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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)
1) \(x^6-x^4-9x^3+9x^2\)
\(=x^2\left(x^4-x^2-9x+9\right)\)
\(=x^2\left[x^2\left(x^2-1\right)-9\left(x-1\right)\right]\)
\(=x^2\left(x-1\right)\left[x^2\left(x+1\right)-9\right]\)
\(=x^2\left(x-1\right)\left(x^3+x^2-9\right)\)
2) \(x^4-4x^3+8x^2-16x+16\)
\(=x^2\left(x^2+4\right)-4x\left(x^2+4\right)+4\left(x^2+4\right)\)
\(=\left(x^2+4\right)\left(x-2\right)^2\)
3) \(x^4-25x^2+20x-4=x^4+5x^3-2x^2-5x^3-25x^2+10x+2x^2+10x-4\)
\(=x^2\left(x^2+5x-2\right)-5x\left(x^2+5x-2\right)+2\left(x^2+5x-2\right)\)
\(=\left(x^2+5x-2\right)\left(x^2-5x+2\right)\)
4) \(5x\left(x-2y\right)+2\left(2y-x\right)^2\)\(=5x\left(x-2y\right)+2\left(x-2y\right)^2=\left(x-2y\right)\left(5x+2x-4y\right)=\left(x-2y\right)\left(7x-4y\right)\)
5) \(x^2\left(x^2-6\right)-x^2+9=x^4-7x^2+9\)
\(=x^4+x^3-3x^2-x^3-x^2+3x-3x^2-3x+9\)
\(=x^2\left(x^2+x-3\right)-x\left(x^2+x-3\right)-3\left(x^2+x-3\right)\)
\(=\left(x^2+x-3\right)\left(x^2-x-3\right)\)
6) \(7x\left(y-4\right)^2-\left(4-y\right)^3=7x\left(y-4\right)^2+\left(y-4\right)^3=\left(y-4\right)^2\left(7x+y-4\right)\)
7) \(x^3+2x^2-6x-27=x^3-3x^2+5x^2-15x+9x-27\)
\(=x^2\left(x-3\right)+5x\left(x-3\right)+9\left(x-3\right)=\left(x-3\right)\left(x^2+5x+9\right)\)
a,\(xy+3x-7y-21\)
\(=x\left(y+3\right)-7\left(y+3\right)\)
\(=\left(y+3\right)\left(x-7\right)\)
\(b,2xy-15-6x+5y\)
\(=\left(2xy-6x\right)+\left(-15+5y\right)\)
\(=2x\left(y-3\right)-5\left(3-y\right)\)
\(=2x\left(y-3\right)+5\left(y-3\right)\)
\(=\left(y-3\right)\left(2x+5\right)\)
a) \([(x-y)3 + (y-z)3]+ (z-x)3\)=\(\left(x-y+y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]-\left(x-z\right)^3\)
\(=\left(x-z\right)\left[\left(\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2-\left(x-z\right)^2\right)\right]\)
\(=\left(x-z\right)\left[\left(x-y\right)\left(x-y-y+z\right)+\left(y-z-x+z\right)\left(y-z+x-z\right)\right]=\left(x-z\right)\left[\left(x-2y+z\right)\left(x+z\right)-\left(x-y\right)\left(x+y-2z\right)\right]\)
\(=\left(x-z\right)\left(x-y\right)\left(x-2y+z-x-y+2z\right)=\left(x-z\right)\left(x-y\right)\left(z-y\right)3\)
b) \(=y^2\left(x^2y-x^3+z^3-z^2y\right)-z^2x^2\left(z-x\right)=y^2\left[-y\left(z^2-x^2\right)-\left(z^3-x^3\right)\right]-z^2x^2\left(z-x\right)\)
\(=y^2\left(z-x\right)\left(-yz-xy-z^2-zx-x^2\right)-z^2x^2\left(z-x\right)=\left(z-x\right)\left(-y^3z-xy^2-z^2y^2-xyz-x^2y^2-z^2x^2\right)\)
đến đây coi như là thành nhân tử rồi nha. em muốn gọn thì ráng ngồi nghĩ rồi tách nha. chỉ cần nhóm mấy cái có ngoặc giống nhau là đc. k khó đâu. chịu khó nghĩ để rèn luyện nha
c) \(x^8+2x^4+1-x^4=\left(x^4+1\right)^2-x^4=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\)
\(\left(9a^3-6a^2\right)+\left(6a^2-4a\right)+\left(-9a+6\right)=3a^2\left(3a-2\right)+2a\left(3a-2\right)-3\left(3a-2\right)=\left(3a-2\right)\left(3a^2+2a-3\right)\)
d) em sửa đề đi. đề sai rồi. đồng nhất hệ số phải có dấu bằng nha.
có gì liên hệ chị. đúng nha ;)
1 ) x3 - 2x2 + x
= x( x2 - 2x + 1 )
= x ( x-1)2
2) 4x3 - 25x
= x ( 4x2 - 25)
= x( 2x-5) ( 2x +5)
11) \(x^2-y^2-4x+4\)
\(=\left(x^2-4x+4\right)-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-y-2\right)\left(x+y-2\right)\)
13) \(x^4+4=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
b) x3y3 + x2y2+ 4 = x3y3- 4xy + (xy)2- 2xy.2 + 22 = xy [ (xy)^2 - 2^2 ] + ( xy - 2)^2
= xy(xy-2)(xy+2)+ (xy-2)^2
= (xy-2) [ xy(xy+2) + ( xy-2) ]
= (xy-2) [ (xy)2 + 2xy + xy - 3 ]
= ( xy - 3) [ (xy)2 + 3xy - 3]
3) (chưa bik làm)
4) x4 +x3 + 6x2 +5x +5
= x4 +x3 + x2 + 5x2 + 5x +5
= x2( x2+x+ 1 ) + 5( x2+x+ 1 )
= ( x2+ 5 ) ( x2+x+ 1 )
5) x4 - 2x3 - 12x2 +12x + 36
= x4 - 2x3 - 6x2 - 6x2 + 12x + 36=
x2 ( x2 - 2x - 6) - 6 ( x2 - 2x - 6)
= (x^2 - 6) ( x2 - 2x - 6) 6) x8y8 + x4y4 + 1 = \(\left[\left(xy\right)^4\right]^2+2x^4y^4+1-x^4y^4\)=\(\left[\left(xy\right)^4+1\right]^2-\left[\left(xy\right)^2\right]^2\)
= \(\left(x^4y^4+1-x^2y^2\right)\left(x^4y^4+1+x^2y^2\right)\)
( mik ko bik đúng hay sai đâu nha) mik thấy nó thành nhân tử thì mik tách thôi
\(1,4x^4+4x^2y^2-8y^4\)
\(=4\left(x^4+x^2y^2-y^4-y^4\right)\)
\(=4\left[\left(x^4-y^4\right)+\left(x^2y^2-y^4\right)\right]\)
\(=4\left[\left(x^2+y^2\right)\left(x^2-y^2\right)+y^2\left(x^2-y^2\right)\right]\)
\(=4\left(x^2-y^2\right)\left(x^2+y^2+y^2\right)\)
\(=4\left(x-y\right)\left(x+y\right)\left(x^2+2y^2\right)\)
\(2,12x^2y-18xy^2-30y^3\)
\(=6y\left(2x^2-3xy-5y^2\right)\)
\(=6y\left[\left(2x^2+2xy\right)-\left(5xy+5y^2\right)\right]\)
\(=6y\left[2x\left(x+y\right)-5y\left(x+y\right)\right]\)
\(=6y\left(x+y\right)\left(2x-5y\right)\)
x^3 - 7x - 6 = x^3 + x^2 - x^2 - 6x - x - 6 = (x^3 + x^2) - (x^2 + x) - (6x + 6)
= x^2(x + 1) - x(x + 1) - 6(x + 1) = (x + 1)(x^2 - x - 6) = (x + 1)(x^2 - 3x + 2x - 6)
= (x + 1){(x^2 - 3x) + (2x - 6)} = (x + 1){(x(x - 3) + 2(x - 3)}
= (x + 1)(x - 3)(x + 2)
ban kia lam dung roi do
k tui nha
thanks