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a) Ở trên
b) \(\left(x^2-8x\right)^2+36\) = \(x^4-16x^2+64+36\)
= \(\left(x^4+20x^2+100\right)-36x^2\)
= \(\left(x^2+10\right)^2-\left(6x\right)^2\)
= \(\left(x^2-6x+10\right)\left(x^2+6x+10\right)\)
a)
12x2 - 23xy + 10y2
= (12x2 - 8xy ) - ( 15xy - 10y2 )
= 4x(3x - 2y) - 5y(3x - 2y)
= (4x - 5y)(3x - 2y)
b)
( x2 - 8)2 + 36 = x4 - 16x2 + 64 +36
= (x4 + 20x2 +100) - 36x2
= (x2 + 10)2 - (6x)2
= (x2 + 10 - 6x)(x2 + 10 + 6x)
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a,x2-4xy+4y2
=(x-2y2
b,4x4+9y2-12x2y
=(2x2)2+(3y)2-12x2y
(2x2-3y)
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câu a đặt chung x ra là xong
câu b
x^3 + 3x^2 - 7x^2 - 21x + 9x+ 27 còn lại tự làm nhé
a) x3 - 2x2 + x - xy2
= x (x2 - 2x + 1 - y2)
= x [(x2 - 2x + 1) - y2]
= x [(x - 1)2 - y2]
= x [(x - 1) + y] [(x - 1) - y]
= x (x - 1 + y) (x - 1 - y)
b) x3 - 4x2 - 12x + 27
= (x3 + 27) - (4x2 + 12x)
= (x3 + 33) - 4x (x + 3)
= (x + 3) (x2 - 3x + 32) - 4x (x + 3)
= (x + 3) [(x2 - 3x + 9) - 4x]
= (x + 3) (x2 - 3x + 9 - 4x)
= (x + 3) (x2 - 7x + 9)
#Học tôt!!!
~NTTH~
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Phân tích đa thức thành nhân tử:
\(36-12x+x^2\)
\(=36-6x-6x+x^2\)
\(=\left(36-6x\right)-\left(6x-x^2\right)\)
\(=6\left(6-x\right)-x\left(6-x\right)\)
\(=\left(6-x\right)\left(6-x\right)=\left(6-x\right)^2\)
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\(x^2-2xy+5x-10y\)
\(=x\left(x-2y\right)+5\left(x-2y\right)\)
\(=\left(x+5\right)\left(x-2y\right)\)
\(x^2-2xy+5x-10y\)
\(=\left(x^2-2xy\right)+\left(5x-10y\right)\)
\(=x\left(x-2y\right)+5\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+5\right)\)
\(x-3\sqrt{x}+\sqrt{xy}-3y\)
\(=\left(x-3\sqrt{x}\right)+\left(\sqrt{xy}-3y\right)\)
\(=\sqrt{x}\left(\sqrt{x}-3\right)+y\left(\sqrt{x}-3\right)\)
\(=\left(\sqrt{x}-3\right)\left(\sqrt{x}+y\right)\)
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Ta có: \(12x^2-23xy+10y^2\) = \(\left(12x^2-8xy\right)-\left(15xy-10y^2\right)\)
= \(4x\left(3x-2y\right)\) \(-5y\left(3x-2y\right)\)
= \(\left(3x-2y\right)\left(4x-5y\right)\)