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b)x2-2x+1=4
⇔(x-1)2=4
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
c)x2-4x+4=9
⇔ (x-2)2=9
\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
d)4x2-4x+1=4
⇔ (2x-1)2=4
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
e)x2-2x-8=0
⇔ x2-4x+2x-8=0
⇔ x(x-4)+2(x-4)=0
⇔(x-4)(x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
f)9x2-6x-8=0
⇔ 9x2-12x+6x-8=0
⇔ 3x(3x-4)+2(3x-4)=0
⇔ (3x-4)(3x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=\dfrac{-2}{3}\end{matrix}\right.\)
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) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)
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
a: \(8x\left(x-2017\right)-2x+4034=0\)
\(\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
`a)16x^2-24x+9=25`
`<=>(4x-3)^2=25`
`+)4x-3=5`
`<=>4x=8<=>x=2`
`+)4x-3=-5`
`<=>4x=-2`
`<=>x=-1/2`
`b)x^2+10x+9=0`
`<=>x^2+x+9x+9=0`
`<=>x(x+1)+9(x+1)=0`
`<=>(x+1)(x+9)=0`
`<=>` \(\left[ \begin{array}{l}x=-9\\x=-1\end{array} \right.\)
`c)x^2-4x-12=0`
`<=>x^2+2x-6x-12=0`
`<=>x(x+2)-6(x+2)=0`
`<=>(x+2)(x-6)=0`
`<=>` \(\left[ \begin{array}{l}x=-2\\x=6\end{array} \right.\)
`d)x^2-5x-6=0`
`<=>x^2+x-6x-6=0`
`<=>x(x+1)-6(x+1)=0`
`<=>(x+1)(x-6)=0`
`<=>` \(\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.\)
`e)4x^2-3x-1=0`
`<=>4x^2-4x+x-1=0`
`<=>4x(x-1)+(x-1)=0`
`<=>` \(\left[ \begin{array}{l}x=1\\x=-\dfrac14\end{array} \right.\)
`f)x^4+4x^2-5=0`
`<=>x^4-x^2+5x^2-5=0`
`<=>x^2(x^2-1)+5(x^2-1)=0`
`<=>(x^2-1)(x^2+5)=0`
Vì `x^2+5>=5>0`
`=>x^2-1=0<=>x^2=1`
`<=>` \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
a) Thực hiện rút gọn VT = -2x – 64
Giải phương trình -2x – 64 = 0 thu được x = -32.
b) Thực hiện rút gọn VT = -62 x +12
Giải phương trình -62x + 12 = -50 thu được x = 1.