Phân tích đa thưc sau thành nhân tử
a] x^2-14x+24
b)36-12x+x^2
c)-18-x^2-9x
d)4-x^4
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\(x^2-2xy+x-2y=x\left(x-2y\right)+x-2y=\left(x-2y\right)\left(x+1\right)\)
\(3x^3+6x+3-3y^2=3\left[\left(x^2+2x+1\right)-y^2\right]=3\left[\left(x+1\right)^2-y^2\right]=3\left(x-y+1\right)\left(x+y+1\right)\)
\(a,Sửa:x^2-xy-13x+13y=x\left(x-y\right)-13\left(x-y\right)=\left(x-13\right)\left(x-y\right)\\ b,=\left(x+y\right)^2-\left(2z\right)^2=\left(x+y-2z\right)\left(x+y+2z\right)\\ c,=\left(x^2-2x\right)-\left(3x-6\right)=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
\(a,=\left(2y^2-1\right)\left(2y^2+1\right)\\ b,=\left(x+y\right)^2-9=\left(x+y+3\right)\left(x+y-3\right)\)
Lời giải:
a. $4y^4-1=(2y^2)^2-1^2=(2y^2-1)(2y^2+1)$
b. $x^2+2xy-9+y^2=(x^2+2xy+y^2)-9$
$=(x+y)^2-3^2=(x+y-3)(x+y+3)$
a) \(7\left(3x-2\right)+y\left(3x-2\right)=\left(3x-2\right)\left(7+y\right)\)
b) \(x\left(y-x\right)-3\left(x-y\right)=x\left(y-x\right)+3\left(y-x\right)=\left(y-x\right)\left(x+3\right)\)
c) \(x^2-6xy+9y^2=\left(x-3y\right)^2\)
Ta có: \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(=\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)-4\)
\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
\(=\left(12x^2+11x\right)^2+\left(12x^2+11x\right)-6\)
\(=\left(12x^2+11x+3\right)\left(12x^2+11x-2\right)\)
\(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\\ =\left(x+4\right)^2\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\\ =\left(x-1\right)\left(x+1\right)\left(x+4-1\right)\left(x+4+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x+5\right)\)
cộng bằng -14 bạn ạ, quên hả
a)x2-14x+24=(x2-2x)-(12x-24)=x(x-2)-12(x-2)=(x-2)(x-12)
b)36-12x+x2=(6-x)2
c)-18-x2-9x=-(x2+9x+18)=-[(x2+3x)+(6x+18)]=-[x(x+3)+6(x+3)]=-(x+3)(x+6)
d)4-x4=(2-x2)(2+x2)