x2 (x+2)+4*(x+2)=0
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26 tháng 10 2021
a: \(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
b: \(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
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
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)
17 tháng 11 2021
\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)
\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)
CM
25 tháng 8 2019
a) x = -1. b) x = 4 hoặc x = 5.
c) x = ± 2 . d) x = 1 hoặc x = 2.
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 |
CM
17 tháng 1 2018
Đáp án: A
Vì x2 + 4 > 0 ∀x ∈ R nên A = ∅.
(x2 - 4)(x2 + 1) = 0 ⇔ (x2 - 4) = 0 ⇔ x = ±2 nên B = {-2; 2}.
|x| < 2 ⇔ -2 < x < 2 nên D = (-2; 2).
=> A ⊂ B = C ⊂ D.
31 tháng 10 2021
\(a,\Leftrightarrow\left(2-x\right)\left(x^2+4\right)>0\Leftrightarrow2-x>0\Leftrightarrow x< 2\\ b,\Leftrightarrow x+3>0\Leftrightarrow x>-3\\ c,\Leftrightarrow\left[{}\begin{matrix}x< -3\\x>4\end{matrix}\right.\)
NT
1
KS
23 tháng 12 2021
(x - 2)(x2 + 2x + 4) + 2(x2 - 4) - 5(x - 2) = 0
(x - 2)(x + 2)2 + 2(x - 2)(x+2) - 5(x - 2) = 0
(x - 2)[(x+2)2 + 2(x+2) - 5]= 0
(x - 2)[(x + 2)2 + 2(x + 2) + 1 - 6] = 0
( x - 2)[(x + 2 + 1)2 - 6] = 0
(x - 2)[(x + 3)2 - 6] = 0
(x - 2)(x + 3 - \(\sqrt{6}\))(x + 3 + \(\sqrt{6}\)) = 0
TH1. x - 2 = 0 <=> x = 2
TH2. x + 3 - \(\sqrt{6}\) = 0 <=> x = \(\sqrt{6}-3\)
TH3. x + 3 + \(\sqrt{6}\) = 0 <=> x = \(-\sqrt{6}-3\)
S = {2; \(\sqrt{6}-3\); \(-\sqrt{6}-3\)}
x 2 ( x + 2 ) + 4 * ( x + 2 ) = 0
=> ( x + 2 ) ( x 2 + 4 ) = 0
=> \(\orbr{\begin{cases}x+2=0\\x^2+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x^2=-4\end{cases}\Leftrightarrow}x=-2}\)
\(x^2\left(x+2\right)+4\left(x+2\right)=0\)
\(\left(x+2\right)\left(x^2+4\right)=0\)
\(x+2=0\left(x^2+4>0\right)\)
\(x=-2\)
Vậy \(x=-2\)