Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\left(1+x^2\right)^2-4x\left(1-x^2\right)\\ =\left(1+x^2\right)^2-\left(\sqrt{4x\left(1-x^2\right)}\right)^2\\ =\left(1+x^2-\sqrt{4x\left(1-x^2\right)}\right)\left(1+x^2+\sqrt{4x\left(1-x^2\right)}\right)\)
\(b,\left(x^2-8\right)^2-36\\ =\left(x^2-8-6\right)\left(x^2-8+6\right)\\ =\left(x^2-14\right)\left(x^2-2\right)\)
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)
\(1,\\ a,\left(x^2+1\right)^2-4x\left(1-x^2\right)\\ =\left(x^2+1\right)^2+4x\left(x^2+1\right)\\ =\left(x^2+1\right)\left(x^2+1+4x\right)\\ b,\left(x^2-8\right)^2+36\\ =x^4-16x^2+64+36\\ =x^4-16x^2+100\\ =x^4-20x^2+100-4x^2\\ =\left(x^2-10\right)^2-4x^2\\ =\left(x^2-10-2x\right)\left(x^2-10+2x\right)\\ c,81x^4+4=81x^4+36x^2+4-36x^2\\ =\left(9x^2+2\right)^2-36x^2\\ =\left(9x^2+2-6x\right)\left(9x^2+2+6x\right)\\ d,x^5+x+1\\ =x^5+x^4+x^3-\left(x^4+x^3+x^2\right)+x^2+x+1\\ =x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+x^2+x+1\\ =\left(x^3-x^2+1\right)\left(x^2+x+1\right)\)
\(2,\\ a,x^3-7x-6\\ =x^3+x^2-x^2-x-6x-6\\ =x^2\left(x+1\right)-x\left(x+1\right)-6\left(x+1\right)\\ =\left(x+1\right)\left(x^2-x-6\right)\\ b,x^3+4x^2-7x-10\\ =x^3-2x^2+6x^2-12x+5x-10\\ =x^2\left(x-2\right)+6x\left(x-2\right)+5\left(x-2\right)\\ =\left(x-2\right)\left(x^2+6x+5\right)\\ =\left(x-2\right)\left(x^2+5x+x+5\right)\\ =\left(x-2\right)\left[x\left(x+5\right)+\left(x+5\right)\right]\\ =\left(x-2\right)\left(x+5\right)\left(x+1\right)\)
A = 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????
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)
\(a,...3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)\(=3\left(x^4+x^2+1\right)-\left(\left(x^4+x^2+1\right)+2\left(x^3+x^2+x\right)\right)\)
\(2\left(x^4+x^2+1\right)-2\left(x^3+x^2+x\right)=2\left(x^4-x^3-x+1\right)\) \(2\left(x^3\left(x-1\right)-\left(x-1\right)\right)=2\left(x-1\right)\left(x^3-1\right)\)
\(2\left(x-1\right)^2\left(x^2+x+1\right)\)
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 , \(81x^2y+18xy^2+27x^2y^2\)\(=9xy\left(9x+2y+3xy\right)\)
b. \(4x^3+x^2+x=x\left(4x^2+x+1\right)\)
c. \(x^6+y^6=\left(x^2\right)^3+\left(y^2\right)^3\)\(=\left(x^2+y^2\right)\left(x^4-x^2y^2+y^4\right)\)
d.
e. \(\left(x+y\right)^3-\left(x-3\right)^3\)\(=x^3+3x^2y+3xy^2+y^3-x^3+9x^2-27x-y^3\)
\(=3x^2y+3xy^2+9x^2-27x\)
\(=3x\left(xy+y^2+3x-9\right)\)
h. \(x^2+x+\frac{1}{4}=\)\(4x^2+4x+1=\left(2x+1\right)^2=\left(2x+1\right)\left(2x+1\right)\)
i.
a) \(x^4+x^2+1\)
\(=x^2\left(x^2+1\right)+\left(x^2+1\right)-x^2\)
\(=\left(x^2+1\right)^2-x^2\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\)
b) \(\left(1+x\right)^2-4x\left(1-x^2\right)\)
\(=\left(1+x\right)^2-4x\left(1-x\right)\left(1+x\right)\)
\(=\left(1+x\right)\left[1+x-4x+4x^2\right]\)
\(=\left(1+x\right)\left[1-3x+4x^2\right]\)
a) x4 + x2 + 1
= [ ( x2 )2 + 2x2 + 1 ] - x2
= ( x2 + 1 ) - x2
= ( x2 + 1 - x2 )( x2 + 1 + x2 )
= 2x2 + 1