a) \({x^2}{y^3}.35xy = 35{x^3}{y^4};\)
\(5.7{x^3}{y^4} = 35{x^3}{y^4}\)
\( \Rightarrow {x^2}{y^3}.35xy = 5.7{x^3}{y^4}\).
Vậy \(\displaystyle {{{x^2}{y^3}} \over 5} = {{7{x^3}{y^4}} \over {35xy}}\)
b) \({x^2}\left( {x + 2} \right).\left( {x + 2} \right) = {x^2}{\left( {x + 2} \right)^2};\)
\(x{\left( {x + 2} \right)^2}.x = {x^2}{\left( {x + 2} \right)^2}\)
\( \Rightarrow {x^2}\left( {x + 2} \right).\left( {x + 2} \right) = x{\left( {x + 2} \right)^2}.x\)
Vậy \(\displaystyle {{{x^2}\left( {x + 2} \right)} \over {x{{\left( {x + 2} \right)}^2}}} = {x \over {x + 2}}\)
c) \(\left( {3 - x} \right)\left( {9 - {x^2}} \right) \)\(\,= \left( {3 - x} \right)\left( {3 - x} \right)\left( {3 + x} \right)\)\(\, = {\left( {3 - x} \right)^2}\left( {3 + x} \right)\)
\(\left( {3 + x} \right)\left( {{x^2} - 6x + 9} \right)\)\(\, = \left( {3 + x} \right)\left( {{x^2} - 2.x.3 + {3^2}} \right) \)\(\,= \left( {3 + x} \right){\left( {x - 3} \right)^2} = \left( {3 + x} \right){\left( {3 - x} \right)^2}\)
\( \Rightarrow \left( {3 - x} \right)\left( {9 - {x^2}} \right) \)\(\,= \left( {3 + x} \right)\left( {{x^2} - 6x + 9} \right)\).
Vậy \(\displaystyle {{3 - x} \over {3 + x}} = {{{x^2} - 6x + 9} \over {9 - {x^2}}}\)
d) \(\left( {{x^3} - 4x} \right).5 = 5{x^3} - 20x;\)
\(\left( {10 - 5x} \right)\left( { - {x^2} - 2x} \right) \)\(\,= - 10{x^2} - 20x + 5{x^3} + 10{x^2}\)\(\, = 5{x^3} - 20x\)
\( \Rightarrow \left( {{x^3} - 4x} \right).5 \)\(\,= \left( {10 - 5x} \right)\left( { - {x^2} - 2x} \right)\)
Vậy \(\displaystyle {{{x^3} - 4x} \over {10 - 5x}} = {{ - {x^2} - 2x} \over 5}\)
