`Q = [(2x – x^2) / (2x^2 + 8) – (2x^2)/(8+ 2x^2 – 4x +x^3)][1-1/x -2/x^2]` 30/10/2021 Bởi Valentina `Q = [(2x – x^2) / (2x^2 + 8) – (2x^2)/(8+ 2x^2 – 4x +x^3)][1-1/x -2/x^2]`
Đáp án: \(\dfrac{{ – \left( {{x^2} + 4x – 4} \right)\left( {x – 2} \right)\left( {x + 1} \right)}}{{2x\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}\) Giải thích các bước giải: \(\begin{array}{l}DK:x \ne 0\\Q = \left[ {\dfrac{{ – {x^2} + 2x}}{{2\left( {{x^2} + 4} \right)}} – \dfrac{{2{x^2}}}{{{x^3} + 4x + 2{x^2} + 8}}} \right].\left( {\dfrac{{{x^2} – x – 2}}{{{x^2}}}} \right)\\ = \left[ {\dfrac{{ – {x^2} + 2x}}{{2\left( {{x^2} + 4} \right)}} – \dfrac{{2{x^2}}}{{x\left( {{x^2} + 4} \right) + 2\left( {{x^2} + 4} \right)}}} \right].\left( {\dfrac{{{x^2} – x – 2}}{{{x^2}}}} \right)\\ = \left[ {\dfrac{{ – {x^2} + 2x}}{{2\left( {{x^2} + 4} \right)}} – \dfrac{{2{x^2}}}{{\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}} \right].\left( {\dfrac{{{x^2} – x – 2}}{{{x^2}}}} \right)\\ = \dfrac{{\left( {x + 2} \right)\left( { – {x^2} + 2x} \right) – 4{x^2}}}{{2\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}.\left( {\dfrac{{{x^2} – x – 2}}{{{x^2}}}} \right)\\ = \dfrac{{ – {x^3} + 2{x^2} – 2{x^2} + 4x – 4{x^2}}}{{2\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}.\dfrac{{\left( {x – 2} \right)\left( {x + 1} \right)}}{{{x^2}}}\\ = \dfrac{{ – {x^3} – 4{x^2} + 4x}}{{2\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}.\dfrac{{\left( {x – 2} \right)\left( {x + 1} \right)}}{{{x^2}}}\\ = \dfrac{{ – x\left( {{x^2} + 4x – 4} \right)}}{{2\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}.\dfrac{{\left( {x – 2} \right)\left( {x + 1} \right)}}{{{x^2}}}\\ = \dfrac{{ – \left( {{x^2} + 4x – 4} \right)\left( {x – 2} \right)\left( {x + 1} \right)}}{{2x\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}\end{array}\) ( bạn xem lại đề bài có nhầm số hay dấu không nhé ) Bình luận
Đáp án:
\(\dfrac{{ – \left( {{x^2} + 4x – 4} \right)\left( {x – 2} \right)\left( {x + 1} \right)}}{{2x\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne 0\\
Q = \left[ {\dfrac{{ – {x^2} + 2x}}{{2\left( {{x^2} + 4} \right)}} – \dfrac{{2{x^2}}}{{{x^3} + 4x + 2{x^2} + 8}}} \right].\left( {\dfrac{{{x^2} – x – 2}}{{{x^2}}}} \right)\\
= \left[ {\dfrac{{ – {x^2} + 2x}}{{2\left( {{x^2} + 4} \right)}} – \dfrac{{2{x^2}}}{{x\left( {{x^2} + 4} \right) + 2\left( {{x^2} + 4} \right)}}} \right].\left( {\dfrac{{{x^2} – x – 2}}{{{x^2}}}} \right)\\
= \left[ {\dfrac{{ – {x^2} + 2x}}{{2\left( {{x^2} + 4} \right)}} – \dfrac{{2{x^2}}}{{\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}} \right].\left( {\dfrac{{{x^2} – x – 2}}{{{x^2}}}} \right)\\
= \dfrac{{\left( {x + 2} \right)\left( { – {x^2} + 2x} \right) – 4{x^2}}}{{2\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}.\left( {\dfrac{{{x^2} – x – 2}}{{{x^2}}}} \right)\\
= \dfrac{{ – {x^3} + 2{x^2} – 2{x^2} + 4x – 4{x^2}}}{{2\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}.\dfrac{{\left( {x – 2} \right)\left( {x + 1} \right)}}{{{x^2}}}\\
= \dfrac{{ – {x^3} – 4{x^2} + 4x}}{{2\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}.\dfrac{{\left( {x – 2} \right)\left( {x + 1} \right)}}{{{x^2}}}\\
= \dfrac{{ – x\left( {{x^2} + 4x – 4} \right)}}{{2\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}.\dfrac{{\left( {x – 2} \right)\left( {x + 1} \right)}}{{{x^2}}}\\
= \dfrac{{ – \left( {{x^2} + 4x – 4} \right)\left( {x – 2} \right)\left( {x + 1} \right)}}{{2x\left( {{x^2} + 4} \right)\left( {x + 2} \right)}}
\end{array}\)
( bạn xem lại đề bài có nhầm số hay dấu không nhé )