Toán Lớp 8: Tính : (x^3 -1) /(x^3 +x^2 +x) – (x^2 -4) /(6*y-3*x*y)
Trả lời Đúng = Câu trả lời hay nhất + Vote 5 sao
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Toán Lớp 8: Tính : (x^3 -1) /(x^3 +x^2 +x) – (x^2 -4) /(6*y-3*x*y)
Trả lời Đúng = Câu trả lời hay nhất + Vote 5 sao
Comments ( 2 )
\dfrac{{{x^3} – 1}}{{{x^3} + {x^2} + x}} – \dfrac{{{x^2} – 4}}{{6y – 3xy}}\\
DKXD:\left\{ \begin{array}{l}
x \ne 1\\
x \ne 2\\
x \ne – 2\\
x \ne 0\\
y \ne 0
\end{array} \right.\\
= \dfrac{{(x – 1)({x^2} + x + 1)}}{{x({x^2} + x + 1)}} – \dfrac{{(x – 2)(x + 2)}}{{ – 3y(x – 2)}}\\
= \dfrac{{x – 1}}{x} – \dfrac{x}{{ – 3y}}\\
= \dfrac{{x – 1}}{x} + \dfrac{x}{{3y}}\\
= \dfrac{{3xy – 3y}}{{3xy}} + \dfrac{{{x^2}}}{{3xy}}\\
= \dfrac{{3xy – 3y + {x^2}}}{{3xy}}
\end{array}\]