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Làm tính chia:
a. \(\left( {5{x^4} – 3{x^3} + {x^2}} \right):3{x^2}\)
b. \(\left( {5x{y^2} + 9xy – {x^2}{y^2}} \right):\left( { – xy} \right)\)
c. \(\left( {{x^3}{y^3} – {1 \over 2}{x^2}{y^3} – {x^3}{y^2}} \right):{1 \over 3}{x^2}{y^2}\)
a. \(\left( {5{x^4} – 3{x^3} + {x^2}} \right):3{x^2}\)
\( = \left( {5{x^4}:3{x^2}} \right) + \left( { – 3{x^3}:3{x^2}} \right) + \left( {{x^2}:3{x^2}} \right) = {5 \over 3}{x^2} – x + {1 \over 3}\)
b. \(\left( {5x{y^2} + 9xy – {x^2}{y^2}} \right):\left( { – xy} \right)\)
\( = \left[ {5x{y^2}:\left( { – xy} \right)} \right] + \left[ {9xy:\left( { – xy} \right)} \right] + \left[ {\left( { – {x^2}{y^2}} \right):\left( { – xy} \right)} \right] = – 5y – 9 + xy\)
c. \(\left( {{x^3}{y^3} – {1 \over 2}{x^2}{y^3} – {x^3}{y^2}} \right):{1 \over 3}{x^2}{y^2}\)
\(\eqalign{& = \left( {{x^3}{y^3}:{1 \over 3}{x^2}{y^2}} \right) + \left( { – {1 \over 2}{x^2}{y^3}:{1 \over 3}{x^2}{y^2}} \right) + \left( { – {x^3}{y^2}:{1 \over 3}{x^2}{y^2}} \right) \cr & = 3xy – {3 \over 2}y – 3x \cr} \)