Toán Lớp 8: (8)/(x-5) – (3)/(x+5) + (2x-20)/(x^2-25)
(6)/(x-5) + (6-2x)/(x^2-25) + (4)/(x+5)
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Toán Lớp 8: (8)/(x-5) – (3)/(x+5) + (2x-20)/(x^2-25)
(6)/(x-5) + (6-2x)/(x^2-25) + (4)/(x+5)
NHỚ TÌM MẪU THỨC CHUNG VÀ LÀM ĐẦY ĐỦ CÁC BƯỚC !
Comments ( 2 )
ĐKXĐ: x \ne +- 5
= {8x+ 40 – 3x + 15 + 2x – 20}/{(x+5)(x-5)}
= { 7x +35}/{(x+5)(x-5)}
= {7(x+5)}/{(x+5)(x-5)}
= 7/{x-5}
= {6x+30 + 6 -2x + 4x – 20}/{(x+5)(x-5)}
= { 8x -16}/{(x+5)(x-5)}
a)\frac{8}{{x – 5}} – \frac{3}{{x + 5}} + \frac{{2x – 20}}{{{x^2} – 25}}\,\,\left( {DK:x \ne 5;\, – 5} \right)\\
= \frac{8}{{x – 5}} – \frac{3}{{x + 5}} + \frac{{2x – 20}}{{\left( {x – 5} \right)\left( {x + 5} \right)}}\\
= \frac{{8\left( {x + 5} \right)}}{{\left( {x + 5} \right)\left( {x – 5} \right)}} – \frac{{3\left( {x – 5} \right)}}{{\left( {x + 5} \right)\left( {x – 5} \right)}} + \frac{{2x – 20}}{{\left( {x + 5} \right)\left( {x – 5} \right)}}\\
= \frac{{8x + 40}}{{\left( {x + 5} \right)\left( {x – 5} \right)}} – \frac{{3x – 15}}{{\left( {x + 5} \right)\left( {x – 5} \right)}} + \frac{{2x – 20}}{{\left( {x + 5} \right)\left( {x – 5} \right)}}\\
= \frac{{8x + 40 – 3x + 15 + 2x – 20}}{{\left( {x + 5} \right)\left( {x – 5} \right)}}\\
= \frac{{7x + 35}}{{\left( {x + 5} \right)\left( {x – 5} \right)}}\\
= \frac{{7\left( {x + 5} \right)}}{{\left( {x + 5} \right)\left( {x – 5} \right)}}\\
= \frac{7}{{x – 5}}\\
b)\frac{6}{{x – 5}} + \frac{{6 – 2x}}{{{x^2} – 25}} + \frac{4}{{x + 5}}\left( {DK:x \ne 5;\, – 5} \right)\\
= \frac{6}{{x – 5}} + \frac{{6 – 2x}}{{\left( {x – 5} \right)\left( {x + 5} \right)}} + \frac{4}{{x + 5}}\\
= \frac{{6\left( {x + 5} \right)}}{{\left( {x + 5} \right)\left( {x – 5} \right)}} + \frac{{6 – 2x}}{{\left( {x + 5} \right)\left( {x – 5} \right)}} + \frac{{4\left( {x – 5} \right)}}{{\left( {x + 5} \right)\left( {x – 5} \right)}}\\
= \frac{{6x + 30}}{{\left( {x + 5} \right)\left( {x – 5} \right)}} + \frac{{6 – 2x}}{{\left( {x + 5} \right)\left( {x – 5} \right)}} + \frac{{4x – 20}}{{\left( {x + 5} \right)\left( {x – 5} \right)}}\\
= \frac{{6x + 30 + 6 – 2x + 4x – 20}}{{\left( {x + 5} \right)\left( {x – 5} \right)}}\\
= \frac{{8x – 16}}{{\left( {x + 5} \right)\left( {x – 5} \right)}}\\
\end{array}\]