phân tích đa thức thành nhân tử: x3-x/49
mk cần gấp...cảm ơn nhiều!!! :)
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= x^2(X-1) - 4(x^2-2x+1)
=x^2(x-1)-4(x-1)^2
=(x-1)(x^2-4x+4)
=(x-1)(x-2)^2
Lời giải:
1.
$x^3+3x^2-16x-48=(x^3+3x^2)-(16x+48)=x^2(x+3)-16(x+3)$
$=(x+3)(x^2-16)=(x+3)(x-4)(x+4)$
2.
$4x(x-3y)+12y(3y-x)=4x(x-3y)-12y(x-3y)=(x-3y)(4x-12y)=4(x-3y)(x-3y)=4(x-3y)^2$
3.
$x^3+2x^2-2x-1=(x^3-x^2)+(3x^2-3x)+(x-1)=x^2(x-1)+3x(x-1)+(x-1)$
$=(x-1)(x^2+3x+1)$
\(x^2-8x-9\)
\(=x^2+x-9x-9\)
\(=\left(x^2+x\right)-\left(9x+9\right)\)
\(=x\left(x+1\right)-9\left(x+1\right)\)
\(=\left(x-9\right)\left(x+1\right)\)
\(x^2-8x-9\)
\(=x^2+x-9x-9\)
\(=\left(x^2+x\right)-\left(9x+9\right)\)
\(=x\left(x+1\right)-9\left(x+1\right)\)
\(=\left(x-9\right)\left(x+1\right)\)
\(x^6-y^6\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
hk
tốt
\(\left(x-y\right)^3-x^3+y^3=\left(x-y\right)^3-\left(x^3-y^3\right)=\left(x-y\right)^3-\left(x-y\right)\left(x^2+xy+y^2\right)=\left(x-y\right)\left(x^2-2xy+y^2-x^2-xy-y^2\right)=-3xy\left(x-y\right)\)
\(\left(x-y\right)^3-x^3+y^3\\ =\left(x-y\right)^3-\left(x^3-y^3\right)\\ =\left(x-y\right)^3-\left(x-y\right)\left(x^2+xy+y^2\right)\\ =\left(x-y\right)\left[\left(x-y\right)^2-\left(x^2+xy+y^2\right)\right]\\ =\left(x-y\right)\left(x^2-2xy+y^2-x^2-xy-y^2\right)\\ =\left(-3xy\right)\left(x-y\right)\)
(3x-1)^2 - 16 = (3x-1)^2 - 4^2
= (3x-1-4)(3x-1+4)
= (3x-5)(3x+3)
\(\left(3x-1\right)^2-16\)
\(=\left(3x-1\right)^2-4^2\)
\(=\left(3x-1-4\right)\left(3x-1+4\right)\)
\(=\left(3x-5\right)\left(3x+3\right)\)
\(=3\left(x+1\right)\left(3x-5\right)\)
(x^2-6x+8)(x^2-8x+15)+1
=(x^2-4x-2x+8)(x^2-5x-3x+15)+1
=(x(x-4)-2(x-4))(x(x-5)-3(x-5))+1
=(x-4)(x-2)(x-5)(x-3)+1
=(x-2)(x-5)(x-3)(x-4)+1
=(x^2-7x+10)(x^2-7x+12)+1
Gọi a=x^2-7x+11, ta có
(a-1)(a+1)+1
= a2 - 1 + 1
= a2
= (x2 - 7x + 11)2