Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^5+x^4+1\)
\(=x^5-x^3-x^2-x^4+x^2+x+x^3-x-1\)
\(=x^2\left(x^2-x-1\right)-x\left(x^3-x-1\right)+\left(x^3-x-1\right)\)
\(=\left(x^2-x+1\right)\left(x^3-x-1\right)\)
\(x^5+x^4+1=x^5+x^4+x^3-x^3+1=x^3\left(x^2+x+1\right)-\left(x^3-1\right)=x^3\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)=\left(x^3-x+1\right)\left(x^2+x+1\right)\)
x^10 + x^5 + 1
= x^10 + x^9 - x^9 + x^8 - x^8 + x^7 - x^7 + x^6 - x^6 + x^5 + x^5 - x^5 + x^4 - x^4 + x^3 - x^3 + x^2 - x^2 + x - x + 1
= (x^10 + x^9 + x^8) - (x^9 + x^8 + x^7) + (x^7 + x^6 + x^5) - (x^6 + x^5 + x^4) + (x^5 + x^4 + x^3) - (x^3 + x^2 + x) + (x^2 + x + 1)
= x^8 (x^2 + x + 1) - x^7 (x^2 + x + 1) + x^5 (x^2 + x + 1) - x^4 (x^2 + x + 1) + x^3 (x^2 + x + 1) - x (x^2 + x + 1) + (x^2 + x + 1)
= (x^2 + x + 1) (x^8 - x^7 + x^5 - x^4 + x^3 - x + 1)
Ta có tổng quát: \(\left(ax^2+bx+c\right)\)\(\left(mx^2+nx+p\right)\)\(\circledast\)
-Nhân ra ta được: \(amx^4+\left(an+bm\right)x^3+\left(ap+bn+cm\right)x^2+\left(bp+cn\right)x+cp\)
-Áp dụng phương pháp hệ số bất định, ta có:
am=1
an+bm=4 (1)
ap+bn+cm=6 (2)
bp+cn=4 (3)
cp=5
-Xét a=m=1 và c=1, p=5
thay vào (1), ta được: n+b=4 (4)
thay vào (3), ta được: n+5b=4 (5)
từ (4),(5)\(\Rightarrow\)n=4 và b=0
giờ thay tất cả vào phương trình (3), ta được: 5+0+1=6 (T/M)
\(\Rightarrow\)Thay vào\(\circledast\), ta được: \(\left(x^2+1\right)\left(x^2+4x+5\right)\)
Cách 2: Ta tách \(6x^2\) thành \(5x^2+x^2\)
ta được: \(x^4+4x^3+5x^2+x^2+4x+5\)
\(\Leftrightarrow x^2\left(x^2+4x+5\right)+\left(x^2+4x+5\right)\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2+4x+5\right)\)
a)\(\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)
b)\(\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)
c)\(\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)
a) =x3-2x2+6x2-12x -12x +24
= x2(x-2)+6x(x-2)-12(x-2)
= (x-2)(x2+6x-12)
mk giải đc câu a thôi, bn zô jup mk lại vs
\(a,x^3+4x^2-24x+24\)
\(=x^3+6x^2-12x-2x^2-12x+24\)
\(=\left(x^3-2x^2\right)+\left(6x^2-12x\right)-\left(12x-24\right)\)
\(=x^2\left(x-2\right)+6x\left(x-2\right)-12\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+6x-12\right)\)
1, \(x^3+8x^2+17x+10=\left(x^3+x^2\right)+\left(7x^2+7x\right)+\left(10x+10\right)\)
\(=x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)\)\(=\left(x+1\right)\left(x^2+7x+10\right)=\left(x+1\right)\left(x+2\right)\left(x+5\right)\)
2. \(2x^3-3x^2+3x-1=\left(2x^3-x^2\right)-\left(2x^2-x\right)+\left(2x-1\right)\)
\(=x^2\left(2x-1\right)-x\left(2x-1\right)+\left(2x-1\right)\)
\(=\left(2x-1\right)\left(x^2-x+1\right)\)
3. \(x^4+x^2+1=\left(x^4+1\right)+x^2=\left(x^2+1\right)^2-2x^2+x^2\)\(=\left(x^2+1\right)^2-x^2=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
4. \(81x^4+4=\left(9x^2\right)^2+2^2=\left(9x^2+2\right)^2-2.9x^2.2=\left(9x^2+2\right)^2-\left(6x\right)^2\)
\(=\left(9x^2+6x+2\right)\left(9x^2-6x+2\right)\)
\(a,\left(x^2+y^2-5\right)^2-4x^2y^2-16xy-16\)
\(=\left(x^2+y^2-5\right)^2-4\left(x^2y^2-4xy-4\right)\)
\(=\left(x^2+y^2-5\right)^2-4\left(xy+2\right)^2\)
\(=\left(x^2+y^2-5\right)^2-\left[2xy+4\right]^2\)
\(=\left(x^2+y^2-5+2xy+4\right)\left(x^2+y^2-5-2xy-4\right)\)
\(=\left[\left(x^2+y^2+2xy\right)-1\right]\left[\left(x^2+y^2-2xy\right)-9\right]\)
\(=\left[\left(x+y\right)^2-1\right]\left[\left(x-y\right)^2-9\right]\)
\(=\left(x+y-1\right)\left(x+y+1\right)\left(x-y-3\right)\left(x-y+3\right)\)
\(b,x^3+5x^2+8x+4\)
\(=x^3+x^2+4x^2+8x+4\)
\(=x^2\left(x+1\right)+4\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)+4\left(x+1\right)^2\)
\(=\left(x+1\right)\left[\left(x^2+4\right)\left(x+1\right)\right]\)
\(=\left(x+1\right)\left(x^2+4x+4\right)\)
\(=\left(x+1\right)\left(x+2\right)^2\)
\(c,x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-x-6\right)\)
\(=\left(x-5\right)\left[x^2+2x-3x-6\right]\)
\(=\left(x-5\right)\left[x\left(x+2\right)-3\left(x+2\right)\right]\)
\(=\left(x-5\right)\left(x-3\right)\left(x+3\right)\)
\(d,125x^3-10x^2+2x-1\)
\(=\left(125x^3-1\right)-\left(10x^2-2x\right)\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)-2x\left(5x-1\right)\)
\(=\left(5x-1\right)\left(25x^2+5x+1-2x\right)\)
\(=\left(5x-1\right)\left(25x^2+3x+1\right)\)
