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a) \(x^2-81=\left(x-9\right)\left(x+9\right)\)
b) \(4x^2-25=\left(2x-5\right)\left(2x+5\right)\)
c) \(x^4-y^4=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
d) \(x^2+6xy+9y^2=\left(x+3y\right)^2\)
e) \(6x-9-x^2=-\left(x^2-6x+9\right)=-\left(x-3\right)^2\)
f) \(x^2-4x^2+4y^2+4xy=\left(x^2+4xy+4y^2\right)-4x^2=\left(x+2y\right)^2-4x^2\\ =\left(x+2y+2x\right)\left(x+2y-2x\right)=\left(3x+2y\right)\left(2y-x\right)\)
g) \(\left(a+b\right)^3+\left(a-b\right)^3=\left(a+b+a-b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2a\left(a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2\right)=2a\left(a^2+3b^2\right)\)
h) \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1+x+1\right)\left(3x+1-x-1\right)\\ =\left(4x+2\right)\cdot2x=4x\left(2x+1\right)\)
\(x^2-y^2+4x+4\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(4x^2-y^2+8\left(y-2\right)\)
\(=4x^2-\left(y^2-8y+16\right)\)
\(=4x^2-\left(y-4\right)^2\)
\(=\left(2x+y-4\right)\left(2x-y+4\right)\)
a. \(-x^3-6x^2+6x+1=-x^3+x^2-7x^2+7x-x+1=\left(1-x\right)\left(x^2+7x+1\right)\)
b. \(x^4-4x^2+4x-1=x^4-1-4x\left(x-1\right)=\left(x-1\right)\left[\left(x+1\right)\left(x^2+1\right)-4x\right]\)
\(=\left(x-1\right)\left(x^3+x^2-3x+1\right)\)
c. \(6x^3-x^2-486x+81=6x^3-54x^2+53x^2-477x-9x+81=\left(x-9\right)\left(6x^2+53x-9\right)\)
\(=\left(x-9\right)\left(x+9\right)\left(6x-1\right)\)
d. \(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)=x^2\left(x^2+8x+16\right)-x^2-8x-16-x^2+1\)
\(=x^4+8x^3+14x^2-8x-15=x^4+5x^3+3x^3+15x^2-x^2-5x-3x-15\)
\(=\left(x+5\right)\left(x^3+3x^3-x-3\right)=\left(x+5\right)\left(x-1\right)\left(x+1\right)\left(x+3\right)\)
Để phân tích nhân tử các dạng này, em cần nhẩm được nghiệm để biết đc nhân tử chung là gì, sau đó tách để xuất hiện nhân tử chung đó. CHÚC EM HỌC TỐT :))
A = 6x4 - 5x3 + 4x2 + 2x - 1
= 6x4 + 3x3 - 8x3 - 4x2 + 8x2 + 4x - 2x - 1
= 3x3. ( 2x + 1 ) - 4x2 ( 2x + 1 ) + 4x ( 2x + 1 ) - ( 2x + 1 )
= ( 2x + 1 ) ( 3x3 - 4x2 + 4x - 1 )
= ( 2x + 1 ) ( 3x3 - x2 - 3x2 + x + 3x - 1 )
= ( 2x + 1 ) [ x2 ( 3x - 1 ) - x ( 3x - 1 ) + ( 3x - 1 ) ]
= ( 2x + 1 ) ( 3x - 1 ) ( x2 - x + 1 )
B = 4x4 + 4x3 + 5x2 + 8x - 6
= 4x4 - 2x3 + 6x3 - 3x2 + 8x2 - 4x + 12x - 6
= 2x3 ( 2x - 1 ) + 3x2 ( 2x - 1 ) + 4x ( 2x - 1 ) + 6 ( 2x - 1 )
= ( 2x - 1 ) ( 2x3 + 3x2 + 4x + 6 )
= ( 2x - 1 ) [ x2 ( 2x + 3 ) + 2 ( 2x + 3 ) ]
= ( 2x - 1 ) ( 2x + 3 ) ( x2 + 2 )
C = x4 + x3 - 5x2 + x - 6
= x4 - 2x3 + 3x3 - 6x2 + x2 - 2x + 3x - 6
= x3 ( x - 2 ) + 3x2 ( x - 2 ) + x ( x - 2 ) + 3 ( x - 2 )
= ( x - 2 ) ( x3 + 3x2 + x + 3 )
= ( x - 2 ) [ x2 ( x + 3 ) + ( x + 3 ) ]
= ( x - 2 ) ( x + 3 ) ( x2 + 1 )
2/ (b4 - 4b2 + 4) - 9a2 = (b2 - 2)2 - 9a2 = (b2 - 2 + 3a)(b2 - 2 - 3a)
a 4x -4y +(x-y)^2
=4(x-y)+(x-y).(x-y)
=(x-y).(4+x-y)
c x^2(x+1)-4(x+1)
(x+1).(x^2-4)
d x^4-(x^2-2x+1)
=x^4-(x-1)^2
=x^2(x-x+1)(x-x-1)
MIK KO BIT DUNG HAY KO CON B THI MIK KO BIET LAM
Câu b dễ thôi
\(x^4-4x^3-8x^2+8x\)
\(=x\left(x^3-4x^2-8x+8\right)\)
\(=x\left(x+2\right)\left(x^2-6x+4\right)\)
a)\(3x^2-8x+4\)
\(=3x^2-2x-6x+4\)
\(=x\left(3x-2\right)-2\left(3x-2\right)\)
\(=\left(x-2\right)\left(3x-2\right)\)
b)\(4x^4+81\)
\(=4x^4+36x^2+81-36x^2\)
\(=\left(2x^2+9\right)^2-36x^2\)
\(=\left(2x^2-6x+9\right)\left(2x^2+6x+9\right)\)
c)\(x^8+98x^4+1\)
\(=\left(x^8+2x^4+1\right)+96x^4\)
\(=\left(x^4+1\right)^2+16x^2\left(x^4+1\right)+64x^4-16x^2\left(x^4+1\right)+32x^4\)
\(=\left(x^4+8x^2+1\right)^2-16x^2\left(x^4-2x^2+1\right)\)
\(=\left(x^4+8x^2+1\right)^2-16x^2\left(x^4-2x^2+1\right)\)
\(=\left(x^4+8x^2+1\right)^2-\left(4x^3-4x\right)^2\)
\(=\left(x^4+4x^3+8x^2-4x+1\right)\left(x^4-4x^3+8x^2+4x+1\right)\)
d)\(x^4+6x^3+7x^2-6x+1\)
\(=x^4+3x^3-x^2+3x^3+9x^2-3x-x^2-3x+1\)
\(=x^2\left(x^2+3x-1\right)+3x\left(x^2+3x-1\right)-\left(x^2+3x-1\right)\)
\(=\left(x^2+3x-1\right)\left(x^2+3x-1\right)\)\(=\left(x^2+3x-1\right)^2\)
a) \(x^4-4x^{3^{ }}+8x+3\)
\(=\left(x^4+x^3\right)-\left(5x^3+5x^2\right)+\left(5x^2+5x\right)+\left(3x+3\right)\)
\(=x^{3^{ }}\left(x+1\right)-5x^{2^{ }}\left(x+1\right)+5x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3-5x^2+5x+3\right)\)
\(=\left(x+1\right)\left[\left(x^3-3x^2\right)-\left(2x^2-6x\right)-\left(x-3\right)\right]\)
\(=\left(x+1\right)\left[x^2\left(x-3\right)-2x\left(x-3\right)-\left(x-3\right)\right]\)
\(=\left(x+1\right)\left(x-3\right)\left(x^2-2x-1\right)\)
\(=\left(x+1\right)\left(x-3\right)\left[\left(x-1\right)^2-2\right]\)
\(=\left(x+1\right)\left(x-3\right)\left(x-1-\sqrt{2}\right)\left(x-1+\sqrt{2}\right)\)
b, \(x^2\left(y^2-4\right)^2-6x\left(y^2-4\right)+9\)
\(=\left[x\left(y^2-4\right)-3\right]^2\)
\(=\left(xy^2-4x-3\right)^2\)