Toán Lớp 8: Thực hiên phép tính:
-1/3x^5y^2 : (-2xy) – (x^2+2x+4) :(x+1)
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Toán Lớp 8: Thực hiên phép tính:
-1/3x^5y^2 : (-2xy) – (x^2+2x+4) :(x+1)
Comments ( 1 )
– \dfrac{1}{3}{x^5}{y^2}:\left( { – 2xy} \right) – \left( {{x^2} + 2x + 4} \right):\left( {x + 1} \right)\\
= \dfrac{{{x^5}{y^2}}}{{3.2xy}} – \dfrac{{{x^2} + x + x + 1 + 3}}{{x + 1}}\\
= \dfrac{{{x^4}y}}{6} – \dfrac{{x\left( {x + 1} \right) + x + 1 + 3}}{{x + 1}}\\
= \dfrac{1}{6}{x^4}y – \dfrac{{\left( {x + 1} \right)\left( {x + 1} \right) + 3}}{{x + 1}}\\
= \dfrac{1}{6}{x^4}y – \left( {x + 1} \right) – \dfrac{3}{{x + 1}}
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