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\(A=\frac{y^3-x^3}{x^3-3x^2y+3xy^2-y^3}\)
\(A=\frac{\left(y-x\right)\left(y^2+xy+x^2\right)}{\left(x-y\right)^3}\)
\(A=\frac{-\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)\left(x-y\right)^2}\)
\(A=\frac{-x^2-xy-y^2}{x^2-2xy+y^2}\)
Ta có: \(\frac{y^2-x^2}{x^3-3x^2y+3xy^2-y^3}\)
= \(\frac{\left(y-x\right)\left(y+x\right)}{\left(x-y\right)^3}\)
=\(-\frac{x+y}{\left(x-y\right)^2}\)
=\(-\frac{x+y}{x^2-2xy+y^2}\)
\(E=\left(x^3+3xy^2+3x^2y+y^3\right)+3\left(x+y\right)-3\left(x^2+2xy+y^2\right)+2016\)
\(=\left(x+y\right)^3+3\left(x+y\right)-3\left(x+y\right)^2+2016\)
\(=21^3+3.21-3.21^2+2016\)
\(=\left(21-1\right)^3+2017=8000+2017=10017\)
Mình không viết lại đề nha ~
\(E=\left(x^3+3xy^2+3x^2y+y^3\right)+\left(3y+3x\right)+\left(3x^2+6xy+3y^2\right)+2016\)
\(E=\left(x+y\right)^3+3\left(x+y\right)+3\left(x+y\right)^2+2016\)
\(E=\left(x+y\right)[\left(x+y\right)^2+3+\left(x+y\right)]+2016\)
\(E=21\left(21^2+3+21\right)+2016\)
\(E=21.465+2016\)
\(E=9765+2016=11781\)
1/
x2 - 3x - 4
= \(x^2-3x+\frac{9}{4}-\frac{9}{4}-4\)
\(=\left(x^2-3x+\frac{9}{4}\right)-\frac{25}{4}\)
\(=\left(x-\frac{3}{2}\right)^2-\left(\frac{5}{2}\right)^2\)
\(=\left(x-\frac{3}{2}-\frac{5}{2}\right)\left(x-\frac{3}{2}+\frac{5}{2}\right)\)
\(=\left(x-4\right)\left(x+1\right)\)
Bài 1 :
\(x^2-3x-4\)
\(=x^2+x-4x-4\)
\(=x\left(x+1\right)-4\left(x+1\right)\)
\(=\left(x+1\right)\left(x-4\right)\)
\(a,A=\dfrac{y^3-x^3}{x^3-3x^2y+3xy^2-y^3}=\dfrac{\left(y-x\right)\left(y^2+xy+x^2\right)}{\left(x-y\right)^3}=-\dfrac{x^2+xy+y^2}{x^2-2xy+y^2}\)b,\(B=\dfrac{x^5+x+1}{x^3+x^2+x}=\dfrac{(x^5-x^2)+x^2+x+1}{x\left(x^2+x+1\right)}=\dfrac{x^2\left(x^3-1\right)+\left(x^2+x+1\right)}{x\left(x^2+x+1\right)}\)\(=\dfrac{x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)}{x\left(x^2+x+1\right)}=\dfrac{\left(x^2+x+1\right)\left(x^3-x^2+1\right)}{x\left(x^2+x+1\right)}\)\(\dfrac{x^3-x^2+1}{x}\)
c, Sửa lại:\(x^2-4x+5\rightarrow x^2-4x-5\)
\(\dfrac{2x^2-x-3}{x^2-4x-5}=\dfrac{2x^2+2x-3x-3}{x^2+x-5x-5}=\dfrac{2x\left(x+1\right)-3\left(x+1\right)}{x\left(x+1\right)-5\left(x+1\right)}=\dfrac{\left(x+1\right)\left(2x-3\right)}{\left(x+1\right)\left(x-5\right)}=\dfrac{2x-3}{x-5}\)
a) \(\dfrac{y^3-x^3}{x^3-3x^2y+3xy^2-y^3}=\dfrac{-\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x-y\right)^3}=\dfrac{-x^2-xy-y^2}{\left(x-y\right)^2}\)b)
\(\dfrac{x^5+x+1}{x^3+x^2+x}=\dfrac{x^5-x^2+x^2+x+1}{x\left(x^2+x+1\right)}=\dfrac{x^2\left(x^3-1\right)+x^2+x+1}{x\left(x^2+x+1\right)}\)\(=\dfrac{x^2\left(x-1\right)\left(x^2+x+1\right)+x^2+x+1}{x\left(x^2+x+1\right)}=\dfrac{\left(x^2+x+1\right)\left(x^3-x^2+1\right)}{x\left(x^2+x+1\right)}=\dfrac{x^3-x^2+1}{x}\)c) \(\dfrac{2x^2-x-3}{x^2-4x-5}=\dfrac{2x^2+2x-3x-3}{x^2+x-5x-5}=\dfrac{2x\left(x+1\right)-3\left(x+1\right)}{x\left(x+1\right)-5\left(x+1\right)}\)\(=\dfrac{\left(x+1\right)\left(2x-3\right)}{\left(x+1\right)\left(x-5\right)}=\dfrac{2x-3}{x-5}\)
Bài giải
a) \(\dfrac{1}{x+2}=\dfrac{x.\left(x-2\right)}{\left(x+2\right)\left(x-2\right).x}=\dfrac{x^2-2x}{x\left(x+2\right)\left(x-2\right)}\)
\(\dfrac{8}{2x-x^2}=\dfrac{8}{x\left(2-x\right)}=-\dfrac{8}{x\left(x-2\right)}=-\dfrac{8.\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}\)
b) \(x^2+1=\dfrac{x^2+1}{1}=\dfrac{\left(x^2+1\right)\left(x^2-1\right)}{x^2-1}=\dfrac{x^4-1}{x^2-1}\)
\(\dfrac{x^4}{x^2-1}\) giữ nguyên.
c) \(\dfrac{x^3}{x^3-3x^2y+3xy^2-y^3}=\dfrac{x^3}{\left(x-y\right)^3}=\dfrac{x^3.y}{\left(x-y\right)^3.y}=\dfrac{x^3y}{y\left(x-y\right)^3}\)
\(\dfrac{x}{y^2-xy}=\dfrac{x}{y.\left(y-x\right)}=-\dfrac{x}{y.\left(x-y\right)}=-\dfrac{x\left(x-y\right)^2}{y.\left(x-y\right).\left(x-y\right)^2}=\dfrac{x\left(x-y\right)^2}{y.\left(x-y\right)^3}\)
\(=\dfrac{\left(y-x\right)\left(y^2+x^2+xy\right)}{\left(x-y\right)^3}=-\dfrac{x^2+y^2+xy}{x^2+y^2-2xy}=-1+\dfrac{-3xy}{x^2+y^2-2xy}\)