Dùng định nghĩa hai phân thức bằng nhau chứng minh các đẳng thức sau:
a. \({{{x^2}{y^3}} \over 5} = {{7{x^3}{y^4}} \over {35xy}}\)
b. \({{{x^2}\left( {x + 2} \right)} \over {x{{\left( {x + 2} \right)}^2}}} = {x \over {x + 2}}\)
c. \({{3 - x} \over {3 + x}} = {{{x^2} - 6x + 9} \over {9 - {x^2}}}\)
d. \({{{x^3} - 4x} \over {10 - 5x}} = {{ - {x^2} - 2x} \over 5}\)
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 \({{{x^2}{y^3}} \over 5} = {{7{x^3}{y^4}} \over {35xy}}\)
b. \({x^2}\left( {x + 2} \right).\left( {x + 2} \right) = {x^2}{\left( {x + 2} \right)^2};x{\left( {x + 2} \right)^2}.x = {x^2}{\left( {x + 2} \right)^2}\)
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\( \Rightarrow {x^2}\left( {x + 2} \right).\left( {x + 2} \right) = x{\left( {x + 2} \right)^2}x\).
Vậy \({{{x^2}\left( {x + 2} \right)} \over {x{{\left( {x + 2} \right)}^2}}} = {x \over {x + 2}}\)
c. \(\left( {3 - x} \right)\left( {9 - {x^2}} \right) = 27 - 3{x^2} - 9x + {x^3}\)
\(\left( {3 + x} \right)\left( {{x^2} - 6x + 9} \right) = 3{x^2} - 18x + 27 + {x^3} - 6{x^2} + 9x = 27 - 3{x^2} - 9x + {x^3}\)
\( \Rightarrow \left( {3 - x} \right)\left( {9 - {x^2}} \right) = \left( {3 + x} \right)\left( {{x^2} - 6x + 9} \right)\).
Vậy \({{3 - x} \over {3 + x}} = {{{x^2} - 6x + 9} \over {9 - {x^2}}}\)
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 \({{{x^3} - 4x} \over {10 - 5x}} = {{ - {x^2} - 2x} \over 5}\)