phân tích đa thức thành nhân tử ( x^10 + y^10)(x^2 + y^2)-(x^8 + y^8)(x^4 + y^4)
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6, x mũ 4 - 4x mũ 3 - 8x mũ 2 + 8x =x (x+2) (x^2-6x+4)
8, x mũ 4 + 2x mũ 3 + x mũ 2 - y mũ 2 = -(y-x^2-x) (y+x^2+x)
10, 4x mũ 2 ( x + y ) -x - y = (2x-1) (2x+1) (y+x)
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a) (x-2)(x+2)(x^2-10)-72=(x^2-4)(x^2-82)
b) x^8+x^6+x^4+x^2+1=x^2 (x^4+x^3+x^2+1+1/x^2)
c)(x+y)^4+x^4+y^4=(x+y)^4+(x+y)^4=2 (x+y)^4
a) (x-2)(x+2)(x^2 - 10) -72
= (x^2 - 4)(x^2 - 10) - 72
= x^4 - 4x^2 -10x^2 + 40 - 72
= x^4 - 14x^2 - 32
= x^4 - 16x^2 + 2x^2 - 32
= x^2(x^2 - 16) + 2(x^2 - 16)
= (x^2 - 16)(x^2 + 2)
= (x-4)(x+4)(x^2 + 2)
c) (x+y)4 + x4 + y4
= 2x4 + 4xy3 + 6x2y2 + 4x3y + 2y3
= 2(y4 + 2xy3 + 3x2y2 + 2x3y + x4)
= 2(y2 + xy + y2)2
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\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
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a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
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a)\(7x\left(y-4\right)^2-\left(4-y\right)^3=7x\left(4-y\right)^2-\left(4-y\right)^3=\left(4-y\right)^2\left(7x-4+y\right)\)
b)\(\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\)
\(=\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)-9\left(4x-8\right)\)
\(=\left(4x-8\right)\left(x^2-x-10\right)=4\left(x-2\right)\left(x^2-x-10\right)\)
a.\(7x.\left(y-4\right)^2-\left(4-y\right)^3\)=\(7x.\left(4-y\right)^2-\left(4-y\right)^3=\left(4-y\right)^2.\left(7x+y-4\right)\)
b.\(\left(4x-8\right).\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9.\left(8-4x\right)\)
=\(\left(4x-8\right)\left(x^2+6-x-7-9\right)=\left(4x-8\right)\left(x^2-x-10\right)\)
(x^10+y^10)(x^2+y^2)-(x^8+y^8)(x^4+y^4)
=x^12+x^10y^2+y^10x^2+y^12-x^12-x^8y^4-x^4y^8-y^12
=x^10y^2+y^10x^2-x^8y^4-x^4y^8
=x^2y^2(x^8+y^8-x^6y^2-x^2y^6)
=x^2y^2[x^6(x^2-y^2)+y^6(y^2-x^2)]
=x^2y^2[x^6(x-y)(x+y)-y^6(x-y)(x+y)]
=x^2y^2(x^6-y^6)(x-y)(x+y)
=x^2y^2(x-y)(x+y)(x^2+xy+y^2)(x^2-xy+y^2)(x-y)(x+y)
=x^2y^2(x-y)^2(x+y)^2(x^2+xy+y^2)(x^2-xy+y^2)