h*) (x + 3)(1 – x) > 0 i*) (x2 – 1)(x2 – 4) < 0 k*) (x2 – 20)(x2 – 30) < 0
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2 tháng 3 2022
a, \(\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
| x-2 | 1 | -1 | 3 | -3 |
| x | 3 | 1 | 5 | -1 |
b, \(3\left(x-2\right)+13⋮x-2\Rightarrow x-2\inƯ\left(13\right)=\left\{\pm1;\pm13\right\}\)
| x-2 | 1 | -1 | 13 | -13 |
| x | 3 | 1 | 15 | -11 |
c, \(x\left(x+7\right)+2⋮x+7\Rightarrow x+7\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
| x+7 | 1 | -1 | 2 | -2 |
| x | -6 | -8 | -5 | -9 |
9 tháng 3 2022
\(\left(x+3\right)\left(1-x\right)>0.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3>0.\\1-x>0.\end{matrix}\right.\\\left\{{}\begin{matrix}x+3< 0.\\1-x< 0.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-3.\\x< 1.\end{matrix}\right.\\\left\{{}\begin{matrix}x< -3.\\x>1.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow-3< x< 1.\)
\(\left(x^2-1\right)\left(x^2-4\right)< 0.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-1< 0.\\x^2-4>0.\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-1>0.\\x^2-4< 0.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2< 1.\\x^2>4.\end{matrix}\right.\\\left\{{}\begin{matrix}x^2>1.\\x^2< 4.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 1.\\x>-1.\end{matrix}\right.\\\left[{}\begin{matrix}x>2.\\x< -2.\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1.\\x< -1.\end{matrix}\right.\\\left[{}\begin{matrix}x< 2.\\x>-2.\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-1< x< 1.\\\left[{}\begin{matrix}x>2.\\x< -2.\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1.\\x< -1.\end{matrix}\right.\\-2< x< 2.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>2.\\x< -2.\\-2< x< -1.\\1< x< 2.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x< -2.\\x>2.\end{matrix}\right.\)
29 tháng 10 2021
a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
CM
25 tháng 8 2019
a) x = -1. b) x = 4 hoặc x = 5.
c) x = ± 2 . d) x = 1 hoặc x = 2.
22 tháng 3 2020
Bài 1)1)\(x^2+5x+6=x^2+3x+2x+6\)=0
=x(x+3)+2(x+3)=(x+2)(x+3)=0
Dễ rồi
2)\(x^2-x-6=0=x^2-3x+2x-6=0\)
=x(x-3)+2(x-3)=0
=(x+2)(x-3)=0
Dễ rồi
3)Phương trình tương đương:\(\left(x^2+1\right)\left(x+2\right)^2=0\)
Vì \(x^2+1>0\)
=>\(\left(x+2\right)^2=0\)
Dễ rồi
4)Phương trình tương đương\(x^2\left(x+1\right)+\left(x+1\right)\)=0
=> \(\left(x^2+1\right)\left(x+1\right)=0Vì\) \(x^2+1>0\)
=>x+1=0
=>..................
5)\(x^2-7x+6=x^2-6x-x+6\) =0
=x(x-6)-(x-6)=0
=(x-1)(x-6)=0
=>.....
6)\(2x^2-3x-5=2x^2+2x-5x-5\)=0
=2x(x+1)-5(x+1)=0
=(2x-5)(x+1)=0
7)\(x^2-3x+4x-12\)=x(x-3)+4(x-3)=(x+4)(x-3)=0
Dễ rồi
Nghỉ đã hôm sau làm mệt
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 3 2020
Bài 1
a/ \(x\left(x^2+1\right)+2\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+1\right)=0\Rightarrow x=-2\)
b/
\(\Leftrightarrow x^3-6x^2+9x+5x^2-30x+45=0\)
\(\Leftrightarrow x\left(x-3\right)^2+5\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-3\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 3 2020
1.
c/ \(\Leftrightarrow x^3+2x^2+2x+x^2+2x+2=0\)
\(\Leftrightarrow x\left(x^2+2x+2\right)+x^2+2x+2=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2+2x+2=0\left(vn\right)\end{matrix}\right.\)
d/
\(\Leftrightarrow x^4+x^3-2x^2-x^3-x^2+2x+4x^2+4x-8=0\)
\(\Leftrightarrow x^2\left(x^2+x-2\right)-x\left(x^2+x-2\right)+4\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x^2-x+4\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+4=0\left(vn\right)\\x^2+x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
A
2
13 tháng 2 2022
e, \(\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2=0\Leftrightarrow\left(x-2\right)\left(x+2+x-2\right)=0\Leftrightarrow x=0;x=2\)
f, \(\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(x-1\right)^2=0\Leftrightarrow x=1;x=-1\)
g, \(x^2\left(x-3\right)+4\left(3-x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\Leftrightarrow x=2;x=-2;x=3\)
h, \(\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\Leftrightarrow\left(x-4\right)\left(3x+2\right)=0\Leftrightarrow x=4;x=-\dfrac{2}{3}\)
22 tháng 5 2022
a: (x-3)(x-2)<0
=>x-2>0 và x-3<0
=>2<x<3
b: \(\left(x+3\right)\left(x+4\right)\left(x^2+2\right)\ge0\)
=>(x+3)(x+4)>=0
=>x+3>=0 hoặc x+4<=0
=>x>=-3 hoặc x<=-4
c: \(\dfrac{x-1}{x-2}\ge0\)
=>x-2>0 hoặc x-1<=0
=>x>2 hoặc x<=1
d: \(\dfrac{x+3}{2-x}>=0\)
=>\(\dfrac{x+3}{x-2}< =0\)
=>x+3>=0 và x-2<0
=>-3<=x<2