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\(\left(1\right)=\dfrac{y}{x\left(2x-y\right)}-\dfrac{4x}{y\left(2x-y\right)}=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(y-2x\right)\left(y+2x\right)}{xy\left(y-2x\right)}=\dfrac{-y-2x}{xy}\\ \left(2\right)=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\\ \left(3\right)=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\\ \left(4\right)=\dfrac{4x^2+15x+4+4x+7+1}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}=\dfrac{4x^2+19x+12}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}\)
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: \(=\dfrac{x^3-1}{x-1}-\dfrac{x^2-1}{x+1}\)
\(=x^2+x+1-x+1=x^2+2\)
`@ 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.
\(\left(\dfrac{x^2}{x+y}+y\right).\left(\dfrac{1}{x^2-xy}-\dfrac{3y^3}{x^4-xy^3}-\dfrac{y}{x^3+x^2y+xy^2}\right)\)
\(=\left(\dfrac{x^2+xy+y^2}{x+y}\right).\left(\dfrac{1}{x\left(x-y\right)}-\dfrac{3y^2}{x\left(x^3-y^3\right)}-\dfrac{y}{x\left(x^2+xy+y^2\right)}\right)\)\(=\left(\dfrac{x^2+xy+y^2}{x+y}\right).\left(\dfrac{x^2+xy+y^2}{x\left(x^3-y^3\right)}-\dfrac{3y^2}{x\left(x^3-y^3\right)}-\dfrac{xy-y^2}{x\left(x^3-y^3\right)}\right)\)
\(=\dfrac{x\left(x^3-y^3\right)}{x^3-xy^2}.\dfrac{x^2+xy+y^2-3y^2-xy+y^2}{x\left(x^3-y^3\right)}\\ =\dfrac{x^2-y^2}{x\left(x^2-y^2\right)}=\dfrac{1}{x}\)
\(=\frac{x^2+xy+y^2}{x+y}.\left(\frac{1}{\left(x-y\right)x}-\frac{3y^2}{x\left(x^3-y^3\right)}-\frac{y}{x\left(x^2+xy+y^2\right)}\right)\)
\(=\frac{x^2+xy+y^2}{x+y}.\frac{x^2+xy+y^2-3y^2-xy+y^2}{x\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\frac{x^2-y^2}{x\left(x-y\right)\left(x+y\right)}=\frac{\left(x-y\right)\left(x+y\right)}{x\left(x-y\right)\left(x+y\right)}=\frac{1}{x}\)
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) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)
\(\dfrac{x^2+xy}{\left(x+y\right)^2}=\dfrac{x\left(x+y\right)}{\left(x+y\right)^2}=\dfrac{x}{x+y}=\dfrac{x\left(x-y\right)}{\left(x+y\right)\left(x-y\right)}\)
\(\dfrac{y^2-xy}{\left(x-y\right)^2}=\dfrac{-y\left(x-y\right)}{\left(x-y\right)^2}=\dfrac{-y}{x-y}=\dfrac{-y\left(x+y\right)}{\left(x-y\right)\left(x+y\right)}\)
\(\dfrac{2xy}{x^2-y^2}=\dfrac{2xy}{\left(x-y\right)\left(x+y\right)}\)
\(\dfrac{\left(x+y\right)^2}{x^2+xy}+\dfrac{\left(x-y\right)^2}{x^2-xy}+7=-\dfrac{\left(x-y\right)^2}{x^2-xy}+\dfrac{\left(x-y\right)^2}{x^2-xy}+\dfrac{7\cdot\left(x^2-xy\right)}{x^2-xy}=7\)