Phân tích đa thức thành nhân tử : ( x + y)^4 + x^4 + y^4
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\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
`#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)`
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`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)`
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`x^3 + 3xy`
`= x(x^2 + 3y)`
\(x^4+y^4\)
\(=x^4+2x^2y^2+y^4-2x^2y^2\)
\(=\left(x^2+y^2\right)^2-\left(\sqrt{2}xy\right)^2\)
\(=\left(x^2+\sqrt{2}xy+y^2\right)\left(x^2-\sqrt{2}xy+y^2\right)\)
\(=\left(3x-3y\right)^2-\left(2x+2y\right)^2=\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\)
\(=\left(x-5y\right)\left(5x-y\right)\)
x^4 - y^4
= (x^2 - y^2)(x^2 + y^2)
= (x - y)(x + y)(x^2 + y^2)
`9(x-y)^2-4(x+y)^2`
`=[3(x-y)]^2-[2(x+y)]^2`
`=(3x-3y)^2-(2x+2y)^2`
`=(3x-3y+2x+2y)(3x-3y-2x-2y)`
`=(5x-y)(x-5y)`
\(9\left(x-y\right)^2-4\left(x+y\right)^2\\ =\left[3\left(x-y\right)\right]^2-\left[2\left(x+y\right)\right]^2\\ =\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\\ =\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\\ =\left(x-5y\right)\left(5x-y\right)\)