Question

Dùng định nghĩa hai phân thức bằng nhau chứng tỏ rằng:

\(a)\dfrac{{3{\rm{x}}}}{2} = \dfrac{{15{\rm{x}}y}}{{10y}}\)                        \(b)\dfrac{{3{\rm{x}} - 3y}}{{2y - 2{\rm{x}}}} = \dfrac{{ - 3}}{2}\)                     \(c)\dfrac{{{x^2} - x + 1}}{x} = \dfrac{{{x^3} + 1}}{{x\left( {x + 1} \right)}}\)

Answer 1

a) Ta có:

\(\begin{array}{l}3{\rm{x}}.10y = 30{\rm{xy}}\\{\rm{2}}{\rm{.15x}}y = 30{\rm{x}}y\end{array}\)

Suy ra: \(3{\rm{x}}.10 = 2.15{\rm{x}}y\) nên \(\dfrac{{3{\rm{x}}}}{2} = \dfrac{{15{\rm{x}}y}}{{10y}}\)

b) Ta có:

\(\begin{array}{l}\left( {3{\rm{x}} - 3y} \right).2 = 2.3\left( {x - y} \right) = 6\left( {x - y} \right)\\\left( { - 3} \right).\left( {2y - 2{\rm{x}}} \right) = \left( { - 3} \right).\left( { - 2} \right)\left( {x - y} \right) = 6\left( {x - y} \right)\end{array}\)

Suy ra: \(2.\left( {3{\rm{x}} - 3y} \right) = \left( { - 3} \right).\left( {2y - 2{\rm{x}}} \right)\) nên \(\dfrac{{3{\rm{x}} - 3y}}{{2y - 2{\rm{x}}}} = \dfrac{{ - 3}}{2}\)

c) Ta có: \(\begin{array}{l}\left( {{x^2} - x + 1} \right).x\left( {x + 1} \right) = x.\left( {x + 1} \right)\left( {{x^2} - x + 1} \right) = x.\left( {{x^3} + 1} \right)\\x.\left( {{x^3} + 1} \right)\end{array}\)

Suy ra: \(\left( {{x^2} - x + 1} \right).x.\left( {x + 1} \right) = x.\left( {{x^3} + 1} \right)\) nên \(\dfrac{{{x^2} - x + 1}}{x} = \dfrac{{{x^3} + 1}}{{x\left( {x + 1} \right)}}\)