Giải thích các bước giải:
Bài 1:
Ta có:
\(\begin{array}{l}
a,\\
M = \dfrac{{{x^4} + 2}}{{{x^6} + 1}} + \dfrac{{{x^2} - 1}}{{{x^4} - {x^2} + 1}} - \dfrac{{{x^2} + 3}}{{{x^4} + 4{x^2} + 3}}\\
= \dfrac{{{x^4} + 2}}{{{{\left( {{x^2}} \right)}^3} + {1^3}}} + \dfrac{{{x^2} - 1}}{{{x^4} - {x^2} + 1}} - \dfrac{{{x^2} + 3}}{{\left( {{x^4} + {x^2}} \right) + \left( {3{x^2} + 3} \right)}}\\
= \dfrac{{{x^4} + 2}}{{\left( {{x^2} + 1} \right).\left( {{x^4} - {x^2} + 1} \right)}} + \dfrac{{{x^2} - 1}}{{{x^4} - {x^2} + 1}} - \dfrac{{{x^2} + 3}}{{{x^2}\left( {{x^2} + 1} \right) + 3\left( {{x^2} + 1} \right)}}\\
= \dfrac{{{x^4} + 2}}{{\left( {{x^2} + 1} \right).\left( {{x^4} - {x^2} + 1} \right)}} + \dfrac{{{x^2} - 1}}{{{x^4} - {x^2} + 1}} - \dfrac{{{x^2} + 3}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 3} \right)}}\\
= \dfrac{{{x^4} + 2}}{{\left( {{x^2} + 1} \right).\left( {{x^4} - {x^2} + 1} \right)}} + \dfrac{{{x^2} - 1}}{{{x^4} - {x^2} + 1}} - \dfrac{1}{{{x^2} + 1}}\\
= \dfrac{{\left( {{x^4} + 2} \right) + \left( {{x^2} - 1} \right).\left( {{x^2} + 1} \right) - \left( {{x^4} - {x^2} + 1} \right)}}{{\left( {{x^2} + 1} \right).\left( {{x^4} - {x^2} + 1} \right)}}\\
= \dfrac{{{x^4} + 2 + {x^4} - 1 - {x^4} + {x^2} - 1}}{{\left( {{x^2} + 1} \right).\left( {{x^4} - {x^2} + 1} \right)}}\\
= \dfrac{{{x^4} + {x^2}}}{{\left( {{x^2} + 1} \right).\left( {{x^4} - {x^2} + 1} \right)}}\\
= \dfrac{{{x^2}\left( {{x^2} + 1} \right)}}{{\left( {{x^2} + 1} \right).\left( {{x^4} - {x^2} + 1} \right)}}\\
= \dfrac{{{x^2}}}{{{x^4} - {x^2} + 1}}\\
b,\\
M = \dfrac{{{x^2}}}{{{x^4} - {x^2} + 1}}\\
\Rightarrow M - 1 = \dfrac{{{x^2}}}{{{x^4} - {x^2} + 1}} - 1 = \dfrac{{{x^2} - \left( {{x^4} - {x^2} + 1} \right)}}{{{x^4} - {x^2} + 1}} = \dfrac{{ - {x^4} + 2{x^2} - 1}}{{\left( {{x^4} - {x^2} + \dfrac{1}{4}} \right) + \dfrac{3}{4}}}\\
= \dfrac{{ - \left( {{x^4} - 2{x^2} + 1} \right)}}{{{{\left( {{x^2} - \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}}} = \dfrac{{ - {{\left( {{x^2} - 1} \right)}^2}}}{{{{\left( {{x^2} - \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}}} \le 0\\
\Rightarrow M - 1 \le 0 \Leftrightarrow M \le 1\\
\Rightarrow {M_{\max }} = 1 \Leftrightarrow {\left( {{x^2} - 1} \right)^2} = 0 \Leftrightarrow {x^2} = 1 \Leftrightarrow x = \pm 1\\
2,\\
\dfrac{{1 - 2x}}{{1 - x}} + \dfrac{{1 - 2y}}{{1 - y}} = 1\\
\Leftrightarrow \dfrac{{\left( {1 - 2x} \right)\left( {1 - y} \right) + \left( {1 - 2y} \right)\left( {1 - x} \right)}}{{\left( {1 - x} \right)\left( {1 - y} \right)}} = 1\\
\Leftrightarrow \dfrac{{1 - y - 2x + 2xy + 1 - x - 2y + 2xy}}{{1 - x - y + xy}} = 1\\
\Leftrightarrow \dfrac{{2 - 3x - 3y + 4xy}}{{1 - x - y + xy}} = 1\\
\Leftrightarrow 2 - 3x - 3y + 4xy = 1 - x - y + xy\\
\Leftrightarrow 1 - 2x - 2y + 3xy = 0\\
\Leftrightarrow 1 - 2x - 2y + 2xy = - xy\\
\Rightarrow M = {x^2} + {y^2} - xy = {x^2} + {y^2} + 1 - 2x - 2y + 2xy\\
= \left( {{x^2} + {y^2} + 2xy} \right) - 2.\left( {x + y} \right) + 1\\
= {\left( {x + y} \right)^2} - 2.\left( {x + y} \right) + 1\\
= {\left( {x + y - 1} \right)^2}
\end{array}\)
