Cho tam giác \(ABC\) vuông tại \(B\) có \(AB = 60{\rm{\;cm}};\,\,AC = 10{\rm{\;dm}}{\rm{.}}\) Độ dài đoạn \(BC\) là

a) \(\frac{{x + y}}{{2y}} + \frac{{2x - y}}{{2y}} = \frac{{x + y + 2x - y}}{{2y}} = \frac{{3x}}{{2y}}.\)

b) \(\frac{{{x^2} - 5x}}{{x - 3}} - \frac{{x - 9}}{{x - 3}} = \frac{{{x^2} - 5x - x + 9}}{{x - 3}} = \frac{{{x^2} - 6x + 9}}{{x - 3}} = \frac{{{{\left( {x - 3} \right)}^2}}}{{x - 3}} = x - 3.\)

c) \(\frac{{x + 2}}{{x - 3}} + \frac{2}{{x + 3}} + \frac{{3{x^2} + x}}{{9 - {x^2}}}\)

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

\( = \frac{{{x^2} + 5x + 6 + 2x - 6 - 3{x^2} - x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \frac{{ - 2{x^2} + 6x}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\)

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