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a) \(4x^3y-12x^2y^3-8x^4y^3\)
\(=4x^2y\left(x-3y^2-2x^2y^2\right)\)
b) \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x-y+1\right)\left(x+y+1\right)\)
c) \(x^3-2x^2+x-xy^2\)
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-y-1\right)\left(x+y-1\right)\)
d) \(x\left(x-2y\right)+3\left(2y-x\right)\)
\(=x\left(x-2y\right)-3\left(x-2y\right)\)
\(=\left(x-3\right)\left(x-2y\right)\)
e) \(x^2+4\)
\(=\left(x^4+4x^2+4\right)-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
f) \(5x^2-7x-6\)
\(=\left(5x^2-10x\right)+\left(3x-6\right)\)
\(=5x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(5x+3\right)\left(x-2\right)\)
bÀI LÀM
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
\(A=\left(x^2+x\right)^2-14\left(x^2+x\right)+24\)
Đặt \(x^2+x=t\), ta có:
\(A=t^2-14t+24\)
\(=t^2-2t-12t+24\)
\(=t\left(t-2\right)-12\left(t-2\right)\)
\(=\left(t-2\right)\left(t-12\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x-12\right)\)
\(B=\left(x^2+x\right)^2+4x^2+4x-12\)
\(=\left(x^2+x\right)^2+4\left(x^2+x\right)-12\)
Đặt \(x^2+x=t\), ta có:
\(B=t^2+4t-12\)
\(=t^2+6t-2t-12\)
\(=t\left(t+6\right)-2\left(t+6\right)\)
\(=\left(t+6\right)\left(t-2\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(C=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)
Đặt \(x^2+5x+4=t\), ta có:
\(C=t\left(t+2\right)+1\)
\(=t^2+2t+1\)
\(=\left(t+1\right)^2\)
\(=\left(x^2+5x+4+1\right)^2\)
\(=\left(x^2+5x+5\right)^2\)
\(D=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(x^2+8x+7=t\), ta có:
\(D=t\left(t+8\right)+15\)
\(=t^2+8t+15\)
\(=t^2+3t+5t+15\)
\(=t\left(t+3\right)+5\left(t+3\right)\)
\(=\left(t+3\right)\left(t+5\right)\)
\(=\left(x^2+8x+7+3\right)\left(x^2+8x+7+5\right)\)
\(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)
\(F=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
Đặt \(x^2+x+1=t\), ta có:
\(F=t\left(t+1\right)-12\)
\(=t^2+t-12\)
\(=t^2+4t-3t-12\)
\(=t\left(t+4\right)-3\left(t+4\right)\)
\(=\left(t+4\right)\left(t-3\right)\)
\(=\left(x^2+x+1+4\right)\left(x^2+x+1-3\right)\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(E=x^4+2x^3+5x^2+4x-12\)
\(=x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12\)
\(=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+3x^2+8x+12\right)\)
\(=\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)\)
\(=\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
a, 4x2 - 12x + 9
= (2x + 3)2
b, 9x4y3 + 3x2y4
= 3x2y3(3x2 + y)
c, ( x - 3 )2 - 2x ( x - 3 )
= (x - 3)(x - 3 - 2x)
= (x - 3)(-x - 3)
d, 3x ( x - 1 ) + 6 ( x - 1 )
= 3(x - 1)(x + 2)
e, 2x ( x + 1 ) - 4x - 4
= 2x(x + 1) - 4(x + 1)
= (x + 1)(2x - 4)
= 2(x + 1)(x - 2)
f, ( 2x - 3 )2 - 4x + 6
= (2x - 3)2 - 2(2x - 3)
= (2x - 3)(2x - 3 - 2)
= (2x - 3)(2x - 5)
bằng phương pháp nào zậy bn????
547675675675678768768789980957457346242645657
b/ 4x4 + 4x3 + 5x2 + 2x + 1
= (4x4 + 4x3 + x2) + 2(2x2 + x) + 1
= (2x2 + x)2 + 2(2x2 + x) + 1
= (2x2 + x + 1)2
c/ x8 + x + 1 = (x2 + x + 1)(x6 - x5 + x3 - x2 + 1)
e/ x4 - 8x + 63 = (x2 - 4x + 7)(x2 + 4x + 9)
\(a,...3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)\(=3\left(x^4+x^2+1\right)-\left(\left(x^4+x^2+1\right)+2\left(x^3+x^2+x\right)\right)\)
\(2\left(x^4+x^2+1\right)-2\left(x^3+x^2+x\right)=2\left(x^4-x^3-x+1\right)\) \(2\left(x^3\left(x-1\right)-\left(x-1\right)\right)=2\left(x-1\right)\left(x^3-1\right)\)
\(2\left(x-1\right)^2\left(x^2+x+1\right)\)
A) 1/2 x(x^2-4)+4(x+2)
=1/2x(x-2)(x+2)+4(x+2)
=(x+2)(1/2x^2-x+4)
b) 21(x-y)^2-7(x-y)^3
= (x-y)^2(21-7x+7y)
=(x-y)^2.7(3-x+y)
c) 1/8x^3-3/4x^2+3/2x-1
=(1/2x)^3-3.(1/2x)^2.1+3.1/2x.1^2-1
=(1/2x-1)^3
a)\(x^3-x^2-x+1=\left(x^3-x\right)-\left(x^2-1\right)=x\left(x^2-1\right)-\left(x^2-1\right)=\left(x-1\right)^2.\left(x+1\right)\)
b)\(x^3+x^2-4x-4=x^2\left(x+1\right)-4\left(x+1\right)=\left(x^2-4\right)\left(x+1\right)=\left(x+2\right)\left(x-2\right)\left(x+1\right)\)
c)\(a^5+27a^2=a^2\left(a^3+27\right)=a^2\left(a+3\right)\left(a^2-3a+9\right)\)
d)\(x^4-8x=x\left(x^3-8\right)=x\left(x-2\right)\left(x^2+2x+4\right)\)
e)\(x^4-4x^3+4x^2=x^2\left(x^2-4x+4\right)=x^2\left(x-2\right)^2\)
f)\(2x^4-32=2\left(x^4-16\right)=2\left(x^2+4\right)\left(x^2-4\right)=2\left(x^2+4\right)\left(x+2\right)\left(x-2\right)\)
a) \(x^3-x^2-x+1\)
\(=x^2\left(x-1\right)-\left(x-1\right)\)
\(=\left(x^2-1\right)\left(x-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-1\right)=\left(x-1\right)^2\left(x+1\right)\)
b) \(x^3+x^2-4x-4\)
\(=x^2\left(x+1\right)-4\left(x+1\right)\)
\(=\left(x^2-4\right)\left(x+1\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x+1\right)\)
c) \(a^5+27a^2=a^2\left(a^3+27\right)\)
\(=a^2\left(a+3\right)\left(a^2-3a+9\right)\)
d) \(x^4-8x=x\left(x^3-8\right)\)
\(=x\left(x-2\right)\left(x^2+2x+4\right)\)
e) \(x^4-4x^3+4x^2\)
\(=\left(x^2\right)^2-2\cdot x^2\cdot2x+\left(2x\right)^2\)
\(=\left(x^2+2x\right)^2\)\(=\left[x\left(x+2\right)\right]^2=x^2\left(x+2\right)^2\)
f) \(2x^4-32=2\left(x^4-16\right)\)
\(=2\left(x^2-4\right)\left(x^2+4\right)\)
\(=2\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\)