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LN
31 tháng 3 2019
Có: \(\left(x+y\right)^4+x^4+y^4\)
\(=\left(x+y\right)^4+\left(x^2+y^2\right)^2-2x^2y^2\)
\(=\left[\left(x+y\right)^4-x^2y^2\right]+\left[\left(x^2+y^2\right)^2-x^2y^2\right]\)
\(=\left[\left(x^2+y^2+2xy\right)^2-\left(xy\right)^2\right]+\left[\left(x^2+y^2\right)^2-\left(xy\right)^2\right]\)
\(=\left(x^2+y^2+xy\right)\left(x^2+y^2+3xy\right)+\left(x^2+y^2+xy\right)\left(x^2+y^2-xy\right)\)
\(=2\left(x^2+y^2+xy\right)\left(x^2+y^2+xy\right)=2\left(x^2+y^2+xy\right)^2\)
TT
0
TV
1
MT
13 tháng 8 2015
x4y4+64=x4y4+16x2y2+64-16x2y2
=(x2y2+8)2-16x2y2
=(x2y2-4xy+8)(x2y2+4xy+8)
12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
F
1
21 tháng 9 2023
`#3107`
`x^4 - 8x + 63`
`= x^4 + 4x^3 + 9x^2 - 4x^3 -16x^2 - 36x + 7x^2 + 28x + 63`
`= (x^4 + 4x^3 + 9x^2) - (4x^3 + 16x^2 + 36x) + (7x^2 + 28x + 63)`
`= x^2(x^2 + 4x + 9) - 4x(x^2 + 4x + 9) + 7(x^2 + 4x + 9)`
`= (x^2 + 4x + 9)(x^2 - 4x + 7)`
____
`64x^4 + y^4`
`= 64x^4 + 16x^2y^2 + y^4 - 16x^2y^2`
`= (64x^4 + 16x^2y^2 + y^4) - (16x^2y^2)`
`= [(8x^2)^2 + 2*8x^2*y^2 + (y^2)^2] - (4xy)^2`
`= (8x^2 + y^2)^2 - (4xy)^2`
`= (8x^2 + y^2 - 4xy)(8x^2 + y^2 + 4xy)`
____
`x^3 + 3xy`
`= x(x^2 + 3y)`
23 tháng 7 2021
a) \(x^6-x^4+2x^3+2x^2\)
\(=x^2\left(x^4-x^2+2x+2\right)\)
\(=x^2\left[x^2\left(x^2-1\right)+2\left(x+1\right)\right]\)
\(=x^2\left(x+1\right)\left[x^2\left(x-1\right)+2\right]\)
\(=x^2\left(x+1\right)\left(x^3-x^2+2\right)\)
b) Ta có: \(4x^4+y^4\)
\(=4x^4+y^4+4x^2y^2-4x^2y^2\)
\(=\left(2x^2+y^2\right)^2-\left(2xy\right)^2\)
\(=\left(2x^2-2xy+y^2\right)\left(2x^2+2xy+y^2\right)\)
23 tháng 7 2021
a, \(x^6-x^4+2x^3+2x^2\)
\(=x^2\left(x^4-x^2+2x+2\right)=x^2\left[x^2\left(x^2-1\right)+2\left(x+1\right)\right]\)
\(=x^2\left[x^2\left(x-1\right)\left(x+1\right)+2\left(x+1\right)\right]=x^2\left(x^3-x^2+2\right)\left(x+1\right)\)
\(=x^2\left(x+1\right)^2\left(x^2-2x+2\right)\)
b, \(4x^4+y^4=\left(2x^2\right)^2+2.2x^2.y^2+\left(y^2\right)^2-4x^2y^2\)
\(=\left(2x^2+y^2\right)^2-\left(2xy\right)^2=\left(2x^2+y^2-2xy\right)\left(2x^2+y^2+2xy\right)\)
NT
2
3 tháng 11 2019
x^4 - y^4
= (x^2 - y^2)(x^2 + y^2)
= (x - y)(x + y)(x^2 + y^2)
TH
1
\(x^4.y^4+4\)
\(=\left(x^4y^4-2x^3y^3+2x^2y^2\right)+\left(2x^3y^3-4x^2y^2+4xy\right)+\left(2x^2y^2-4xy+4\right)\)
\(=x^2y^2\left(x^2y^2-2xy+2\right)+2xy\left(x^2y^2-2xy+2\right)+2\left(x^2y^2-2xy+2\right)\)
= (x2y2 + 2xy + 2)(x2y2 - 2xy + 2)
Dùng cách này cho nhanh :v
Đặt xy = t cho dễ nhìn. \(t^4+4=\left(t^4+2t^2.2+4\right)-\left(2t\right)^2\)
\(=\left(t^2+2\right)^2-\left(2t\right)^2=\left(t^2-2t+2\right)\left(t^2+2t+2\right)\)
\(=\left(x^2y^2-2xy+2\right)\left(x^2y^2+2xy+2\right)\)