Toán Lớp 8: x-3/2x+1×(3×1/x^2-9-1/3-x) rút gpnj p
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Toán Lớp 8: x-3/2x+1×(3×1/x^2-9-1/3-x) rút gpnj p
Comments ( 2 )
ĐKXĐ : x\ne ±3,x\ne (-1)/2
(x-3)/(2x+1) ((3x-1)/(x^2-9) -1/(3-x))
=(x-3)/(2x+1) . (3x-1 +x+3)/((x-3)(x+3))
=1/(2x+1) . (4x+2)/(x+3)
=1/(2x+1) . (2(2x+1))/(x+3)
= 2/(x+3)
Giải đáp:$P = \dfrac{2}{{x + 3}}$
Lời giải và giải thích chi tiết:
$\begin{array}{l}
Dkxd:x \ne – \dfrac{1}{2};x \ne 3;x \ne – 3\\
P = \dfrac{{x – 3}}{{2x + 1}}.\left( {\dfrac{{3x – 1}}{{{x^2} – 9}} – \dfrac{1}{{3 – x}}} \right)\\
= \dfrac{{x – 3}}{{2x + 1}}.\left( {\dfrac{{3x – 1}}{{\left( {x – 3} \right)\left( {x + 3} \right)}} + \dfrac{1}{{x – 3}}} \right)\\
= \dfrac{{x – 3}}{{2x + 1}}.\dfrac{{3x – 1 + x + 3}}{{\left( {x + 3} \right)\left( {x – 3} \right)}}\\
= \dfrac{1}{{2x + 1}}.\dfrac{{4x + 2}}{{x + 3}}\\
= \dfrac{{2\left( {2x + 1} \right)}}{{\left( {2x + 1} \right)\left( {x + 3} \right)}}\\
= \dfrac{2}{{x + 3}}
\end{array}$