\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)

\(=\left(\frac{x}{y\left(x-y\right)}+\frac{y-2x}{x\left(x-y\right)}\right):\left(\frac{y}{xy}+\frac{x}{xy}\right)\)

\(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\)

\(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\)

0,7320721383

bn ơi là giải chi tiết(dũng não ko phải tay)   ^^

\(\left(1-\frac{1}{x+1}\right)\left(1-\frac{1}{x+2}\right)\left(1-\frac{1}{x+3}\right)...\left(1-\frac{1}{x+2018}\right)\)

\(=\frac{x}{x+1}.\frac{x+1}{x+2}.\frac{x+2}{x+3}....\frac{x+2017}{x+2018}\)

\(=\frac{x}{x+2018}\)

\(\frac{x^2+3x+9}{2x+10}.\frac{x+5}{x^3-27}\)

\(=\frac{x^2+3x+9}{2\left(x+5\right)}.\frac{x+5}{\left(x-3\right)\left(x^2+3x+9\right)}\)

\(=\frac{\left(x+5\right)\left(x^2+3x+9\right)}{2\left(x+5\right)\left(x-3\right)\left(x^2+3x+9\right)}\)

\(=\frac{1}{2\left(x-3\right)}\)

\(\left(\frac{6x+1}{x^2-6x}+\frac{6x-1}{x^2+6x}\right)\left(\frac{x^2-36}{x^2+1}\right)\)

\(=\left[\frac{6x+1}{x\left(x-6\right)}+\frac{6x-1}{x\left(x+6\right)}\right]\left[\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\right]\)

\(=\frac{\left(6x+1\right)\left(x+6\right)+\left(6x-1\right)\left(x-6\right)}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)

\(=\frac{6x^2+36x+x+6+6x^2-36x-x+6}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)

\(=\frac{12x^2+12}{x\left(x-6\right)\left(x+6\right)}.\frac{\left(x-6\right)\left(x+6\right)}{x^2+1}\)

\(=\frac{12\left(x^2+1\right).\left(x-6\right)\left(x+6\right)}{x\left(x-6\right)\left(x+6\right)\left(x^2+1\right)}\)

\(=\frac{12}{x}\)

để A nhỏ nhất => x²+1 nhỏ nhất và lớn hơn 0 (vì 2>0 và không đổi)

ta có: \(x^2+1\ge1\)

dấu = xảy ra khi x²=0

=> x=0

Vậy Min A=\(\frac{1}{2}\)khi x=0