Toán Lớp 8: A = [(x^2/x^3 – 4x) + (6/6-3x) + (1/x+2)] : [x-2+(10-x/x+2)]
a, Rút gọn
b, Tìm x để A=2
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Toán Lớp 8: A = [(x^2/x^3 – 4x) + (6/6-3x) + (1/x+2)] : [x-2+(10-x/x+2)]
a, Rút gọn
b, Tìm x để A=2
Comments ( 1 )
a) – \dfrac{6}{{\left( {x – 2} \right)\left( {{x^2} – x + 6} \right)}}\\
b)x = 1,564099733
\end{array}\)
a)DK:x \ne \left\{ { – 2;0;2} \right\}\\
A = \left[ {\dfrac{{{x^2}}}{{{x^3} – 4x}} + \dfrac{6}{{6 – 3x}} + \dfrac{1}{{x + 2}}} \right]:\left( {x – 2 + \dfrac{{10 – x}}{{x + 2}}} \right)\\
= \left[ {\dfrac{{{x^2}}}{{x\left( {x – 2} \right)\left( {x + 2} \right)}} + \dfrac{2}{{2 – x}} + \dfrac{1}{{x + 2}}} \right]:\dfrac{{{x^2} – 4 + 10 – x}}{{x + 2}}\\
= \dfrac{{x – 2\left( {x + 2} \right) + x – 2}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}.\dfrac{{x + 2}}{{{x^2} – x + 6}}\\
= \dfrac{{ – 6}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}.\dfrac{{x + 2}}{{{x^2} – x + 6}}\\
= – \dfrac{6}{{\left( {x – 2} \right)\left( {{x^2} – x + 6} \right)}}\\
b)A = 2\\
\to – \dfrac{6}{{\left( {x – 2} \right)\left( {{x^2} – x + 6} \right)}} = 2\\
\to \left( {x – 2} \right)\left( {{x^2} – x + 6} \right) = – 3\\
\to {x^3} – {x^2} + 6x – 2{x^2} + 2x – 12 = – 3\\
\to {x^3} – 3{x^2} + 8x – 9 = 0\\
\to x = 1,564099733
\end{array}\)