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NT
6
MH
1 tháng 11 2021
1.
a) \(2x^4-4x^3+2x^2\)
\(=2x^2\left(x^2-2x+1\right)\)
\(=2x^2\left(x-1\right)^2\)
b) \(2x^2-2xy+5x-5y\)
\(=\left(2x^2-2xy\right)+\left(5x-5y\right)\)
\(=2x\left(x-y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\cdot\left(2x+5\right)\)
KJ
1 tháng 11 2021
2 .
a,
\(4x\left(x-3\right)-x+3=0\)
⇒\(4x\left(x-3\right)-\left(x-3\right)=0\)
⇒\(\left(x-3\right)\left(4x-1\right)=0\)
⇒\(\left[{}\begin{matrix}x-3=0\\4x-1=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=3\\4x=1\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\end{matrix}\right.\)
vậy \(x\in\left\{3;\dfrac{1}{4}\right\}\)
b,
\(\)\(\left(2x-3\right)^2-\left(x+1\right)^2=0\)
⇒\(\left(2x-3-x-1\right)\left(2x-3+x+1\right)\) = 0
⇒\(\left(x-4\right)\left(3x-2\right)=0\)
⇔\(\left[{}\begin{matrix}x-4=0\\3x-2=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=4\\3x=2\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x=4\\x=\dfrac{2}{3}\end{matrix}\right.\)
vậy \(x\in\left\{4;\dfrac{2}{3}\right\}\)
Y
26 tháng 1 2023
\(2x^2+5x+3=0\)
\(\Leftrightarrow2x^2+2x+3x+3=0\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=-1\end{matrix}\right.\)
Vậy \(S=\left\{-1;-\dfrac{3}{2}\right\}\)
\(\left(x-\sqrt{2}\right)-3\left(x^2-2\right)=0\)
\(\Leftrightarrow x-\sqrt{2}-3x^2+6=0\)
\(\Leftrightarrow-3x^2+x+6-\sqrt{2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{1}{6}-\dfrac{\sqrt{73-3\sqrt{32}}}{6}\\x_2=\dfrac{\sqrt{73-3\sqrt{32}}}{6}+\dfrac{1}{6}\end{matrix}\right.\)
23 tháng 10 2021
\(a,\Leftrightarrow\left(2x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow x^3-27-x^3+4x=1\\ \Leftrightarrow4x=28\Leftrightarrow x=7\\ c,\Leftrightarrow4x^2-4x-8=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ d,\Leftrightarrow2x^2+6x+x+3=0\\ \Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
19 tháng 2 2020
\(a,\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
\(\left(x-1\right)\left(5x+3-3x+8\right)=0\)
\(\left(x-1\right)\left(2x+11\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\2x=-11\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-\frac{11}{2}\end{cases}}}\)
19 tháng 2 2020
\(b,3x\left(25x+15\right)-35\left(5x+3\right)=0\)
\(15x\left(5x+3\right)-35\left(5x+3\right)=0\)
\(\left(5x+3\right).5\left(3x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x+3=0\\5\left(3x-7\right)=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-3\\3x-7=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\x=\frac{7}{3}\end{cases}}}\)
NN
2
Y
11 tháng 4 2022
\(\dfrac{x+2}{x-5}-3< 0\)
\(\Leftrightarrow\dfrac{x+2-3\left(x-5\right)}{x-5}< 0\)
\(\Leftrightarrow x+2-3x+15< 0\)
\(\Leftrightarrow-2x+17< 0\)
\(\Leftrightarrow-2x< -17\)
\(\Leftrightarrow x>\dfrac{17}{2}\)
20 tháng 10 2021
a: \(\left(x-4\right)^2-\left(x-3\right)\left(x+3\right)=5\)
\(\Leftrightarrow x^2-8x+16-x^2+9=5\)
\(\Leftrightarrow-8x=-20\)
hay \(x=\dfrac{5}{2}\)
NT
2
23 tháng 6 2017
a) \(3\left(x-1\right)+2x-2x^2=0\)
\(\Leftrightarrow3x-3+2x-2x^2=0\)
\(\Leftrightarrow-2x^2+5x-3=0\)
\(\Leftrightarrow-2x^2+2x+3x-3=0\)
\(\Leftrightarrow-2x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\-2x+3=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy..
b) \(x^2+8x+15=0\)
\(\Leftrightarrow x^2+3x+5x+15=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
Vậy..
23 tháng 6 2017
Tìm x :
a) 3(x - 1 ) + 2x - 2x2 = 0
\(\Leftrightarrow3\left(x-1\right)-2x^2+2x=0\)
\(\Leftrightarrow\) 3\(\left(x-1\right)-2x\left(x-1\right)=0\)
\(\Leftrightarrow\) (x - 1 )( 3-2x) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\-2x=-3\Rightarrow x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy....
b) x2 + 8x + 15 = 0
\(\Leftrightarrow x^2+3x+5x+15=0\)
\(\Leftrightarrow\) (x2 + 3x ) + ( 5x + 15 ) =0
\(\Leftrightarrow x\left(x+3\right)+5\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
Vậy....
NN
1
15 tháng 3 2022
a: =>7-x=0
hay x=7
b: \(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(x+5\right)\left(3x-8\right)=0\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2};-5;\dfrac{8}{3}\right\}\)
\(2x^2-x\left(x-2\right)-3=0\)
\(2x^2-x^2+2x-3=0\)
\(x^2+2x-3=0\)
\(x^2+3x-x-3=0\)
\(x\left(x+3\right)-\left(x+3\right)=0\)
\(\left(x+3\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=1\end{cases}}}\)