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\(4x^4-21x^2y^2+y^4\)
\(=\left(4x^4+4x^2y^2+y^4\right)-25x^2y^2\)
\(=\left(2x^2+y^2\right)^2-\left(5xy\right)^2\)
\(=\left(2x^2+y^2-5xy\right)\left(2x^2+y^2+5xy\right)\)
\(a,4x^4-21x^2y^2+y^4=\left(2x^2\right)^2+4x^2y^2+y^4-4x^2y^2-21x^2y^2\)
\(=\left(2x^2+y^2\right)^2-25x^2y^2\)
\(=\left(2x^2+y^2-5xy\right)\left(2x^2+y^2+5xy\right)\)
\(b,x^5-5x^3+4x=x\left(x^4-5x^2+4\right)\)
\(=x\left(x^4-4x^2-x^2+4\right)\)
\(=x\left[x^2\left(x^2-4\right)-\left(x^2-4\right)\right]\)
\(=x\left(x^2-4\right)\left(x^2-1\right)\)
\(=x\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)
\(c,x^3+5x^2+3x-9=x^3-x^2+6x^2-6x+9x-9\)
\(=x^2\left(x-1\right)+6x\left(x-1\right)+9\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+6x+9\right)\)
\(=\left(x-1\right)\left(x^2+3x+3x+9\right)\)
\(=\left(x-1\right)\left[x\left(x+3\right)+3\left(x+3\right)\right]\)
\(=\left(x-1\right)\left(x+3\right)\left(x+3\right)\)
\(=\left(x-1\right)\left(x+3\right)^2\)
\(d,x^{16}+x^8-2=x^{16}+2x^8-x^8-2\)
\(=x^8\left(x^8-1\right)+2\left(x^8-1\right)\)
\(=\left(x^8-1\right)\left(x^8+2\right)\)
\(x^2-y^2\)
\(=x^2+xy-xy-y^2\)
\(=x\left(x+y\right)-y\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y\right)\)
a) \(x^3+3x^2+3x+1=\left(x+1\right)^3\)
b) \(x^3-6x^2+12x-8=\left(x-2\right)^3\)
c) \(x^2-2xy+y^2-16=\left(x-y\right)^2-4^2=\left(x-y+4\right)\left(x-y-4\right)\)
d) \(49-x^2+2xy-y^2=7^2-\left(x-y\right)^2=\left(7+x-y\right)\left(7-x+y\right)\)
\(x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2\right)^2-4x^2=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\)
\(4x^8+1=\left(2x^4\right)^2+1=\left(2x^4\right)^2-2.2x^4+1+2.2.x^4=\left(2x^4+1\right)^2-4x^4\)
\(=\left(2x^4+2x^2+1\right)\left(4x^4-2x^2+1\right)\)
\(x^2-8x-9==x^2+x-9x-9=x\left(x+1\right)-9\left(x+1\right)=\left(x+1\right)\left(x-9\right)\)
\(x^2+14x+48=x^2+6x+8x+48=x\left(x+6\right)+8\left(x+6\right)=\left(x+6\right)\left(x+8\right)\)
a) \(x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2\right)^2-4x^2=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\)
b) \(4x^8+1=\left(2x^4\right)^2+1=\left(2x^4\right)^2-2.2x^4+1+2.2.x^4=\left(2x^4+1\right)^2-4x^4\)
c) \(x^2-8x-9==x^2+x-9x-9=x\left(x+1\right)-9\left(x+1\right)=\left(x+1\right)\left(x-9\right)\)
d) \(x^2+14x+48=x^2+6x+8x+48=x\left(x+6\right)+8\left(x+6\right)=\left(x+6\right)\left(x+8\right)\)
\(x^2+2xy+x+2y\)
\(=x\left(x+1\right)+2y\left(x+1\right)\)
\(=\left(x+1\right)\left(2y+x\right)\)
\(7x^2-7xy-5x+5y\)
\(=7x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(7x-5\right)\)
a)x2+2xy+x+2y
=(2xy+x2)+(2y+x)
=x(2y+x)+(2y+x)
=(x+1)(2y+x)
b)7x2-7xy-5x+5y
=(5y-7xy)+(7x2-5x)
=y(5-7x)-x(5-7x)
=(5-7x)(y-x)
c)x2-6x+9-9y2
=(x2+3xy-3x)-(3xy+9y2-9y)-(3x+9y-9)
=x(x+3y-3)-3y(x+3y-3)-3(x+3y-3)
=(x-3y-3)(x+3y-3)
d)x3-3x2+3x-1+2(x2-x)
Ta thấy x=1 là nghiệm của đa thức
=>đa thức có 1 hạng tử là x-1
=(x-1)(x2+1)
e) (x+y)(y+z)(z+x)+xyz
đề sai
f)x(y2-z2)+y(z2-x2)
=(xy2+yz2)+(x2y+xz2)
=y(xy+z2)-x(xy+z2)
=(y-x)(xy+z2)
x3-x2+x+3=x3+1-x2+1+x+1
=(x+1)(x2+x+1)-(x2-1)+(x+1)
=(x+1)(x2+x+1)-(x+1)(x-1)+(x+1)
=(x+1)[(x2+x+1)-(x-1)+1]
=(x+1)(x2+x+1-x+1+1)
=(x+1)(x2+3)
ồ cuk dễ nhỉ
Nếu các bn thích thì ...........
cứ cho NTN này nhé !
=(y+x-6)(y+x-2)
Đặt \(x+y=u\)
Biểu thức trở thành \(u^2-8u+12\)
\(=u^2-2u-6u+12\)
\(=u\left(u-2\right)-6\left(u-2\right)\)
\(=\left(u-6\right)\left(u-2\right)\)
Thay ngược trở lại, ta được:
\(\left(x+y\right)^2-8\left(x+y\right)+12=\left(x+y-6\right)\left(x+y-2\right)\)