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13 tháng 11 2021
\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
T
1
18 tháng 8 2023
a: =>2x^2+9x-6x-27=0
=>x(2x+9)-3(2x+9)=0
=>(2x+9)(x-3)=0
=>x=3 hoặc x=-9/2
b: =>-10x^2+6x-5x+3=0
=>-2x(5x-3)-(5x-3)=0
=>(5x-3)(-2x-1)=0
=>x=-1/2 hoặc x=5/3
c: =>-x^3+2x^2-x^2+4=0
=>-x^2(x-2)-(x-2)(x+2)=0
=>(x-2)(-x^2-x-2)=0
=>x-2=0
=>x=2
d: =>(x^3+8)-4x(x+2)=0
=>(x+2)(x^2-2x+4)-4x(x+2)=0
=>(x+2)(x^2-6x+4)=0
=>x=-2 hoặc \(x=3\pm\sqrt{5}\)
AH
Akai Haruma
Giáo viên
25 tháng 10 2021
a.
$x^2-y^2-2x+2y=(x^2-y^2)-(2x-2y)=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)$
b.
$x^2(x-1)+16(1-x)=x^2(x-1)-16(x-1)=(x-1)(x^2-16)=(x-1)(x-4)(x+4)$
c.
$x^2+4x-y^2+4=(x^2+4x+4)-y^2=(x+2)^2-y^2=(x+2-y)(x+2+y)$
d.
$x^3-3x^2-3x+1=(x^3+1)-(3x^2+3x)=(x+1)(x^2-x+1)-3x(x+1)$
$=(x+1)(x^2-4x+1)$
AH
Akai Haruma
Giáo viên
25 tháng 10 2021
e.
$x^4+4y^4=(x^2)^2+(2y^2)^2+2.x^2.2y^2-4x^2y^2$
$=(x^2+2y^2)^2-(2xy)^2=(x^2+2y^2-2xy)(x^2+2y^2+2xy)$
f.
$x^4-13x^2+36=(x^4-4x^2)-(9x^2-36)$
$=x^2(x^2-4)-9(x^2-4)=(x^2-9)(x^2-4)=(x-3)(x+3)(x-2)(x+2)$
g.
$(x^2+x)^2+4x^2+4x-12=(x^2+x)^2+4(x^2+x)-12$
$=(x^2+x)^2-2(x^2+x)+6(x^2+x)-12$
$=(x^2+x)(x^2+x-2)+6(x^2+x-2)=(x^2+x-2)(x^2+x+6)$
$=[x(x-1)+2(x-1)](x^2+x+6)=(x-1)(x+2)(x^2+x+6)$
h.
$x^6+2x^5+x^4-2x^3-2x^2+1$
$=(x^6+2x^5+x^4)-(2x^3+2x^2)+1$
$=(x^3+x^2)^2-2(x^3+x^2)+1=(x^3+x^2-1)^2$
C
1
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 12 2020
a.
\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)
b.
\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
c.
\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)
\(=\left(x+3\right)^3\)
d.
\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
e.
\(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
f.
\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
CM
0
CM
0
C
3
7 tháng 8 2021
a) Ta có: \(36x^3-4x=0\)
\(\Leftrightarrow4x\left(9x^2-1\right)=0\)
\(\Leftrightarrow x\left(3x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=\dfrac{-1}{3}\end{matrix}\right.\)
b) Ta có: \(3x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{3}\end{matrix}\right.\)
\(x^6+2x^3+1=0\)
\(\Leftrightarrow\left(x^3\right)^2+2x^3+1=0\)
\(\Leftrightarrow\left(x^3+1\right)^2=0\)
\(\Leftrightarrow x^3=\left(-1\right)^3\)
\(\Leftrightarrow x=-1\)
___________
\(x\left(x-5\right)=4x-20\)
\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
_____________
\(x^4-2x^2=8-4x^2\)
\(\Leftrightarrow x^2\left(x^2-2\right)+\left(4x^2-8\right)=0\)
\(\Leftrightarrow x^2\left(x^2-2\right)+4\left(x^2-2\right)=0\)
\(\Leftrightarrow\left(x^2-2\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow x^2=2\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
_______________
\(\left(x^3-x^2\right)-4x^2+8x-4\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)