Toán Lớp 8: P=(x-3/x-x/x-3+9/x^2+3x):2x-2/x
nhanh dùm vote 5 sao lun
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Toán Lớp 8: P=(x-3/x-x/x-3+9/x^2+3x):2x-2/x
nhanh dùm vote 5 sao lun
Comments ( 1 )
DK:x \ne \left\{ { – 3;0;1;3} \right\}\\
P = \left( {\dfrac{{x – 3}}{x} – \dfrac{x}{{x – 3}} + \dfrac{9}{{{x^2} + 3x}}} \right):\dfrac{{2x – 2}}{x}\\
= \dfrac{{{{\left( {x – 3} \right)}^2}\left( {x + 3} \right) – {x^2}\left( {x + 3} \right) + 9\left( {x – 3} \right)}}{{x\left( {x – 3} \right)\left( {x + 3} \right)}}.\dfrac{x}{{2\left( {x – 1} \right)}}\\
= \dfrac{{\left( {{x^2} – 6x + 9} \right)\left( {x + 3} \right) – {x^3} – 3{x^2} + 9x – 27}}{{x\left( {x – 3} \right)\left( {x + 3} \right)}}.\dfrac{x}{{2\left( {x – 1} \right)}}\\
= \dfrac{{{x^3} + 3{x^2} – 6{x^2} – 18x + 9x + 27 – {x^3} – 3{x^2} + 9x – 27}}{{x\left( {x – 3} \right)\left( {x + 3} \right)}}.\dfrac{x}{{2\left( {x – 1} \right)}}\\
= \dfrac{{ – 6{x^2}}}{{x\left( {x – 3} \right)\left( {x + 3} \right)}}.\dfrac{x}{{2\left( {x – 1} \right)}}\\
= – \dfrac{{3{x^2}}}{{\left( {{x^2} – 9} \right)\left( {x – 1} \right)}}
\end{array}\)