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1/
\(\dfrac{\left(x-y\right)^3-3xy\left(x+y\right)+y^3}{x-6y}\)
\(=\dfrac{x^3-3x^2y+3xy^2-y^3-3x^2y-3xy^2+y^3}{x-6y}\)
\(=\dfrac{x^3-6x^2y}{x-6y}\)
\(=\dfrac{x^2\left(x-6y\right)}{x-6y}\)
\(=x^2\)
\(2\)/
\(\dfrac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\)
\(=\dfrac{\left(x-y+z^{ }\right)^2}{\left(x-y\right)^2-z^2}\)
\(=\dfrac{\left(x-y+z\right)^2}{\left(x-y-z\right)\left(x-y+z\right)}\)
\(=\dfrac{x-y+z}{x-y-z}\)
3/
\(\dfrac{\left(n+1\right)!}{n!\left(n+2\right)}\)
\(=\dfrac{n!\left(n+1\right)}{n!\left(n+2\right)}\)
\(=\dfrac{n+1}{n+2}\)
4/
\(\dfrac{n!}{\left(n+1\right)!-n!}\)
\(=\dfrac{n!}{n!\left(n+1\right)-n!}\)
\(=\dfrac{n!}{n!\left[\left(n+1\right)-1\right]}\)
\(=\dfrac{n!}{n!.n}\)
\(=\dfrac{1}{n}\)
5/
\(\dfrac{\left(n+1\right)!-\left(n+2\right)!}{\left(n+1\right)!+\left(n+2\right)!}\)
\(=\dfrac{\left(n+1\right)!-\left(n+1\right)!\left(n+2\right)}{\left(n+1\right)!+\left(n+1\right)!\left(n+2\right)}\)
\(=\dfrac{\left(n+1\right)!\left(-n-1\right)}{\left(n+1\right)!\left(n+3\right)}\)
\(=\dfrac{-n-1}{n+3}\)
1: \(=\dfrac{\left(x^2+2xy+y^2\right)-1}{\left(x^2+2x+1\right)-y^2}\)
\(=\dfrac{\left(x+y+1\right)\left(x+y-1\right)}{\left(x+1-y\right)\left(x+1+y\right)}=\dfrac{x+y-1}{x-y+1}\)
2: \(=\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}\)
\(=\dfrac{\left(x-y\right)\left(x^2+y^2\right)}{x^2-xy+y^2}\)
3: \(=\dfrac{\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz}{2x^2+2y^2+2z^2-2xy-2yz-2xz}\)
\(=\dfrac{\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)}{2\left(x^2+y^2+z^2-xy-yz-xz\right)}\)
\(=\dfrac{x+y+z}{2}\)
Bài 1:
a: \(A=\dfrac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\dfrac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}\)
\(=\dfrac{\left(x+1\right)\left(x^3+1\right)}{\left(x^2-x+1\right)\left(x^2+1\right)}=\dfrac{\left(x+1\right)^2}{x^2+1}\)
Để A=0 thì x+1=0
hay x=-1
b: \(B=\dfrac{x^4-5x^2+4}{x^4-10x^2+9}=\dfrac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-9\right)}=\dfrac{x^2-4}{x^2-9}\)
Để B=0 thi (x-2)(x+2)=0
=>x=2 hoặc x=-2
1) \(\dfrac{A\left(x-5\right)}{\left(x+1\right)\left(x-5\right)}=\dfrac{3x\left(x+3\right)}{\left(x+1\right)\left(x+3\right)}\)
\(\Rightarrow A=3x\)
2) \(\dfrac{\left(x+3\right)\left(x-2\right)}{A\left(x-3\right)}=\dfrac{\left(5x-1\right)\left(x-2\right)}{\left(5x-1\right)\left(x^2+3\right)}\)
\(\Leftrightarrow\dfrac{\left(x+3\right)}{A\left(x-3\right)}=\dfrac{1}{\left(x^2+3\right)}\)
\(\Rightarrow A=\dfrac{\left(x^2+3\right)\left(x+3\right)}{x-3}\)
3) \(\dfrac{\left(x-5\right)\left(x+5\right)}{\left(x+5\right)\left(2x-3\right)}=\dfrac{\left(x-5\right)A}{\left(2x-3\right)\left(x+2\right)}\)
\(\Leftrightarrow1=\dfrac{A}{\left(x+2\right)}\)
\(\Leftrightarrow A=x+2\)
a) \(\dfrac{x^2-4}{2x^2-4x}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{2x\left(x-2\right)}\)
\(=\dfrac{x+2}{2x}\)
b) \(\dfrac{2x-x^2}{x^2-4x+4}\)
\(=\dfrac{x\left(2-x\right)}{\left(x-2\right)^2}\)
\(=\dfrac{x\left(2-x\right)}{\left(2-x\right)^2}\)
\(=\dfrac{x}{2-x}\)
c) \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)
\(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}\)
\(=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}\)
\(=\dfrac{x-y}{x+y}\)
d) \(\dfrac{5x^2+10x+5}{x+x^2}\)
\(\dfrac{5\left(x^2+2x+1\right)}{x\left(1+x\right)}\)
\(=\dfrac{5\left(x+1\right)^2}{x\left(x+1\right)}\)
\(=\dfrac{5\left(x+1\right)}{x}\)
e) \(\dfrac{3x^2+3x}{\left(x+1\right)\left(2x+6\right)}\)
\(=\dfrac{3x\left(x+1\right)}{\left(x+1\right).2\left(x+3\right)}\)
\(=\dfrac{3x}{2\left(x+3\right)}\)
f) \(\dfrac{\left(2-3x\right)\left(x+1\right)}{x^2+2x+1}\)
\(=\dfrac{\left(2-3x\right)\left(x+1\right)}{\left(x+1\right)^2}\)
\(=\dfrac{2-3x}{x+1}\)
\(1.\text{ }\text{ }\text{ }\dfrac{\left(x^2+2\right)^2-4x^2}{y\left(x^2+2\right)-2xy-\left(x-1\right)^2-1}\\ =\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{x^2y+2y-2xy-x^2+2x-1-1}\\ =\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{\left(x^2y-x^2\right)-\left(2xy-2x\right)+\left(2y-2\right)}\\ =\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{x^2\left(y-1\right)-2x\left(y-1\right)+2\left(y-1\right)}\\ =\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{\left(x^2-2x+2\right)\left(y-1\right)}\\ =\dfrac{x^2+2x+2}{y-1}\)
\(2.\text{ }\text{ }\text{ }\text{ }\dfrac{x^2+5x+6}{x^2+3x+2}\\ =\dfrac{x^2+3x+2x+6}{x^2+2x+x+2}\\ =\dfrac{\left(x^2+3x\right)+\left(2x+6\right)}{\left(x^2+2x\right)+\left(x+2\right)}\\ =\dfrac{x\left(x+3\right)+2\left(x+3\right)}{x\left(x+2\right)+\left(x+2\right)}\\ =\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)\left(x+1\right)}\\ =\dfrac{x+3}{x+1}\)
\(3.\text{ }\text{ }\text{ }\dfrac{x^2+y^2-z^2-2zt+2xy-t^2}{x^2-y^2+z^2-2yt+2xz-t^2}\text{ ( Chữa đề ) }\\ =\dfrac{\left(x^2+2xy+y^2\right)-\left(z^2+2zt+t^2\right)}{\left(x^2+2xz+z^2\right)-\left(y^2+2yt+t^2\right)}\\ =\dfrac{\left(x+y\right)^2-\left(z+t\right)^2}{\left(x+z\right)^2-\left(y+t\right)^2}\\ =\dfrac{\left(x+y+z+t\right)\left(x+y-z-t\right)}{\left(x+z+y+t\right)\left(x+z-y-t\right)}\\ =\dfrac{x+y-z-t}{x+z-y-t}\)
\(4.\text{ }\text{ }\text{ }\dfrac{\left(n+1\right)!}{\left(n+1\right)!+\left(n+2\right)!}=\dfrac{\left(n+1\right)!}{\left(n+1\right)!\left(1+n+2\right)}=\dfrac{1}{n+3}\)
\(5.\text{ }\text{ }\text{ }\dfrac{x^2+5x+4}{x^2-1}\\ =\dfrac{x^2+x+4x+4}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{\left(x^2+x\right)+\left(4x+4\right)}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{x\left(x+1\right)+4\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{\left(x+1\right)\left(x+4\right)}{\left(x+1\right)\left(x-1\right)}\\ =\dfrac{x+4}{x-1}\)
\(6.\text{ }\text{ }\text{ }\dfrac{x^2-3x}{2x^2-7x+3}\\ =\dfrac{x\left(x-3\right)}{2x^2-6x-x+3}\\ =\dfrac{x\left(x-3\right)}{\left(2x^2-6x\right)-\left(x-3\right)}\\ =\dfrac{x\left(x-3\right)}{2x\left(x-3\right)-\left(x-3\right)}\\ =\dfrac{x\left(x-3\right)}{\left(2x-1\right)\left(x-3\right)}\\ =\dfrac{x}{2x-1}\)