=-4\end{cases}}\)
vậy \(\orbr{\begin{cases}x=9\\x=-4\end{cases}}\)
b) \(\left|7x+3\right|=66\)
\(\Rightarrow\orbr{\begin{cases}7x+3=66\\7x+3=-66\end{cases}}\Rightarrow\orbr{\begin{cases}7x=63\\7x=-63\end{cases}}\Rightarrow\orbr{\begin{cases}x=9\\x=-9\end{cases}}\)
vậy \(\orbr{\begin{cases}x=9\\x=-9\end{cases}}\)
c) \(\left|5x-2\right|\le13\)
\(\Leftrightarrow\hept{\begin{cases}5x-2\le3\\5x-2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}5x\le5\\5x>2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le1\\x>\frac{2}{5}\end{cases}}\)
vậy \(\frac{2}{5}< x\le1\)
a, => 2x-5=13 hoặc 2x-5=-13
=> x=9 hoặc x=-4
b, => 7x+3=66 hoặc 7x+3=-66
=> x=9 hoặc x=-69/7
c, => -13 < = 5x-2 < = 13
=> -13+2 < = 5x < = 13+2
=> -11 < = 5x < = 15
=> -11/5 < = x < =3
Tk mk nha
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm x thuoc z:
1) \(26-\left|x+9\right|=-13\)
\(\Leftrightarrow\left|x+9\right|=26-\left(-13\right)\)
\(\Leftrightarrow\left|x+9\right|=39\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=39\\x+9=-39\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=39-9=30\\x=-39-9=-48\end{matrix}\right.\)
Vậy: \(x\in\left\{30;-48\right\}\)
2) \(\left|x+7\right|-13=25\)
\(\Leftrightarrow\left|x+7\right|=25+13=38\)
\(\Leftrightarrow x+7\in\left\{38;-38\right\}\)
\(\Leftrightarrow x\in\left\{31;-45\right\}\)
Vậy:.................
tim x biet
\(1)123-3.\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=123-23\)
\(\Leftrightarrow3\left(x+4\right)=100\)
\(\Leftrightarrow x+4=\frac{100}{3}\)
\(\Leftrightarrow x=\frac{100}{3}-4=\frac{100-12}{3}=\frac{88}{3}\)
Vậy:................
2) Tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
a) Ta có: x-3 là ước của 13
\(\Leftrightarrow x-3\inƯ\left(13\right)\)
\(\Leftrightarrow x-3\in\left\{1;-1;13;-13\right\}\)
hay \(x\in\left\{4;2;16;-10\right\}\)(thỏa mãn)
Vậy: \(x\in\left\{4;2;16;-10\right\}\)
b) Ta có: \(x^2-7\) là ước của \(x^2+2\)
\(\Leftrightarrow x^2+2⋮x^2-7\)
\(\Leftrightarrow x^2-7+9⋮x^2-7\)
mà \(x^2-7⋮x^2-7\)
nên \(9⋮x^2-7\)
\(\Leftrightarrow x^2-7\inƯ\left(9\right)\)
\(\Leftrightarrow x^2-7\in\left\{1;-1;3;-3;9;-9\right\}\)
mà \(x^2-7\ge-7\forall x\)
nên \(x^2-7\in\left\{1;-1;3;-3;9\right\}\)
\(\Leftrightarrow x^2\in\left\{8;6;10;4;16\right\}\)
\(\Leftrightarrow x\in\left\{2\sqrt{2};-2\sqrt{2};-\sqrt{6};\sqrt{6};\sqrt{10};-\sqrt{10};2;-2;4;-4\right\}\)
mà \(x\in Z\)
nên \(x\in\left\{2;-2;4;-4\right\}\)
Vậy: \(x\in\left\{2;-2;4;-4\right\}\)
Câu 2:
a) Ta có: \(2\left(x-3\right)-3\left(x-5\right)=4\left(3-x\right)-18\)
\(\Leftrightarrow2x-6-3x+15=12-4x-18\)
\(\Leftrightarrow-x+9+4x+6=0\)
\(\Leftrightarrow3x+15=0\)
\(\Leftrightarrow3x=-15\)
hay x=-5
Vậy: x=-5
a) => 2x-5 = 13
2x-5 = -13
=> 2x = 13+5 = 18
2x = -13+5 = -8
=> x= -18:2 = -9
x= -8:2 = -4
b) 7x+3 = 66
=> 7x+3 = 66
7x+3 = -66
7x = 66-3 = 33
7x = -66 - 3 = -69
=> x= 33:7 = ????
x= -69:7 = ???? sai đề hay sao ý
c) => 5x-2 = 13
=> 5x -2 = -13
5x = 13+2 = 15
5x = -13+2 = -11
=> x= 15:5 = 3
x= -11: 5 = -2,2
duyệt đi
ý câu b) sai chỗ này: 7x = 66-3 = 63
7x = -66 - 3 = -69
=> x= 63:7 = 9
x= -69 :7 = -9,857.........
bài lạ duyệt đi