Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a/\)
\(4x-4y+x^2-2xy+y^2\)
\(=\left(4x-4y\right)+\left(x^2-2xy+y^2\right)\)
\(=4\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(4+x-y\right)\)
\(b/\)
\(x^4-4x^3-8x^2+8x\)
\(=\left(x^4+8x\right)-\left(4x^3+8x^2\right)\)
\(=x\left(x^3+8\right)-4x^2\left(x+2\right)\)
\(=x\left(x+2\right)\left(x^2-2x+4\right)-4x^2\left(x+2\right)\)
\(=x\left(x+2\right)\left(x^2-2x+4-4x\right)\)
\(=x\left(x+2\right)\left(x^2-6x-4\right)\)
\(d/\)
\(x^4-x^2+2x-1\)
\(=x^4-\left(x-1\right)^2\)
\(=\left(x^2+x-1\right)\left(x^2-x+1\right)\)
\(e/\)(Xem lại đề)
\(x^4+x^3+x^2+2x+1\)
\(=\left(x^4+x^3\right)+\left(x^2+2x+1\right)\)
\(=x^3\left(x+1\right)+\left(x+1\right)^2\)
\(=\left(x+1\right)\left(x^3+x+1\right)\)
\(f/\)
\(x^3-4x^2+4x-1\)
\(=x\left(x^2-4x+4\right)-1^2\)
\(=x\left(x-2\right)^2-1\)
\(=[\sqrt{x}\left(x-2\right)]^2-1\)
\(=[\sqrt{x}\left(x-2\right)-1][\sqrt{x}\left(x-2\right)+1]\)
\(c/\)
\(x^3+x^2-4x-4\)
\(=\left(x^3-2x^2\right)+\left(3x^2-6x\right)+\left(2x-4\right)\)
\(=x^2\left(x-2\right)+3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+3x+2\right)\)
\(=\left(x-2\right)[\left(x^2+x\right)+\left(2x+2\right)]\)
\(=\left(x-2\right)\left(x+1\right)\left(x+2\right)\)
a) \(4x-4y+x^2-2xy+y^2\)
\(=4\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(4+x-y\right)\)
b) \(x^4-4x^3-8x^2+8x\)
\(=x^4+2x^3-6x^3-12x^2+4x^2+8x\)
\(=x^3\left(x+2\right)-6x^2\left(x+2\right)+4x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^3-6x^2+4x\right)\)
\(=x\left(x+2\right)\left(x^2-6x+4\right)\)
c) \(x^3+x^2-4x-4\)
\(=x^3-2x^2+3x^2-6x+2x-4\)
\(=x^2\left(x-2\right)+3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+3x+2\right)\)
\(=\left(x-2\right)\left(x^2+2x+x+2\right)\)
\(=\left(x-2\right)\left[x\left(x+2\right)+\left(x+2\right)\right]\)
\(=\left(x-2\right)\left(x+2\right)\left(x+1\right)\)
d) \(x^4-x^2+2x-1\)
\(=x^4-\left(x^2-2x+1\right)\)
\(=x^4-\left(x-1\right)^2\)
\(=\left(x^2\right)^2-\left(x-1\right)^2\)
\(=\left(x^2-x+1\right)\left(x^2+x-1\right)\)
e)Sửa đề \(x^4+x^3+x^2-1\)
\(=x^3\left(x+1\right)+\left(x-1\right)\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+x-1\right)\)
f) \(x^3-4x^2+4x-1\)
\(=x^3-x^2-3x^2+3x+x-1\)
\(=x^2\left(x-1\right)-3x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-3x+1\right)\)
a,sửa đề : \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x^2-4}\right)\)
\(=\left(\frac{1}{\left(x+2\right)^2}-\frac{1}{\left(x-2\right)^2}\right):\left(\frac{x-2+1}{\left(x+2\right)\left(x-2\right)}\right)\)
\(=\left(\frac{x^2-4x+4-x^2-4x-4}{\left(x+2\right)^2\left(x-2\right)^2}\right):\left(\frac{x-1}{\left(x+2\right)\left(x-2\right)}\right)\)
\(=\frac{-8x\left(x+2\right)\left(x-2\right)}{\left(x+2\right)^2\left(x-2\right)^2\left(x-1\right)}=\frac{-8x}{\left(x-1\right)\left(x^2-4\right)}\)
b, \(\left(\frac{2x}{2x-y}-\frac{4x^2}{4x^2+4xy+y^2}\right):\left(\frac{2x}{4x^2-y^2}+\frac{1}{y-2x}\right)\)
\(=\left(\frac{2x}{2x-y}-\frac{4x^2}{\left(2x+y\right)^2}\right):\left(\frac{2x}{\left(2x-y\right)\left(2x+y\right)}-\frac{1}{2x-y}\right)\)
\(=\left(\frac{2x\left(2x+y\right)^2-4x^2\left(2x-y\right)}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{2x-\left(2x+y\right)}{\left(2x-y\right)\left(2x+y\right)}\right)\)
\(=\left(\frac{8x^3+8x^2y+2xy^2-8x^3+4x^2y}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{-y}{\left(2x-y\right)\left(2x+y\right)}\right)\)
\(=-\left(\frac{12x^2y+xy^2}{2x+y}\right)=\frac{-12x^2y-xy^2}{2x+y}\)
\(\left(\dfrac{x+1}{4x-4}\right)-\left(\dfrac{x-1}{4x+4}\right)+\left(\dfrac{x}{1-x^2}\right)\)
\(\left(\dfrac{x+1}{4.\left(x-1\right)}\right)-\left(\dfrac{x-1}{4\left(x+1\right)}\right)-\left(\dfrac{x}{\left(x-1\right)\left(x+1\right)}\right)\)
\(\left(\dfrac{\left(x+1\right)\left(x+1\right)}{4\left(x-1\right)\left(x+1\right)}\right)-\left(\dfrac{\left(x-1\right)\left(x-1\right)}{4\left(x+1\right)\left(x-1\right)}\right)-\left(\dfrac{4x}{4\left(x+1\right)\left(x-1\right)}\right)\)
\(\left(\dfrac{\left(x+1\right)^2}{4\left(x-1\right)\left(x+1\right)}\right)-\left(\dfrac{\left(x-1\right)^2}{4\left(x+1\right)\left(x-1\right)}\right)-\left(\dfrac{4x}{4\left(x+1\right)\left(x-1\right)}\right)\)
\(\dfrac{\left(x+1\right)^2-\left(x-1\right)^2-4x}{4\left(x+1\right)\left(x-1\right)}\)
\(\dfrac{\left(x+1-x+1\right).\left(x+1+x-1\right)-4x}{4.\left(x+1\right)\left(x-1\right)}\)
\(\dfrac{2.2x-4x}{4.\left(x+1\right)\left(x-1\right)}\)