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x6+3x4y2-8x3y3+3x2y4+y6= x6+3x4y2+3x2y4+y6-8x3y3=(x2+y2)3-(2xy)3
= (x2+y2-2xy)[(x2+y2)2+2xy(x2+y2)+(2xy)2]= (x-y)2(x4+6x2y2+y4+2x3y+2xy3)
(x2+y2-5)2-4x2y2-16xy-16=(x2+y2-5)2-(4x2y2+16xy+16)=(x2+y2-5)2-(2xy+4)2
=(x2+y2-5+2xy+4)(x2+y2-5-2xy-4)=(x2+2xy+y2-1)(x2-2xy+y2-9)=[(x+y)2-1][(x-y)2-32]=(x+y-1)(x+y+1)(x-y-3)(x-y+3)
x4+324=x4+36x2+324-36x2=(x2+18)2-(6x)2=(x2+18-6x)(x2+18+6x)
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\(A=3x^2-14x^2+4x+3\)
Giả sử:
\(A=\left(3x+a\right)\left(x^2+bx+c\right)\)
\(=3x^3+3bx^2+3cx+ax^{2\:}+abx+ac\)
\(=3x^3+\left(3b+a\right)x^2+\left(3c+ab\right)x+ac\)
Ta có:
\(\begin{cases}3b+a=-14\\3c+ab=4\\ac=3\end{cases}\)\(\Rightarrow\begin{cases}a=1\\b=-5\\c=3\end{cases}\)
Vậy \(A=\left(3x+1\right)\left(x^2-5x+3\right)\)
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Hình như đề bài hơi sai sai í bn '^' Mk nghĩ sửa 12x thành 12xy nha
TL :
4x2 - 12xy + 5y2
= 4x2 - 2xy - 10xy + 5y2
= 2x( 2x - y) - 5y( 2x - y )
= ( 2x - 5y )( 2x - y )
Thế này là xog òi đó bn :)
Nếu đề bài là đúng như bn vt trên kia thỳ xl ạ :>
Mk k phân tích đc :)
Đồng ý với bạn Linh nhé :3
4x2 - 12xy + 5y2
= 4x2 - 2xy - 10xy + 5y2
= 2x( 2x - y) - 5y( 2x - y)
= ( 2x - y)(2x - 5y )
Mk nghĩ như v ms đúng :)
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a) \(4\left(x^2-y^2\right)+4x+1\)
\(=4x^2-4y^2+4x+1\)
\(=\left[\left(2x\right)^2+2\cdot2x\cdot1+1^2\right]-\left(2y\right)^2\)
\(=\left(2x+1\right)^2-\left(2y\right)^2\)
\(=\left(2x-2y+1\right)\left(2x+2y+1\right)\)
b) \(x^2+1-x^3-x^2\)
\(=1-x^3\)
\(=\left(1-x\right)\left(1+x+x^2\right)\)
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\(8x^3+4x^2-y^3-y^2\)
\(=\left(8x^3-y^3\right)+\left(4x^2-y^2\right)\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)+\left(2x-y\right)\left(2x+y\right)\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2+2x+y\right)\)
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4x3+4x4−x2−x
=4x3(x+1)−x(x+1)
=(x+1)(4x3−1)
ĐÂY NHÉ. T.I.C.K MÌNH VỚI
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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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b )=x4-2x3-2x3+4x2+4x2-8x-8x+16
=x3(x-2)-2x2(x-2)+4x(x-2)-8(x-2)
=(x-2)(x3-2x2+4x-8)
=(x-2)[x2(x-2)+4(x-2)]
=(x-2)2(x2+4)
a) đề thiếu ko bn?
b) \(x^4-4x^3+8x^2-16x+16=\left(x^4-4x^2\right)-\left(4x^3-12x^2+8x\right)-\left(8x-16\right)\)
\(=x^2\left(x-2\right)\left(x+2\right)-4x\left(x^2-3x+2\right)-8\left(x-2\right)\)
\(=x^2\left(x-2\right)\left(x+2\right)-4x\left(x-2\right)\left(x-1\right)-8\left(x-2\right)\)
\(=\left(x-2\right)\left[x^2\left(x+2\right)-4x\left(x-1\right)-8\right]=\left(x-2\right)\left(x^3-2x^2+4x-8\right)\)
\(=\left(x-2\right)\left[\left(x^3-8\right)-\left(2x^2-4x\right)\right]=\left(x-2\right)\left[\left(x-2\right)\left(x^2+2x+4\right)-2x\left(x-2\right)\right]\)
\(=\left(x-2\right)\left(x-2\right)\left(x^2+2x+4-2x\right)=\left(x-2\right)^2\left(x^2+4\right)\)
\(4x^3y-16xy^2=4xy\left(x^2-4y^2\right)=4xy\left(x-2y\right)\left(x+2y\right)\)
phân tích đa thức sau thành nhân tử
x3-x+3x2y+3xy2+y3-y
giúp câu này nữa nha