pt đa thức thành nhân tử A=2(x+y)^4-3(x+y)^2-5
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12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
24 tháng 9 2021
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
3 tháng 9 2016
a) \(3x-3+5\left(1-x\right)\)
\(=3x-3+5-5x\)
\(=3x-5x+2\)
\(=x\left(3-5\right)+2\)
\(=-2x+2\)
\(=2\left(1-x\right)\)
b) \(12a^2-3ab+8ac-2bc\)
\(=3a\left(4a-b\right)+2c\left(4a-b\right)\)
\(=\left(4a-b\right)\left(3a+2c\right)\)
c) \(x^2-25+y^2-2xy\)
\(=x^2-2xy+y^2-25\)
\(=\left(x-y\right)^2-5^2\)
\(=\left(x-y-5\right)\left(x-y+5\right)\)
KT
17 tháng 7 2018
d) mk chỉnh lại đề
\(8xy^2-5xyz-24y+15z\)
\(=xy\left(8y-5z\right)-3\left(8y-5z\right)\)
\(=\left(8y-5z\right)\left(xy-3\right)\)
e) \(x^4-x^3-x+1=\left(x-1\right)^2\left(x^2+x+1\right)\)
f) \(x^4+x^2y^2+y^4=\left(x^2-xy+y^2\right)\left(x^2+xy-y^2\right)\)
g) \(x^3+3x-4=\left(x-1\right)\left(x^2+x+4\right)\)
h) \(x^3-3x^2+2=\left(x-1\right)\left(x^2-2x-2\right)\)
i) \(2x^3+x^2-4x-12=\left(x-2\right)\left(2x^2+5x+6\right)\)
k) \(25x^2\left(x-5\right)-x+y=\left(1-5x\right)\left(1+5x\right)\left(y-x\right)\)
12 tháng 9 2020
a) x4 - x5 = x4( x - 1 )
b) -8x2y2 - 12xy3 - 4xy2
= -4xy( 2xy + 3y2 + y )
c) ( x - y )3 - x3 + y3
= x3 - 3x2y + 3xy2 - y3 - x3 + y3
= 3xy2 - 3x2y
= 3xy( y - x )
21 tháng 1 2022
b: \(=\dfrac{12\left(y-z\right)^4+3\left(y-z\right)^5}{6\left(y-z\right)^2}=2\left(y-z\right)^2+\dfrac{1}{2}\left(y-z\right)^3\)
MN
4
5 tháng 9 2021
b: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-3\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-3\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+21\)
\(=\left(x^2+5x+3\right)\left(x^2+5x+7\right)\)
Ta có: \(A=2\left(x+y\right)^4-3\left(x+y\right)^2-5\)
\(=2\left(x+y\right)^4-5\left(x+y\right)^2+2\left(x+y\right)^2-5\)
\(=\left(x+y\right)^2\left[2\left(x+y\right)^2-5\right]+\left[2\left(x+y\right)^2-5\right]\)
\(=\left[2\left(x+y\right)^2-5\right]\left[\left(x+y\right)^2+1\right]\)