`@ x+y+z=1`.

`<=>` \(\left\{{}\begin{matrix}x=1-y-z\\y=1-z-x\\z=1-x-y\end{matrix}\right.\)

`P=(x+y)^2/(xy+1-x-y).(y+z)^2/(yz-y-z+1).(x+z)^2/(xy-x-y+1)`.

`<=> ((1-z)^2(1-y)^2(1-x)^2)/((1-x)(1-y)(1-y)(1-z)(1-z)(1-x).`

`=1.`

Vậy `P` không phụ thuộc vào giá trị của biến.

`@ x+y+z=1`.

`<=>` \(\left\{{}\begin{matrix}x=1-y-z\\y=1-z-x\\z=1-x-y\end{matrix}\right.\)

`P=(x+y)^2/(xy+1-x-y).(y+z)^2/(yz-y-z+1).(x+z)^2/(xy-x-y+1)`.

`<=> ((1-z)^2(1-y)^2(1-x)^2)/((1-x)(1-y)(1-y)(1-z)(1-z)(1-x).`

`=1.`

Vậy `P` không phụ thuộc vào giá trị của biến.

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: \(=\dfrac{x^3-1}{x-1}-\dfrac{x^2-1}{x+1}\)

\(=x^2+x+1-x+1=x^2+2\)

\(\begin{array}{l}a) A = \left( {\frac{1}{{x - 1}} + \frac{1}{{x + 1}}} \right)\left( {x - \frac{1}{x}} \right)\\ = \left( {\frac{{x + 1 + x - 1}}{{{x^2} - 1}}} \right).\left( {\frac{{{x^2} - 1}}{x}} \right)\\ = \frac{{2x}}{{{x^2} - 1}}.\frac{{{x^2} - 1}}{x} = \frac{{2x.\left( {{x^2} - 1} \right)}}{{x\left( {{x^2} - 1} \right)}} = 2\end{array}\)

Vậy A = 2 không phụ thuộc vào giá trị của các biến

\(\begin{array}{l}b) B = \left( {\dfrac{x}{{xy - {y^2}}} + \dfrac{{2{\rm{x}} - y}}{{xy - {x^2}}}} \right).\dfrac{{{x^2}y - x{y^2}}}{{{{\left( {x - y} \right)}^2}}}\\= \dfrac{x}{{y\left( {x - y} \right)}}.\dfrac{{{x^2}y - x{y^2}}}{{{{\left( {x - y} \right)}^2}}} + \dfrac{{2{\rm{x}} - y}}{{x\left( {y - x} \right)}}.\dfrac{{{x^2}y - x{y^2}}}{{{{\left( {x - y} \right)}^2}}}\\= \dfrac{x}{{y\left( {x - y} \right)}}.\dfrac{{xy\left( {x - y} \right)}}{{{{\left( {x - y} \right)}^2}}} + \dfrac{{2{\rm{x}} - y}}{{ - x\left( {x - y} \right)}}.\dfrac{{xy\left( {x - y} \right)}}{{{{\left( {x - y} \right)}^2}}}\\= \dfrac{{{x^2}}}{{{{\left( {x - y} \right)}^2}}} - \dfrac{{\left( {2{\rm{x}} - y} \right)y}}{{{{\left( {x - y} \right)}^2}}}\\= \dfrac{{{x^2} - \left( {2{\rm{x}} - y} \right)y}}{{{{\left( {x - y} \right)}^2}}} = \dfrac{{{x^2} - 2{\rm{x}}y + {y^2}}}{{{{\left( {x - y} \right)}^2}}} = \dfrac{{{{\left( {x - y} \right)}^2}}}{{{{\left( {x - y} \right)}^2}}} = 1\end{array}\)

Vậy B = 1 không phụ thuộc vào giá trị của biến x

a: ĐKXĐ: \(x\notin\left\{0;1;-1\right\}\)

\(A=\left(\dfrac{1}{x-1}+\dfrac{1}{x+1}\right)\cdot\left(x-\dfrac{1}{x}\right)\)

\(=\dfrac{x+1+x-1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x^2-1}{x}\)

\(=\dfrac{2x}{x^2-1}\cdot\dfrac{x^2-1}{x}=\dfrac{2x}{x}=2\)

b: ĐKXĐ: \(\left\{{}\begin{matrix}x\ne y\\x\ne0\\y\ne0\end{matrix}\right.\)

\(B=\left(\dfrac{x}{xy-y^2}+\dfrac{2x-y}{xy-x^2}\right)\cdot\dfrac{x^2y-xy^2}{\left(x-y\right)^2}\)

\(=\left(\dfrac{x}{y\left(x-y\right)}-\dfrac{2x-y}{x\left(x-y\right)}\right)\cdot\dfrac{xy\left(x-y\right)}{\left(x-y\right)^2}\)

\(=\left(\dfrac{x^2-y\left(2x-y\right)}{xy\left(x-y\right)}\right)\cdot\dfrac{xy}{x-y}\)

\(=\dfrac{x^2-2xy+y^2}{xy\left(x-y\right)^2}\cdot xy=\dfrac{\left(x-y\right)^2}{\left(x-y\right)^2}=1\)

d)

\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+99\right)\left(x+100\right)}\)=\(\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+.....-\dfrac{1}{x+99}+\dfrac{1}{x+100}\)=\(\dfrac{1}{x}-\dfrac{1}{x+100}\)

=\(\dfrac{x+100}{x\left(x+100\right)}-\dfrac{x}{x\left(x+100\right)}\)

=\(\dfrac{x+100-x}{x\left(x+100\right)}=\dfrac{100}{x\left(x+100\right)}\)

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: \(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{x+1-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\right)\)

\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{2}{\left(x-1\right)}=\dfrac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x-1}{x^2+1}\)

a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)

b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)

c) \(=\dfrac{2x\left(x+1\right)}{x+1}=2x\)

d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)

e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)