phân tích da thức thành nhân tử :
(x^2+y^2-5)^2 -4x^2y^2-16xy-16
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(=\left(x^2+y^2-5\right)^2-4\left(xy-2\right)^2=\left(x^2+y^2-5+2xy-4\right)\left(x^2+y^2-5-2xy+4\right)\)
\(=\left(\left(x+y\right)^2-9\right)\left(\left(x-y\right)^2-1\right)=\left(x+y-3\right)\left(x+y+3\right)\left(x-y+1\right)\left(x-y-1\right)\)
câu a nek
(4x^2 -7x-50)^2 -4x^2 (4x^2+14x+49/4)
= (4x^2 -7x-50)^2 -(2x)^2 (2x+7/2)^2
= (4x^2 -7x-50)^2 - (4x^2+7x)^2
= (4x^2 -7x-50 +4x^2+7x) (4x^2 -7x-50-4x^2-7x)
= (8x^2-50) (-14x-50)
=2(4x^2-25)* (-2)(7x+25)
=-4 (2x-5)(2x+5)(7x+25)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
x6+3x4y2-8x3y3+3x2y4+y6= x6+3x4y2+3x2y4+y6-8x3y3=(x2+y2)3-(2xy)3
= (x2+y2-2xy)[(x2+y2)2+2xy(x2+y2)+(2xy)2]= (x-y)2(x4+6x2y2+y4+2x3y+2xy3)
(x2+y2-5)2-4x2y2-16xy-16=(x2+y2-5)2-(4x2y2+16xy+16)=(x2+y2-5)2-(2xy+4)2
=(x2+y2-5+2xy+4)(x2+y2-5-2xy-4)=(x2+2xy+y2-1)(x2-2xy+y2-9)=[(x+y)2-1][(x-y)2-32]=(x+y-1)(x+y+1)(x-y-3)(x-y+3)
x4+324=x4+36x2+324-36x2=(x2+18)2-(6x)2=(x2+18-6x)(x2+18+6x)