cong phan thuc
a)x2+2/x2+4+5/x+2
b)x+y/2+x+2/2x2+4
c)8/(x2+3)(x2-1)+2/x2+3+1/X+1
Quảng cáo
2 câu trả lời 334
\[\begin{array}{l}
a)\frac{{{x^2} + 2}}{{{x^2} + 4}} + \frac{x}{{x + 2}}\\
dk:x \ne - 2\\
= \frac{{\left( {{x^2} + 2} \right)\left( {x + 2} \right) + x\left( {{x^2} + 4} \right)}}{{\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}\\
= \frac{{{x^3} + 2{{\rm{x}}^2} + 2{\rm{x}} + 4 + {x^3} + 4{\rm{x}}}}{{\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}\\
= \frac{{2{{\rm{x}}^3} + 2{{\rm{x}}^2} + 6{\rm{x + 4}}}}{{\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}\\
b)\frac{{x + y}}{{2 + x}} + \frac{2}{{2{{\rm{x}}^2} + 4}}\\
dk:x \ne - 2\\
= \frac{{x + y}}{{2 + x}} + \frac{2}{{2\left( {{x^2} + 2} \right)}}\\
= \frac{{x + y}}{{2 + x}} + \frac{1}{{{x^2} + 2}}\\
= \frac{{\left( {x + y} \right)\left( {{x^2} + 2} \right) + 2 + x}}{{\left( {2 + x} \right)\left( {{x^2} + 2} \right)}}\\
= \frac{{{x^3} + 2{\rm{x}} + {x^2}y + 2y + 2 + x}}{{\left( {2 + x} \right)\left( {{x^2} + 2} \right)}}\\
= \frac{{{x^3} + {x^2}y + 3{\rm{x}} + 2y + 2}}{{\left( {2 + x} \right)\left( {{x^2} + 2} \right)}}\\
c)\frac{8}{{\left( {{x^2} + 3} \right)\left( {{x^2} - 1} \right)}} + \frac{2}{{{x^2} + 3}} + \frac{1}{{x + 1}}\\
dk:x \ne \pm 1\\
= \frac{8}{{\left( {{x^2} + 3} \right)\left( {x - 1} \right)\left( {x + 1} \right)}} + \frac{2}{{{x^2} + 3}} + \frac{1}{{x + 1}}\\
= \frac{{8 + 2\left( {{x^2} - 1} \right) + \left( {{x^2} + 3} \right)\left( {x - 1} \right)}}{{\left( {{x^2} + 3} \right)\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{{8 + 2{{\rm{x}}^2} - 2 + {x^3} - {x^2} + 3{\rm{x}} - 3}}{{\left( {{x^2} + 3} \right)\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{{{x^3} + {x^2} + 3{\rm{x}} + 3}}{{\left( {{x^2} + 3} \right)\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{{{x^2}\left( {x + 1} \right) + 3\left( {x + 1} \right)}}{{\left( {{x^2} + 3} \right)\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{{\left( {x + 1} \right)\left( {{x^2} + 3} \right)}}{{\left( {{x^2} + 3} \right)\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \frac{1}{{x - 1}}
\end{array}\]
Quảng cáo
Bạn cần hỏi gì?
Câu hỏi hot cùng chủ đề
-
Đã trả lời bởi chuyên gia
16474 -
Đã trả lời bởi chuyên gia
8211 -
Đã trả lời bởi chuyên gia
7760 -
Đã trả lời bởi chuyên gia
6998
