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a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
a) \(\Rightarrow x\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
b) \(\Rightarrow x\left(x^2-4\right)=0\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)
Bài 1:
a)x2-10x+9
=x2-x-9x+9
=x(x-1)-9(x-1)
=(x-9)(x-1)
b)x2-2x-15
=x2+3x-5x-15
=x(x+3)-5(x+3)
=(x-5)(x+3)
c)3x2-7x+2
=3x2-x-6x+2
=x(3x-1)-2(3x-1)
=(x-2)(3x-1)x^3-12+x^2
d)x3-12+x2
=x3+3x2+6x-2x2-6x-12
=x(x2+3x+6)-2(x2+3x+6)
=(x-2)(x2+3x+6)
a. \(8x\left(x-2007\right)-2x+4034=0\)
\(\Rightarrow\left(x-2017\right)\left(4x-1\right)\)
\(\Rightarrow\left[{}\begin{matrix}x-2017=0\\4x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2017\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy x=2017 hoặc x=1/4
b.\(\dfrac{x}{2}+\dfrac{x^2}{8}=0\)
\(\Rightarrow\dfrac{x}{2}\left(1+\dfrac{x}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=0\\1+\dfrac{x}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{x}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
Vậy x=0 hoặc x=-4
c.\(4-x=2\left(x-4\right)^2\)
\(\Rightarrow\left(4-x\right)-2\left(x-4\right)^2=0\)
\(\Rightarrow\left(4-x\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4-x=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{2}\end{matrix}\right.\)
Vậy x=4 hoặc x=7/2
d.\(\left(x^2+1\right)\left(x-2\right)+2x=4\)
\(\Rightarrow\left(x-2\right)\left(x^2+3\right)=0\)
Nxet: (x2+3)>0 với mọi x
=> x-2=0 <=>x=2
Vậy x=2
a, 8\(x\).(\(x-2007\)) - 2\(x\) + 4034 = 0
4\(x\)(\(x\) - 2007) - \(x\) + 2017 = 0
4\(x^2\) - 8028\(x\) - \(x\) + 2017 = 0
4\(x^2\) - 8029\(x\) + 2017 = 0
4(\(x^2\) - 2. \(\dfrac{8029}{8}\) \(x\) +( \(\dfrac{8029}{8}\))2) - (\(\dfrac{8029}{4}\))2 + 2017 = 0
4.(\(x\) + \(\dfrac{8029}{8}\))2 = (\(\dfrac{8029}{4}\))2 - 2017
\(\left[{}\begin{matrix}x=-\dfrac{8029}{8}+\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\\x=-\dfrac{8029}{8}-\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\end{matrix}\right.\)
a: \(8x\left(x-2017\right)-2x+4034=0\)
\(\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)
a)4x2-9=0
⇔ (2x-3)(2x+3)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b)(x+5)2-(x-1)2=0
⇔ (x+5-x+1)(x+5+x-1)=0
⇔ 12(x+2)=0
⇔ x=-2
c)x2-6x-7=0
⇔ x2-7x+x-7=0
⇔ x(x-7)+(x-7)=0
⇔ (x-7)(x+1)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)
d)(x+1)2-(2x-1)2=0
⇔ (x+1-2x+1)(x+1+2x-1)=0
⇔3x(2-x)=0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
a, 4x2 - 9 = 0
<=> 4x2 = 9
<=> x2 = \(\dfrac{9}{4}\) => x = \(\sqrt{\dfrac{9}{4}}\)
b, (x + 5 )2 - ( x - 1 )2 = 0
<=> ( x+5-x+1 )(x+5+x-1) = 0
<=> 6(2x+4) = 0
<=> 12x+24=0
<=> 12x = -24
<=> x = -2
c, x2-6x-7=0
<=> x2+x-7x-7=0
<=> x(x+1)-7(x+1)=0
<=> (x-7)(x+1)=0
=> x+7=0 hoặc x+1=0
+ x-7=0 => x=7
+ x+1=0 => x=-1
d, \(\left(x+1\right)^2-\left(2x-1\right)^2=0\)
<=> \(\left(x+1-2x+1\right)\left(x+1+2x-1\right)=0\)
<=> (-x+2).3x=0
=> x=0 hoặc (-x+2).3=0
+ (-x+2).3=0 => -3x+6=0 => x=-2