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H9
HT.Phong (9A5)
CTVHS
18 tháng 8 2023
\(x^6+2x^3+1=0\)
\(\Leftrightarrow\left(x^3\right)^2+2x^3+1=0\)
\(\Leftrightarrow\left(x^3+1\right)^2=0\)
\(\Leftrightarrow x^3=\left(-1\right)^3\)
\(\Leftrightarrow x=-1\)
___________
\(x\left(x-5\right)=4x-20\)
\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
_____________
\(x^4-2x^2=8-4x^2\)
\(\Leftrightarrow x^2\left(x^2-2\right)+\left(4x^2-8\right)=0\)
\(\Leftrightarrow x^2\left(x^2-2\right)+4\left(x^2-2\right)=0\)
\(\Leftrightarrow\left(x^2-2\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow x^2=2\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
_______________
\(\left(x^3-x^2\right)-4x^2+8x-4\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
C
2
7 tháng 6 2021
`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`
`<=>` \(\left[ \begin{array}{l}x=-2\\x=6\end{array} \right.\)
7 tháng 6 2021
`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`
`<=>` \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
HN
10 tháng 10 2021
a) ⇔ \(4x^2+4x-x-1=0\)
⇔ \(4x^2+3x-1=0\)
⇔ \(4x(x+1)-(x+1)=0\)
⇔ \((x+1)(4x-1)=0\)
⇒ \(\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy...
b) \(x^3-4x^2+4x=0\)
⇔ \(x^2(x-2)-2x(x-2)=0\)
⇔ \((x-2)(x^2-2x)=0\)
⇒ \(\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
Vậy...
c) \(x^2-3x+2=0\)
⇔ \(x(x-2)-(x-2)=0\)
⇔ \((x-1)(x-2)=0\)
⇒ \(\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy...
13 tháng 2 2022
\(a,\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\\ b,4x^2-1=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(c,x^2-4x+3=0\\ \Leftrightarrow x^2-3x-x+3=0\\ \Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
\(d,9x^2-6x+1=0\\ \Leftrightarrow\left(3x-1\right)^2=0\\ \Leftrightarrow3x-1=0\\ \Leftrightarrow x=\dfrac{1}{3}\)
13 tháng 11 2021
\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
PT
1
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 12 2020
a.
\(\left(4x^2+4x+1\right)-y^2=\left(2x+1\right)^2-y^2=\left(2x+1-y\right)\left(2x+1+y\right)\)
b.
\(\Leftrightarrow2x^2+2x-x-1-2x^2-3x+1=0\)
\(\Leftrightarrow-2x=0\)
\(\Leftrightarrow x=0\)
\(4x^2+4x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
\(\Rightarrow2x+1=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy...
\(4x^2+4x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
<=> 2x + 1 = 0
<=> x = -1/2