Toán Lớp 8: Cho A = ( x/x+1 + 2/2x+8 ) : x+1/x-1 x khác ± 1
a, Rút gọn A
b, Tìm x biết A=1/3
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Toán Lớp 8: Cho A = ( x/x+1 + 2/2x+8 ) : x+1/x-1 x khác ± 1
a, Rút gọn A
b, Tìm x biết A=1/3
Comments ( 1 )
a)A = \dfrac{{{x^3} + 4{x^2} – 4x – 1}}{{{x^3} + 6{x^2} + 9x + 4}}\\
b)\left[ \begin{array}{l}
x = – 4,970716961\\
x = 2,279597682\\
x = – 0,3088807215
\end{array} \right.
\end{array}\)
a)A = \left( {\dfrac{x}{{x + 1}} + \dfrac{2}{{2x + 8}}} \right):\dfrac{{x + 1}}{{x – 1}}\\
= \left( {\dfrac{x}{{x + 1}} + \dfrac{1}{{x + 4}}} \right).\dfrac{{x – 1}}{{x + 1}}\\
= \dfrac{{x\left( {x + 4} \right) + x + 1}}{{\left( {x + 1} \right)\left( {x + 4} \right)}}.\dfrac{{x – 1}}{{x + 1}}\\
= \dfrac{{{x^2} + 4x + x + 1}}{{\left( {x + 1} \right)\left( {x + 4} \right)}}.\dfrac{{x – 1}}{{x + 1}}\\
= \dfrac{{{x^2} + 5x + 1}}{{\left( {x + 1} \right)\left( {x + 4} \right)}}.\dfrac{{x – 1}}{{x + 1}}\\
= \dfrac{{{x^3} – {x^2} + 5{x^2} – 5x + x – 1}}{{\left( {x + 4} \right)\left( {{x^2} + 2x + 1} \right)}}\\
= \dfrac{{{x^3} + 4{x^2} – 4x – 1}}{{{x^3} + 2{x^2} + 4{x^2} + 8x + x + 4}}\\
= \dfrac{{{x^3} + 4{x^2} – 4x – 1}}{{{x^3} + 6{x^2} + 9x + 4}}\\
b)A = \dfrac{1}{3}\\
\to \dfrac{{{x^3} + 4{x^2} – 4x – 1}}{{{x^3} + 6{x^2} + 9x + 4}} = \dfrac{1}{3}\\
\to 3{x^3} + 12{x^2} – 12x – 3 = {x^3} + 6{x^2} + 9x + 4\\
\to 2{x^3} + 6{x^2} – 21x – 7 = 0\\
\to \left[ \begin{array}{l}
x = – 4,970716961\\
x = 2,279597682\\
x = – 0,3088807215
\end{array} \right.
\end{array}\)