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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)
bằng phương pháp nào zậy bn????
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a)x4-4(x2+5)-25=x4-4x2-45=(x4-9x2)+(5x2-45)=x2(x2-9)+5(x2-9)=(x2-9)(x2+5)=(x-3)(x+3)(x2+5)
b)a2-b2-2a+1=(a2-2a+1)-b2=(a-1)2-b2=(a-b-1)(a+b-1)
c)x2-2x-4y2-4y=(x2-2x+1)-(4y2+4y+1)=(x-1)2-(2y+1)2=(x-1-2y-1)(x-1+2y+1)=(x-2y-2)(x+2y)
d)x2+4x-y2+4=(x2+4x+4)-y2=(x+2)2-y2=(x-y+2)(x+y+2)
\(1,\\ a,\left(x^2+1\right)^2-4x\left(1-x^2\right)\\ =\left(x^2+1\right)^2+4x\left(x^2+1\right)\\ =\left(x^2+1\right)\left(x^2+1+4x\right)\\ b,\left(x^2-8\right)^2+36\\ =x^4-16x^2+64+36\\ =x^4-16x^2+100\\ =x^4-20x^2+100-4x^2\\ =\left(x^2-10\right)^2-4x^2\\ =\left(x^2-10-2x\right)\left(x^2-10+2x\right)\\ c,81x^4+4=81x^4+36x^2+4-36x^2\\ =\left(9x^2+2\right)^2-36x^2\\ =\left(9x^2+2-6x\right)\left(9x^2+2+6x\right)\\ d,x^5+x+1\\ =x^5+x^4+x^3-\left(x^4+x^3+x^2\right)+x^2+x+1\\ =x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+x^2+x+1\\ =\left(x^3-x^2+1\right)\left(x^2+x+1\right)\)
\(2,\\ a,x^3-7x-6\\ =x^3+x^2-x^2-x-6x-6\\ =x^2\left(x+1\right)-x\left(x+1\right)-6\left(x+1\right)\\ =\left(x+1\right)\left(x^2-x-6\right)\\ b,x^3+4x^2-7x-10\\ =x^3-2x^2+6x^2-12x+5x-10\\ =x^2\left(x-2\right)+6x\left(x-2\right)+5\left(x-2\right)\\ =\left(x-2\right)\left(x^2+6x+5\right)\\ =\left(x-2\right)\left(x^2+5x+x+5\right)\\ =\left(x-2\right)\left[x\left(x+5\right)+\left(x+5\right)\right]\\ =\left(x-2\right)\left(x+5\right)\left(x+1\right)\)
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)
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)
\(a,...3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)\(=3\left(x^4+x^2+1\right)-\left(\left(x^4+x^2+1\right)+2\left(x^3+x^2+x\right)\right)\)
\(2\left(x^4+x^2+1\right)-2\left(x^3+x^2+x\right)=2\left(x^4-x^3-x+1\right)\) \(2\left(x^3\left(x-1\right)-\left(x-1\right)\right)=2\left(x-1\right)\left(x^3-1\right)\)
\(2\left(x-1\right)^2\left(x^2+x+1\right)\)
\(a,16x-5x^2-3\)
\(=-5x^2+15x+x-3\)
\(=-5x\left(x-3\right)+\left(x-3\right)\)
\(=\left(x-3\right)\left(-5x+1\right)\)
\(b,x^2-4x-5\)
\(=x^2+x-5x-5\)
\(=x\left(x+1\right)-5\left(x+1\right)\)
\(=\left(x+1\right)\left(x-5\right)\)
x2 - 4x - 5
= x2 - x + 5x - 5
= x ( x - 1 ) + 5 ( x - 1 )
= ( x - 1 ) ( x + 5 )
\(a,\left(1+x^2\right)^2-4x\left(1-x^2\right)\\ =\left(1+x^2\right)^2-\left(\sqrt{4x\left(1-x^2\right)}\right)^2\\ =\left(1+x^2-\sqrt{4x\left(1-x^2\right)}\right)\left(1+x^2+\sqrt{4x\left(1-x^2\right)}\right)\)
\(b,\left(x^2-8\right)^2-36\\ =\left(x^2-8-6\right)\left(x^2-8+6\right)\\ =\left(x^2-14\right)\left(x^2-2\right)\)
Theo mk là trừ 36 nhé