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Bài 1:
Để E nguyên thì \(x+5⋮x-2\)
\(\Leftrightarrow x-2\in\left\{1;-1;7;-7\right\}\)
hay \(x\in\left\{3;1;9;-5\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2^x=8\)
⇔ \(2^x=2^3\)
⇒ \(x=3\)
b) \(3^x=27\)
⇔ \(3^x=3^3\)
⇒ \(x=3\)
c) \(\left(-\dfrac{1}{2}\right)x=\left(-\dfrac{1}{2}\right)^4\)
⇔ \(x=\left(-\dfrac{1}{2}\right)^4\div\left(-\dfrac{1}{2}\right)\)
⇔ \(x=\left(-\dfrac{1}{2}\right)^3\)
d) \(x\div\left(-\dfrac{3}{4}\right)=\left(-\dfrac{3}{4}\right)^2\)
⇔ \(x=\left(-\dfrac{3}{4}\right)^2\cdot\left(-\dfrac{3}{4}\right)\)
⇔ \(x=\left(-\dfrac{3}{4}\right)^3=-\dfrac{27}{64}\)
d) \(\left(x+1\right)^3=-125\)
⇔ \(\left(x+1\right)^3=\left(-5\right)^3\)
⇔ \(x+1=-5\)
⇔ \(x=-5-1=-6\)
2:
a: (x-1,2)^2=4
=>x-1,2=2 hoặc x-1,2=-2
=>x=3,2(loại) hoặc x=-0,8(loại)
b: (x-1,5)^2=9
=>x-1,5=3 hoặc x-1,5=-3
=>x=-1,5(loại) hoặc x=4,5(loại)
c: (x-2)^3=64
=>(x-2)^3=4^3
=>x-2=4
=>x=6(nhận)
![](https://rs.olm.vn/images/avt/0.png?1311)
a,x.y=3=1x3=3x1=-1x(-3)=-3x(-1).
Vậy (x,y)=(1,3)=(3,1)=(-1,-3)=(-3,-1)
b,x.(y+1)=5=1x5=5x1=-1x(-5)=-5x(-1)
=>
x | 1 | 5 | -1 | -5 |
y+1 | 5 | 1 | -5 | -1 |
y | 4 | 0 | -6 | -2 |
Vậy (x,y)=(1,4)=(5,0)=(-1,-6)=(-1,-2).
c,(x-2)(y+3)=7=1x7=7x1=-1x(-7)=-7(-1)
=>
x-2 | 1 | 7 | -1 | -7 |
y+3 | 7 | 1 | -7 | -1 |
x | 3 | 9 | 1 | -5 |
y | 4 | -2 | -10 | -4 |
Vậy (x,y)=(3,4)=(9,-2)=(1,-10)=(-5,-4).
![](https://rs.olm.vn/images/avt/0.png?1311)
1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/ Ta có \(\frac{1}{3}< \frac{9}{x}< \frac{1}{2}\)
\(\Rightarrow\frac{9}{27}< \frac{9}{x}< \frac{9}{18}\)
\(\Rightarrow27>x>18\)
Vì \(x\in Z\Rightarrow x\in\left\{19,20,...,26\right\}\)
Vậy....
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b) \(\left(x^2+5\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5=0\\x^2-25=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2=-5\\x^2=25\end{matrix}\right.\) \(\Leftrightarrow x^2=25\) \(\Leftrightarrow x=\pm5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
b) ĐKXĐ: \(x\ne3\)
Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-3}\)
\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=100\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\left(nhận\right)\\x=-7\left(nhận\right)\end{matrix}\right.\)
Vậy: \(x\in\left\{13;-7\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 8 - |x + 2| = 5
-|x + 2| = 5 - 8
-|x + 2| = -3
|x + 2| = 3
x + 2 = 3; -3
x + 2 = 3 hoặc x + 2 = -3
x = 3 - 2 x = -3 - 2
x = 1 x = -5
=> x = 1 hoặc x = -5
Có : VT = |x+2| + |3-x| >= |x+2+3-x| = 5 = VP
Dấu "=" xảy ra <=> (x+2).(3-x) >= 0
<=> -2 < = x < = 3
Vậy -2 < = x < = 3
Tk mk nha