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\(\dfrac{x^2+xy}{\left(x+y\right)^2}=\dfrac{x\left(x+y\right)}{\left(x+y\right)^2}=\dfrac{x}{x+y}=\dfrac{x\left(x-y\right)}{\left(x+y\right)\left(x-y\right)}\)
\(\dfrac{y^2-xy}{\left(x-y\right)^2}=\dfrac{-y\left(x-y\right)}{\left(x-y\right)^2}=\dfrac{-y}{x-y}=\dfrac{-y\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}\)
\(\dfrac{2xy}{x^2-y^2}=\dfrac{2xy}{\left(x-y\right)\left(x+y\right)}\)
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\(=\left(\dfrac{x\left(x+y\right)}{x^2\left(x+y\right)+y^2\left(x+y\right)}+\dfrac{y}{x^2+y^2}\right):\left(\dfrac{1}{x-y}-\dfrac{2xy}{x^2\left(x-y\right)+y^2\left(x-y\right)}\right)\)
\(=\dfrac{x+y}{x^2+y^2}:\left(\dfrac{1}{x-y}-\dfrac{2xy}{\left(x-y\right)\left(x^2+y^2\right)}\right)\)
\(=\dfrac{x+y}{x^2+y^2}:\dfrac{x^2+y^2-2xy}{\left(x-y\right)\left(x^2+y^2\right)}\)
\(=\dfrac{x+y}{x^2+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+y^2\right)}{\left(x-y\right)^2}\)
\(=\dfrac{x+y}{x-y}\)
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\(\left(1\right)=\dfrac{y}{x\left(2x-y\right)}-\dfrac{4x}{y\left(2x-y\right)}=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(y-2x\right)\left(y+2x\right)}{xy\left(y-2x\right)}=\dfrac{-y-2x}{xy}\\ \left(2\right)=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\\ \left(3\right)=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\\ \left(4\right)=\dfrac{4x^2+15x+4+4x+7+1}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}=\dfrac{4x^2+19x+12}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}\)
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\(A=\dfrac{x^2+y^2}{xy}+\dfrac{2xy}{x^2+y^2}=\dfrac{x^2+y^2}{2xy}+\dfrac{x^2+y^2}{2xy}+\dfrac{2xy}{x^2+y^2}\)
\(A\ge\dfrac{2xy}{2xy}+2\sqrt{\left(\dfrac{x^2+y^2}{2xy}\right)\left(\dfrac{2xy}{x^2+y^2}\right)}=3\)
Dấu "=" xảy ra khi \(x=y\)
\(B=\dfrac{\left(x+y\right)^2-4xy}{xy}+\dfrac{4xy}{\left(x+y\right)^2}=\dfrac{\left(x+y\right)^2}{xy}+\dfrac{4xy}{\left(x+y\right)^2}-4\)
\(B=\dfrac{\left(x+y\right)^2}{4xy}+\dfrac{4xy}{\left(x+y\right)^2}+\dfrac{3}{4}.\dfrac{\left(x+y\right)^2}{xy}-4\)
\(B\ge2\sqrt{\dfrac{\left(x+y\right)^2.4xy}{4xy.\left(x+y\right)^2}}+\dfrac{3}{4}.\dfrac{4xy}{xy}-4=1\)
\(B_{min}=1\) khi \(x=y\)
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a: \(=\dfrac{4x^2+4x+1-\left(4x^2-4x+1\right)}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x}{2x+1}\cdot\dfrac{5}{4x}=\dfrac{10}{2x+1}\)
c: \(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)
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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 >.< ^^
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\(=\dfrac{x\left(x+y\right)}{\left(x+y\right)\left(x^2+y^2\right)}\cdot\left(\dfrac{1}{x-y}-\dfrac{2xy}{\left(x-y\right)\left(x^2+y^2\right)}\right)\)
\(=\dfrac{x}{x^2+y^2}\cdot\dfrac{x^2+y^2-2xy}{\left(x-y\right)\left(x^2+y^2\right)}\)
\(=\dfrac{x}{x^2+y^2}\cdot\dfrac{x-y}{x^2+y^2}=\dfrac{x\left(x-y\right)}{\left(x^2+y^2\right)^2}\)
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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,\(\frac{x^2+y}{y}:(\frac{z}{x^2}:\frac{xy}{x^2y})=\frac{x^2+y}{ y}:(\frac{z}{x^2}\frac{x^2y}{xy})=\frac{x^2+y}{y}\frac{z}{x} \)
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Ta có: \(\dfrac{y}{x-y}-\dfrac{x^3-xy^2}{x^2+y^2}\cdot\left(\dfrac{x}{x^2-2xy+y^2}-\dfrac{y}{x^2-y^2}\right)\)
\(=\dfrac{y}{x-y}-\dfrac{x\left(x^2-y^2\right)}{x^2+y^2}\cdot\left(\dfrac{x\left(x+y\right)}{\left(x-y\right)^2\cdot\left(x+y\right)}-\dfrac{y\cdot\left(x-y\right)}{\left(x-y\right)^2\cdot\left(x+y\right)}\right)\)
\(=\dfrac{y}{x-y}-\dfrac{x\left(x-y\right)\left(x+y\right)}{x^2+y^2}\cdot\dfrac{x^2+xy-xy+y^2}{\left(x-y\right)^2\left(x+y\right)}\)
\(=\dfrac{y}{x-y}-\dfrac{x\cdot\left(x^2+y^2\right)}{\left(x^2+y^2\right)\cdot\left(x-y\right)}\)
\(=\dfrac{y}{x-y}-\dfrac{x}{x-y}\)
\(=\dfrac{y-x}{x-y}=\dfrac{-\left(x-y\right)}{x-y}=-1\)