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a,7x= -14
x= -14 /7
x= -2
b,6x-(-5)=17
6x=17+(-5)
6x=12
x=12/6
x=2
c,(x+2)(x-9)=o
suy ra(x+2)=0 hoặc (x-9)=0
nếu x+2=0 suy ra x= -2
nếu x-9=0 suy ra x=9
a) 7x=-14
x=-2
b) 6x-(-5) =17
6x . =12
x . =2
c) (x+2)(x-9)=0
=) * x+2=0=) x=-2
* x-9 =0=) x=9
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-5< 0\\x+2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-5>0\\x+2< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 5\\x>-2\end{matrix}\right.\\\left\{{}\begin{matrix}x>5\\x< -2\end{matrix}\right.\end{matrix}\right.\Rightarrow-2< x< 5\\ \Rightarrow x\in\left\{-1;0;1;2;3;4\right\}\\ b,\Rightarrow5< x^2< 14\\ \Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 7x=14
x = 14 : 7
x = 2
b ) 6x - (-5) = 17
6x = 17+(-5)
6x = 12
x = 12 :6
x = 2
c) (x+2) ( x-9) = 0
=> x +2 = 0 hoặc x-9=0
=> x= -2 => x=9
mình làm rùi , đi
![](https://rs.olm.vn/images/avt/0.png?1311)
`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)
\(x+\left(x+2\right)+\left(x+4\right)+...+\left(x+98\right)=0\)
\(x+x+2+x+4+...+x+98=0\)
\(50x+\left(98+2\right).\left[\left(98-2\right):2+1\right]:2=0\)
\(50x+100.49:2=0\)
\(50x+49.50=0\)
\(50x=0-49.50\)
\(50x=-2450\)
\(x=-2450:50\)
\(x=-49\)
b)
\(\left(x-5\right)+\left(x-4\right)+\left(x-3\right)+...+\left(x+11\right)+\left(x+12\right)=99\)
\(x+x+x+...+x-5-4-3-...+11+12=99\)
\(18x+6+7\text{+ 8 + 9 + 10 + 11 + 12 = 99}\)
\(18x+63=99\)
\(18x=99-63\)
\(18x=36\)
\(x=36:18\)
\(x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 7x = -14
=> x = -14/7
=> x = -2
b) 6x - (-5) = 17
=> 6x + 5 = 17
=> 6x = 17 +5
=> 6x = 22
=> x = 22/6 = 11/3
c) (x + 2) . (x-9) = 0
=> x + 2 = 0 hoặc x-9 = 0
=> x = -2 hoặc x=9
vậy x = -2; x= 9
a.7x=-14
\(\Leftrightarrow x=\dfrac{-14}{7}\)
\(\Leftrightarrow x=-2\)
Vậy x=-2.
b.6x-(-5)=17
\(\Leftrightarrow6x=17-5\)
\(\Leftrightarrow6x=12\)
\(\Leftrightarrow x=2\)
Vậy x=2
c.(x+2).(x-9)=0
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=9\end{matrix}\right.\)
Vậy x=-2 hoặc x=9.
![](https://rs.olm.vn/images/avt/0.png?1311)
b) A=\(\frac{5x-2}{x-3}=\frac{5x-15+13}{x-3}=\frac{5x-15}{x-3}+\frac{13}{x-3}=\frac{5\left(x-3\right)}{x-3}+\frac{13}{x-3}=5+\frac{13}{x-3}\)
Để A thuộc Z thì \(5+\frac{13}{x-3}\in Z\)
=>13 chia hết cho x-3
=>x-3 \(\in\)Ư(13)={-1;1;-13;13}
x-3=-1 x-3=1 x-3 =-13 x-3=13
x =-1+3 x =1+3 x =-13+3 x =13+3
x=2 x =4 x=-10 x=16
Vậy x=2;4;-10;16 thì A thuộc Z
c)B=\(\frac{6x-1}{3x+2}=\frac{6x+4-5}{3x+2}=\frac{6x+4}{3x+2}+\frac{-5}{3x+2}=\frac{2\left(3x+2\right)}{3x+2}+\frac{-5}{3x+2}=2+\frac{-5}{3x+2}\)
Để B thuộc Z thì \(2+\frac{-5}{3x+2}\in Z\)
=>-5 chia hết cho 3x+2
=>3x+2\(\in\)Ư(-5)={-1;1;-5;5}
3x+2=-1 3x+2=1 3x+2=-5 3x+2=5
3x =-3 3x =-1 3x =-7 3x =3
x =-1 x =-1/3 x =-7/3 x =1
Vậy x=-1;-1/3;-7/3;1 thì B thuộc Z
d) C=\(\frac{10x}{5x-2}=\frac{10x-4+4}{5x-2}=\frac{10-4}{5x-2}+\frac{4}{5x-2}=\frac{2\left(5x-2\right)}{5x-2}+\frac{4}{5x-2}=2+\frac{4}{5x-2}\)
Để C thuộc Z thì \(2+\frac{4}{5x-2}\in Z\)
=> 4 chia hết cho 5x-2
=>5x-2\(\in\)Ư(4)={-1;1;-2;2;-4;4}
5x-2=-1 5x-2=1 5x-2=2 5x-2=-2 5x-2=4 5x-2=-4
bạn tự giải tìm x như các bài trên nhé
d) bạn ghi đề mjk ko hjeu
e)E=\(\frac{4x+5}{x-3}=\frac{4x-12+17}{x-3}=\frac{4x-12}{x-3}+\frac{17}{x-3}=\frac{4\left(x-3\right)}{x-3}+\frac{17}{x-3}=4+\frac{17}{x-3}\)
Để E thuộc Z thì\(4+\frac{17}{x-3}\in Z\)
=>17 chia hết cho x-3
=>x-3 \(\in\)Ư(17)={1;-1;17;-17}
x-3=1 x-3=-1 x-3=17 x-3=-17
bạn tự giải tìm x nhé
điều cuối cùng cho mjk ****
a) 7x = -14
=> x = -2
b) 6x - (-5) = 17
=> 6x = (-5) + 17
=> 6x = 12
=> x = 12
c ) ( x + 2 ) . ( x - 9 ) = 0
=> x + 2 = 0 hoặc x - 9 = 0
=> x = -2 ; x = 9
a, x = -2
b, 6x = 12
x = 2
c, \(\orbr{\begin{cases}x=-2\\x=9\end{cases}}\)