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\(A=\left(x^2+x\right)^2-14\left(x^2+x\right)+24\)
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:
\(F=t\left(t+1\right)-12\)
\(=t^2+t-12\)
\(=t^2+4t-3t-12\)
\(=t\left(t+4\right)-3\left(t+4\right)\)
\(=\left(t+4\right)\left(t-3\right)\)
\(=\left(x^2+x+1+4\right)\left(x^2+x+1-3\right)\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(E=x^4+2x^3+5x^2+4x-12\)
\(=x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12\)
\(=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+3x^2+8x+12\right)\)
\(=\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)\)
\(=\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
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)
bằng phương pháp nào zậy bn????
547675675675678768768789980957457346242645657
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)
A = x2(x - 1) + 6(1 - x)
A = x3 - x2 + 6 - 6x
A = (x3 - 6x) - (x2 - 6)
A = x.(x2 - 6) - (x2 - 6)
A = (x - 1)(x2 - 6)
C = x2 + 2xy + y2 - yz - xz
C = (x + y)2 - z.(x + y)
C = (x + y - z).(x + y)
\(1,\\ a,\left(x^2+1\right)^2-4x\left(1-x^2\right)\\ =\left(x^2+1\right)^2+4x\left(x^2+1\right)\\ =\left(x^2+1\right)\left(x^2+1+4x\right)\\ b,\left(x^2-8\right)^2+36\\ =x^4-16x^2+64+36\\ =x^4-16x^2+100\\ =x^4-20x^2+100-4x^2\\ =\left(x^2-10\right)^2-4x^2\\ =\left(x^2-10-2x\right)\left(x^2-10+2x\right)\\ c,81x^4+4=81x^4+36x^2+4-36x^2\\ =\left(9x^2+2\right)^2-36x^2\\ =\left(9x^2+2-6x\right)\left(9x^2+2+6x\right)\\ d,x^5+x+1\\ =x^5+x^4+x^3-\left(x^4+x^3+x^2\right)+x^2+x+1\\ =x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+x^2+x+1\\ =\left(x^3-x^2+1\right)\left(x^2+x+1\right)\)
\(2,\\ a,x^3-7x-6\\ =x^3+x^2-x^2-x-6x-6\\ =x^2\left(x+1\right)-x\left(x+1\right)-6\left(x+1\right)\\ =\left(x+1\right)\left(x^2-x-6\right)\\ b,x^3+4x^2-7x-10\\ =x^3-2x^2+6x^2-12x+5x-10\\ =x^2\left(x-2\right)+6x\left(x-2\right)+5\left(x-2\right)\\ =\left(x-2\right)\left(x^2+6x+5\right)\\ =\left(x-2\right)\left(x^2+5x+x+5\right)\\ =\left(x-2\right)\left[x\left(x+5\right)+\left(x+5\right)\right]\\ =\left(x-2\right)\left(x+5\right)\left(x+1\right)\)
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)
\(a,...3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)\(=3\left(x^4+x^2+1\right)-\left(\left(x^4+x^2+1\right)+2\left(x^3+x^2+x\right)\right)\)
\(2\left(x^4+x^2+1\right)-2\left(x^3+x^2+x\right)=2\left(x^4-x^3-x+1\right)\) \(2\left(x^3\left(x-1\right)-\left(x-1\right)\right)=2\left(x-1\right)\left(x^3-1\right)\)
\(2\left(x-1\right)^2\left(x^2+x+1\right)\)
\(a,\left(1+x^2\right)^2-4x\left(1-x^2\right)\\ =\left(1+x^2\right)^2-\left(\sqrt{4x\left(1-x^2\right)}\right)^2\\ =\left(1+x^2-\sqrt{4x\left(1-x^2\right)}\right)\left(1+x^2+\sqrt{4x\left(1-x^2\right)}\right)\)
\(b,\left(x^2-8\right)^2-36\\ =\left(x^2-8-6\right)\left(x^2-8+6\right)\\ =\left(x^2-14\right)\left(x^2-2\right)\)
Theo mk là trừ 36 nhé
a) = (9x2)2 +22 = (9x2)2 + 2×9x2×2 + 22 - 36x2 = (9x2+2)2 -(6x)2 = (9x2+2+6x)(9x2+2-6x)
b) = x4 - 16x2 + 64+36= x4-16x2+100=x4+20x2+100-36x2= (x2+10)2 - (6x)2= (x2-6x+10)(x2-6x+10)
d) Đặt a=x2+x,ta có: a2+4a-12 = a2-2a+6a-12= a(a-2)+6(a-2) = (a+6)(a-2) tương đương (x2+x+6)(x2+x-2)=(x2+x+6)(x-1)(x+2)
e)Đặt a=x2+x+1(a>0), ta có: a(a+1)-12= a2+a-12= a2 -3a+4a -12= a(a-3) +4(a-3)=(a+4)(a-3) tương đương (x2+x+1+4)(x2+x+1-3) =( x2 +x+4)(x2+x-2)=(x2+x+4)(x-1)(x+2)
ta có: 81x^4 + 4
= (3x)^4 + 2^2
= (9x^2)^2 + 2.9x^2.2+2^2- 36x^2
= (9x^2)^2 + 36x^2 + 2^2 -36x^2
= (9x^2+2)^2 - (6x)^2
= (9x^2+2-6x)( 9x^2 +2 +6x)