thực hiện phép tính: a, 11x+13/3x-3 + 15x+17/4-4x b,2x+1/2x^2-2 + 32x^2/1-4x^2 + 1-2x/2x^2+x c,1/x^2+x+1 + 1/x^2-x ++2x/1-x^3 d,x+9/x^2-9 – 3/x^2+3x

Đáp án:

 b) -8

Giải thích các bước giải:

\(\begin{array}{l}
a)\dfrac{{11x + 13}}{{3\left( {x – 1} \right)}} + \dfrac{{15x + 17}}{{4\left( {1 – x} \right)}}\\
 = \dfrac{{4\left( {11x + 13} \right) – 3\left( {15x + 17} \right)}}{{12\left( {x – 1} \right)}}\\
 = \dfrac{{ – x + 1}}{{12\left( {x – 1} \right)}} =  – \dfrac{1}{{12}}\\
b)\dfrac{{2x + 1}}{{2{x^2} – x}} + \dfrac{{32{x^2}}}{{1 – 4{x^2}}} + \dfrac{{1 – 2x}}{{2{x^2} + x}}\\
 = \dfrac{{2x + 1}}{{x\left( {2x – 1} \right)}} + \dfrac{{32{x^2}}}{{\left( {1 – 2x} \right)\left( {1 + 2x} \right)}} + \dfrac{{1 – 2x}}{{x\left( {2x + 1} \right)}}\\
 = \dfrac{{ – {{\left( {2x + 1} \right)}^2} + 32{x^3} + {{\left( {1 – 2x} \right)}^2}}}{{x\left( {1 – 2x} \right)\left( {1 + 2x} \right)}}\\
 = \dfrac{{ – 4{x^2} – 4x – 1 + 32{x^3} + 1 – 4x + 4{x^2}}}{{x\left( {1 – 2x} \right)\left( {1 + 2x} \right)}}\\
 = \dfrac{{32{x^3} – 8x}}{{x\left( {1 – 2x} \right)\left( {1 + 2x} \right)}}\\
 = \dfrac{{8\left( {4{x^2} – 1} \right)}}{{1 – 4{x^2}}} =  – 8\\
c)\dfrac{1}{{{x^2} + x + 1}} + \dfrac{1}{{{x^2} – x}} + \dfrac{1}{{1 – {x^3}}}\\
 = \dfrac{1}{{{x^2} + x + 1}} + \dfrac{1}{{x\left( {x – 1} \right)}} – \dfrac{1}{{\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\\
 = \dfrac{{x\left( {x – 1} \right) + {x^2} + x + 1 – x}}{{x\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\\
 = \dfrac{{2{x^2} – x + 1}}{{x\left( {x – 1} \right)\left( {{x^2} + x + 1} \right)}}\\
d)\dfrac{{x + 9}}{{{x^2} – 9}} – \dfrac{3}{{{x^2} + 3x}}\\
 = \dfrac{{x\left( {x + 9} \right) – 3\left( {x – 3} \right)}}{{x\left( {x + 3} \right)\left( {x – 3} \right)}}\\
 = \dfrac{{{x^2} + 9x – 3x + 9}}{{x\left( {x + 3} \right)\left( {x – 3} \right)}}\\
 = \dfrac{{{x^2} + 6x + 9}}{{x\left( {x + 3} \right)\left( {x – 3} \right)}} = \dfrac{{{{\left( {x + 3} \right)}^2}}}{{x\left( {x + 3} \right)\left( {x – 3} \right)}}\\
 = \dfrac{{x + 3}}{{x\left( {x – 3} \right)}}
\end{array}\)