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Ta có : x3 - 7x + 6
= x3 - x - 6x + 6
= x(x2 - 1) - 6(x - 1)
= x(x + 1)(x - 1) - 6(x - 1)
= (x - 1) [x(x + 1) - 6]
= (x - 1) (x2 + x - 6) .
CÁC Ý SAU TƯƠNG TỰ
a) \(x^5+4x+5=\left(x^5+x^4\right)-\left(x^4+x^3\right)+\left(x^3+x^2\right)-\left(x^2+x\right)+\left(5x+5\right)=x^4\left(x+1\right)-x^3\left(x+1\right)+x^2\left(x+1\right)-x\left(x+1\right)+5\left(x+1\right)=\left(x^4-x^3+x^2-x+5\right)\left(x+1\right)\)
b) \(x^4+6x^3+11x^2+6x+1=\left(x^4+3x^3+x^2\right)+\left(3x^3+9x^2+3x\right)+\left(x^2+3x+1\right)=x^2\left(x^2+3x+1\right)+3x\left(x^2+3x+1\right)+\left(x^2+3x+1\right)=\left(x^2+3x+1\right)^2\)
c) \(64x^4+1=\left[\left(8x^2\right)^2+16x^2+1\right]-16x^2=\left(8x^2+1\right)^2-\left(4x\right)^2=\left(8x^2-4x+1\right)\left(8x^2+4x+1\right)\)d) \(81x^4+4=\left[\left(9x^2\right)^2+36x^2+2^2\right]-36x^2=\left(9x^2+2\right)^2-\left(6x\right)^2=\left(9x^2-6x+2\right)\left(9x^2+6x+2\right)\)
1
x3-7x+6
=x3+0x2-7x +6
= x3-x2+x2-x-6x+6
=(x3-x2)+(x2-x)-(6x-6)
=x2(x-1)+x(x-1)-6(x-1)
=(x-1)(x2+x-6)
=(x-1)(x2+3x-2x-6)
=(x-1)[x(x+3)-2(x+3)]
=(x-1)(x-2)(x+3)
7) (x+2)(x+3)(x+4)(x+5)-24
=(x+2)(x+5) (x+3)(x+4)-24
=[x(x+5)+2(x+5)][x(x+4)+3(x+4)]-24
=[x2+5x+2x+10][x2+4x+3x+12]-24
=[x2+7x+10][x2+7x+12]-24
đặt a=x2+7x+10
=>x2+7x+12=a+2
=a(a+2)-24
=a2+2a-24
=a2+6a-4a-24
=(a2+6a)-(4a+24)
=a(a+6)-4(a+6)
=(a+6)(a-4)
thay a= x2+7x+10 vào ta được
(x2+7x+10+6)(x2+7x+10-4)
=(x2+7x+16)(x2+7x+6)
a)\(x^3+3xy+y^3-1\)
\(=x^3+3x^2y+3xy^2+y^3-1-3x^2y-3xy^2+3xy\)
\(=\left(x+y\right)^3-1^3-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1\right)-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1-3xy\right)\)
\(=\left(x+y-1\right)\left(x^2-xy+y^2+x+y+1\right)\)
b) Đặt \(B=3x^2+22xy+11x+37y+7y^2+10\)
Giả sử \(B=\left(ax+by+c\right)\left(mx+ny+p\right)\)
\(=amx^2+anxy+apx+bmxy+bny^2+bpy+cmx+cny+cp\)
\(=amx^2+\left(an+bm\right)xy+\left(ap+cm\right)x+bny^2+\left(bp+cn\right)y+cp\)
Ta được hệ: \(\left\{{}\begin{matrix}am=3;an+bm=22\\ap+cm=11;bn=7\\bp+cn=37;cp=10\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=3;b=1\\c=5;m=1\\n=7;p=2\end{matrix}\right.\)
Vậy B phân tích được thành \(\left(3x+y+5\right)\left(x+7y+2\right)\).
a/ =(x+y)3-1-3xy(x+y-1)
=(x+y-1)(x2+2xy+y2+xy+1)-3xy(x+y-1)
=(x+y-1)(x2+y2+1)
mơn nha
\(x^8+x+1\)
\(=x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)
\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)
a)\(2x^3-x^2+5x+3=2x^3-2x^2+x^2+6x-x+3\)
\(=\left(2x^3-2x^2+6x\right)+\left(x^2-x+3\right)\)
\(=2x\left(x^2-x+3\right)+\left(x^2-x+3\right)\)
\(=\left(x^2-x+3\right)\left(2x+1\right)\)
b)\(27x^3-27x^2+18x-4=27x^3-18x^2-9x^2+12x+6x-4\)
\(=3x\left(9x^2-6x+4\right)-\left(9x^2-6x+4\right)\)
\(=\left(9x^2-6x+4\right)\left(3x-1\right)\)
c)\(4x^4-32x^2+1=4x^4+4x^2-36x^2+1\)
\(=\left(4x^4+4x^2+1\right)-36x^2\)
\(=\left(2x^2+1\right)^2-\left(6x\right)^2\)
\(=\left(2x^2+1+6x\right)\left(2x^2+1-6x\right)\)
a)x^2-(a+b)x+ab
= x^2 - ax - bx + ab
= (x^2 - ax) - (bx - ab)
= x(x-a) - b(x-a)
= (x-b)(x-a)
b)7x^3-3xyz-21x^2+9z
=
c)4x+4y-x^2(x+y)
= 4(x + y) - x^2(x+y)
= (4-x^2) (x+y)
= (2-x)(2+x)(x+y)
d) y^2+y-x^2+x
= (y^2 - x^2) + (x+y)
= (y-x)(y+x)+ (x+y)
= (y-x+1) (x+y)
e)4x^2-2x-y^2-y
= [(2x)^2 - y^2] - (2x +y)
= (2x-y)(2x+y) - (2x+y)
= (2x -y -1)(2x+y)
f)9x^2-25y^2-6x+10y
=
a,
\(3(x^{4}+x^{2}+1)-(x^{2}+x+1)^{2}\\=3[(x^{4}+2x^{2}+1)-x^{2}]-(x^{2}+x+1)^{2}\\=3[(x^{2}+1)^{2}-x^{2}]-(x^{2}+x+1)^{2}\\=3(x^{2}+x+1)(x^{2}-x+1)-(x^{2}+x+1)^{2}\\=(x^{2}+x+1)(3x^{2}-3x+3-x^{2}-x-1)\\=(x^{2}+x+1)(2x^{2}-4x+2)\\=2(x^{2}+x+1)(x^{2}-2x+1)\\=2(x^{2}+x+1)(x-1)^{2}\)
bạn còn làm đk câu nào k giúp mình với