Đáp án:
f) \(\dfrac{1}{{x + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)8{x^3} - 4{x^2} + 4{x^2} - 2x + 2x - 1\\
= 8{x^3} - 1\\
b)\dfrac{x}{{2\left( {x - 1} \right)}} + \dfrac{{ - {x^2} - 1}}{{2\left( {1 - x} \right)\left( {1 + x} \right)}} - 1\\
= \dfrac{{x\left( {x + 1} \right) + {x^2} + 1 - 2\left( {{x^2} - 1} \right)}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{{x^2} + x + {x^2} + 1 - 2{x^2} + 2}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{x + 3}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
c)\dfrac{{3{x^4} + {x^3} - {x^2} - 6{x^2} - 2x + 2}}{{3{x^2} + x - 1}}\\
= \dfrac{{{x^2}\left( {3{x^2} + x - 1} \right) - 2\left( {3{x^2} + x - 1} \right)}}{{3{x^2} + x - 1}}\\
= \dfrac{{\left( {3{x^2} + x - 1} \right)\left( {{x^2} - 2} \right)}}{{3{x^2} + x - 1}}\\
= {x^2} - 2\\
d)4{x^3} + 2{x^2}y + x{y^2} - 8{x^2}y - 4x{y^2} - 2{y^3}\\
= 4{x^3} - 6{x^2}y - 3x{y^2} - 2{y^3}\\
e)\dfrac{{4\left( { - {a^2} + {b^2}} \right) - 2a\left( {3a - 2b} \right) - {b^2}}}{{12ab}}\\
= \dfrac{{ - 4{a^2} - 4{b^2} - 6{a^2} + 4ab - {b^2}}}{{12ab}}\\
= \dfrac{{ - 10{a^2} - 5{b^2} + 4ab}}{{12ab}}\\
f)\dfrac{{{{\left( {x + 1} \right)}^2} - {x^2} - 3}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{2x + 1 - 3}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}} = \dfrac{{2x - 2}}{{2\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{1}{{x + 1}}
\end{array}\)
