Phân Tích Thành Nhân Tử:
a, x^11+x^4+1
b, x^7+x^4+1
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a: \(x^4+3x^3+x^2+3x\)
\(=x\left(x^3+3x^2+x+3\right)\)
\(=x\left(x+3\right)\left(x^2+1\right)\)
c: \(x^2-xy-x+y\)
\(=x\left(x-y\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left(x-1\right)\)
a) x2+x-2
= x2-x+2x-2
= x(x-1)+2(x-1)
= (x+2)(x-1)
b) 2x2+5x+3
= 2x2+2x+3x+3
= 2x(x+1)+3(x+1)
= (2x+3)(x+1)
c) 3x2+5x-2
= 3x2+6x-1x-2
= 3x(x+2)-1(x+2)
= (3x-1)(x+2)
\(x-y=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\)
a: \(=25x^2-4y^2+4y-1\)
\(=25x^2-\left(2y-1\right)^2\)
\(=\left(5x-2y+1\right)\left(5x+2y-1\right)\)
(3-x)4+(4-x)4-(7-2x)4
Đặt 3-x=a, 4-x=b=> 7-2x=a+b
=> (3-x)4+(4-x)4-(7-2x)4
=a4+b4-(a+b)4
=-(4a3b+6a2b2+4ab3)
=-2ab(a2+3ab+b2)
\(a,=\left(2y^2-1\right)\left(2y^2+1\right)\\ b,=\left(x+y\right)^2-9=\left(x+y+3\right)\left(x+y-3\right)\)
Lời giải:
a. $4y^4-1=(2y^2)^2-1^2=(2y^2-1)(2y^2+1)$
b. $x^2+2xy-9+y^2=(x^2+2xy+y^2)-9$
$=(x+y)^2-3^2=(x+y-3)(x+y+3)$
\(x^4-x^3-x+1=\left(x^4-x^3\right)-\left(x-1\right)=x^3\left(x-1\right)-\left(x-1\right)=\left(x^3-1\right)\left(x-1\right)=\left(x-1\right)^2.\left(x^2+x+1\right)\)
x4 - x3 - x + 1
= (x4 - x3) - (x - 1)
= x3(x - 1) - (x - 1)
= (x3 - 1)(x - 1)
a) = x * x * x * x * x * x *x *x * x * x * x
x^11+x^4+1
=x^11-x^2+x^4-x+x^2+x+1
=x^2(x^9-1)+x(x^3-1)+(x^2+x+1)
=x^2[(x^3-1)(x^6+x^3+1)]+x(x-1)(x^2+x+1)+(x^2+x+1)
=x^2(x-1)(x^2+x+1)(x^6+x^3+1)+x(x-1)(x^2+x+1)+(x^2+x+1)
=(x^2+x+1)[x^2(x-1)(x^6+x^3+1)+x(x-1)+1]
=(x^2+x+1)(x^9-x^8+x^6-x^5+x^3-x+1)