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PT
1
25 tháng 10 2020
a) 4x4 + 1
= 4x4 + 4x2 + 1 - 4x2
= ( 4x4 + 4x2 + 1 ) - 4x2
= ( 2x2 + 1 )2 - ( 2x )2
= ( 2x2 - 2x + 1 )( 2x2 + 2x + 1 )
b) x3 + 2x - x2 - 2
= ( x3 - x2 ) + ( 2x - 2 )
= x2( x - 1 ) + 2( x - 1 )
= ( x - 1 )( x2 + 2 )
c) x4 + x2 - 27x - 9
= ( x4 - 27x ) + ( x2 - 9 )
= x( x3 - 27 ) + ( x - 3 )( x + 3 )
= x( x - 3 )( x2 + 3x + 9 ) + ( x - 3 )( x + 3 )
= ( x - 3 )[ x( x2 + 3x + 9 ) + x + 3 ]
= ( x - 3 )( x3 + 3x2 + 9x + x + 3 )
= ( x - 3 )( x3 + 3x2 + 10x + 3 )
14 tháng 8 2021
\(8m^3+36m^2n+54mn^2+27n^3\)
\(=\left(2m\right)^3+3\left(2m\right)^2.3n+3.2m\left(3n\right)^2+\left(3n\right)^2\)
\(=\left(2m+3n\right)^3\)
\(a^4-b^4\)
\(=\left(a^2+b^2\right)\left(a-b\right)\left(a+b\right)\)
\(x^3-3x^2+3x-1\)
\(=\left(x^3-1\right)-3x\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)-3x\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-2x+1\right)\)
\(=\left(x-1\right)^3\)
\(x^3+9x^2y+27xy^2+27y^3\)
\(=x^2+3x^2.3y+3x\left(3y\right)^2+\left(3y\right)^3\)
\(=\left(x+3y\right)^3\)
H
19 tháng 12 2023
Bài `1`
\(a,5x^2-10xy=5x\left(x-2y\right)\\ b,3x\left(x-y\right)-6\left(x-y\right)=\left(x-y\right)\left(3x-6\right)\\ =3\left(x-y\right)\left(x-2\right)\\ c,2x\left(x-y\right)-4y\left(y-x\right)=2x\left(x-y\right)+4y\left(x-y\right)\\ =\left(x-y\right)\left(2x+4y\right)=2\left(x-y\right)\left(x+2y\right)\\ d,9x^2-9y^2=\left(3x\right)^2-\left(3y\right)^2=\left(3x-3y\right)\left(3x+3y\right)\\ f,xy-xz-y+z=\left(xy-xz\right)-\left(y-z\right)\\ =x\left(y-z\right)-\left(y-z\right)=\left(y-z\right)\left(x-1\right)\)
Bài `3`
\(a,3x^2+8x=0\\ \Leftrightarrow x\left(3x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{8}{3}\end{matrix}\right.\)
\(b,9x^2-25=0\\ \Leftrightarrow\left(3x\right)^2-5^2=0\\ \Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-5=0\\3x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=5\\3x=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
\(c,x^3-16x=0\\ \Leftrightarrow x\left(x^2-16\right)=0\\ \Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
\(d,x^3+x=0\\ \Leftrightarrow x\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1\in\varnothing\\x=0\end{matrix}\right.\Rightarrow x=0\)
TM
2
27 tháng 11 2021
3 . (x+3) - x2+9
= 3.(x+3)-(x2-9)
=3.(x+3)-[(x-3).(x+3)]
=(x+3).[3-(x-3)]
=(x+3).(3-x+3)
=(x+3).(9-x)
12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
HN
4
N
19 tháng 11 2021
\(=\left(x-3\right)\left(x+3\right)+\left(x-3\right)\left(2x-5\right)\\ =\left(x-3\right)\left(x+3+2x-5\right)\\ =\left(x-3\right)\left(3x-2\right)\)
BD
1
BD
0
3(x+3) - x² + 9
= 3(x+3) - (x² - 3²)
= 3(x+3) - (x-3)(x+3)
= (3 - x + 3)(x+3)
= (6 -x)(x+3)
k mình đi bạn
3(x + 3) - x2 + 9
= 3(x + 3) - (x2 - 32)
= 3(x + 3) - (x + 3)(x - 3)
= (x + 3)(- x + 6)