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\(=\left(\dfrac{x\left(x+y\right)}{x^2\left(x+y\right)+y^2\left(x+y\right)}+\dfrac{y}{x^2+y^2}\right):\left(\dfrac{1}{x-y}-\dfrac{2xy}{x^2\left(x-y\right)+y^2\left(x-y\right)}\right)\)
\(=\dfrac{x+y}{x^2+y^2}:\left(\dfrac{1}{x-y}-\dfrac{2xy}{\left(x-y\right)\left(x^2+y^2\right)}\right)\)
\(=\dfrac{x+y}{x^2+y^2}:\dfrac{x^2+y^2-2xy}{\left(x-y\right)\left(x^2+y^2\right)}\)
\(=\dfrac{x+y}{x^2+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+y^2\right)}{\left(x-y\right)^2}\)
\(=\dfrac{x+y}{x-y}\)
a: \(=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)
\(=\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\dfrac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{3x\left(x+3\right)}{3x-9-x^2}\)
\(=\dfrac{3}{x-3}\cdot\dfrac{-\left(x^2-3x+9\right)}{x^2-3x+9}=\dfrac{-3}{x-3}\)
b: \(=\dfrac{x+1}{x+2}:\left(\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x+3\right)^2}\right)\)
\(=\dfrac{x+1}{x+2}\cdot\dfrac{\left(x+3\right)^2}{\left(x+2\right)\left(x+1\right)}=\dfrac{\left(x+3\right)^2}{\left(x+2\right)^2}\)
c: \(=\dfrac{x^2-2xy+y^2+x^2+2xy+y^2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{x^2+2xy+y^2}{2xy}\cdot\dfrac{xy}{x^2+y^2}\)
\(=\dfrac{2\left(x^2+y^2\right)}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x+y\right)^2}{x^2+y^2}\cdot\dfrac{1}{2}\)
\(=\dfrac{\left(x+y\right)}{x-y}\)
a) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
\(=\left(\dfrac{\left(2x+1\right)\left(2x+1\right)}{2x^2-1}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{2x^2-1}\right):\dfrac{4x}{10x-5}\)
\(=\left(\dfrac{\left(2x+1\right)^2-\left(2x-1\right)^2}{2x^2-1}\right):\dfrac{4x}{10x-5}\)
\(=\left(\dfrac{\left(2x+1-2x-1\right)\left(2x+1+2x-1\right)}{2x^2-1}\right):\dfrac{4x}{10x-5}\)
\(=\dfrac{4x}{2x^2-1}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{5}{2x+1}\)
b) \(\left(\dfrac{1}{x^2+1}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1}{x^2+1}-\dfrac{x\left(2-x\right)}{x\left(x+1\right)}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1-2x+x^2}{x^2+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\left(\dfrac{1-2x+x^2}{x^2+1}\right):\left(\dfrac{1}{x}+\dfrac{x^2}{x}-\dfrac{2x}{x}\right)\)
\(=\left(\dfrac{1-2x+x^2}{x^2+1}\right):\left(\dfrac{x^2-2x+1}{x}\right)\)
\(=\dfrac{\left(x-1\right)^2}{x^2+1}.\dfrac{x}{\left(x-1\right)^2}\)
\(=\dfrac{x}{x^2+1}\)
c) d) Tự làm đi mình làm biếng quass >.< ^^
a: \(=\dfrac{1}{x-y}-\dfrac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\dfrac{x-y}{x^2+xy+y^2}\)
\(=\dfrac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2x^2-4xy+2y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)
d: \(=\dfrac{x^3-1}{x-1}-\dfrac{x^2-1}{x+1}\)
\(=x^2+x+1-x+1=x^2+2\)
a,\(\frac{x^2+y^2-xy}{x^2-y^2}:\frac{x^3+y^3}{x^2+y^2-2xy} =\frac{x^2+y^2-xy}{(x-y)(x+y)}\frac{(x+y)^2}{(x+y) (x^2-xy+y^2)}=\frac{1}{x-y} \)
b,\(\frac{x^3y+xy^3}{x^4y}:(x^2+y^2)=\frac{xy(x^2+y^2)}{x^4y(x^2+y^2)}=\frac{1}{x^3} \)
c,\(\frac{x^2-xy}{y}:\frac{x^2-xy}{xy+y}:\frac{x^2-1}{x^2+y} =\frac{x(x-y)y(x+y)(x^2+y)}{yx(x-y)(x^2-1)} =\frac{(x^2+y)(x+y)}{x^2-1} \)
d,\(\frac{x^2+y}{y}:(\frac{z}{x^2}:\frac{xy}{x^2y})=\frac{x^2+y}{ y}:(\frac{z}{x^2}\frac{x^2y}{xy})=\frac{x^2+y}{y}\frac{z}{x} \)
1) \(\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}-\dfrac{1}{1-x}+1\)
\(=\dfrac{x^2}{x+1}+\dfrac{2x}{x^2-1}+\dfrac{1}{x-1}+1\)
\(=\dfrac{x^2}{x+1}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}+\dfrac{1}{x-1}+1\) MTC: \(\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{x^2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}+\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2\left(x-1\right)+2x+\left(x+1\right)+\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^3-x^2+2x+x+1+x^2-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x^2+3\right)}{\left(x-1\right)\left(x+1\right)}\)
b) \(\dfrac{1}{x^3-x}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{x^2-1}\)
\(=\dfrac{1}{x\left(x^2-1\right)}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{x\left(x-1\right)\left(x+1\right)}-\dfrac{1}{\left(x-1\right)x}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\) MTC: \(x\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{1}{x\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{x\left(x-1\right)\left(x+1\right)}+\dfrac{2x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1-\left(x+1\right)+2x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1-x-1+2x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x}{x\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)
Ta có: \(\dfrac{y}{x-y}-\dfrac{x^3-xy^2}{x^2+y^2}\cdot\left(\dfrac{x}{x^2-2xy+y^2}-\dfrac{y}{x^2-y^2}\right)\)
\(=\dfrac{y}{x-y}-\dfrac{x\left(x^2-y^2\right)}{x^2+y^2}\cdot\left(\dfrac{x\left(x+y\right)}{\left(x-y\right)^2\cdot\left(x+y\right)}-\dfrac{y\cdot\left(x-y\right)}{\left(x-y\right)^2\cdot\left(x+y\right)}\right)\)
\(=\dfrac{y}{x-y}-\dfrac{x\left(x-y\right)\left(x+y\right)}{x^2+y^2}\cdot\dfrac{x^2+xy-xy+y^2}{\left(x-y\right)^2\left(x+y\right)}\)
\(=\dfrac{y}{x-y}-\dfrac{x\cdot\left(x^2+y^2\right)}{\left(x^2+y^2\right)\cdot\left(x-y\right)}\)
\(=\dfrac{y}{x-y}-\dfrac{x}{x-y}\)
\(=\dfrac{y-x}{x-y}=\dfrac{-\left(x-y\right)}{x-y}=-1\)