1)Rút gọn P=2/2x+3+3/2x+1-6x+5/(2x+3).(2x-3) Nếu thấy sai đề bài thì sửa giúp mình nhé Mình đang cần gấp 16/07/2021 Bởi Ariana 1)Rút gọn P=2/2x+3+3/2x+1-6x+5/(2x+3).(2x-3) Nếu thấy sai đề bài thì sửa giúp mình nhé Mình đang cần gấp
Đáp án: \(\dfrac{{8{x^2} – 24x – 38}}{{\left( {4{x^2} – 9} \right)\left( {2x + 1} \right)}}\) Giải thích các bước giải: \(\begin{array}{l}DK:x \ne \left\{ { – \dfrac{3}{2}; – \dfrac{1}{2};\dfrac{3}{2}} \right\}\\P = \dfrac{2}{{2x + 3}} + \dfrac{3}{{2x + 1}} – \dfrac{{6x + 5}}{{\left( {2x + 3} \right)\left( {2x – 3} \right)}}\\ = \dfrac{{2\left( {2x – 3} \right)\left( {2x + 1} \right) + 3\left( {4{x^2} – 9} \right) – \left( {6x + 5} \right)\left( {2x + 1} \right)}}{{\left( {2x + 3} \right)\left( {2x – 3} \right)\left( {2x + 1} \right)}}\\ = \dfrac{{\left( {2x + 1} \right)\left( {4x – 6} \right) + 12{x^2} – 27 – 12{x^2} – 16x – 5}}{{\left( {2x + 3} \right)\left( {2x – 3} \right)\left( {2x + 1} \right)}}\\ = \dfrac{{8{x^2} – 8x – 6 – 16x – 32}}{{\left( {2x + 3} \right)\left( {2x – 3} \right)\left( {2x + 1} \right)}}\\ = \dfrac{{8{x^2} – 24x – 38}}{{\left( {4{x^2} – 9} \right)\left( {2x + 1} \right)}}\end{array}\) Bình luận
Đáp án:
\(\dfrac{{8{x^2} – 24x – 38}}{{\left( {4{x^2} – 9} \right)\left( {2x + 1} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne \left\{ { – \dfrac{3}{2}; – \dfrac{1}{2};\dfrac{3}{2}} \right\}\\
P = \dfrac{2}{{2x + 3}} + \dfrac{3}{{2x + 1}} – \dfrac{{6x + 5}}{{\left( {2x + 3} \right)\left( {2x – 3} \right)}}\\
= \dfrac{{2\left( {2x – 3} \right)\left( {2x + 1} \right) + 3\left( {4{x^2} – 9} \right) – \left( {6x + 5} \right)\left( {2x + 1} \right)}}{{\left( {2x + 3} \right)\left( {2x – 3} \right)\left( {2x + 1} \right)}}\\
= \dfrac{{\left( {2x + 1} \right)\left( {4x – 6} \right) + 12{x^2} – 27 – 12{x^2} – 16x – 5}}{{\left( {2x + 3} \right)\left( {2x – 3} \right)\left( {2x + 1} \right)}}\\
= \dfrac{{8{x^2} – 8x – 6 – 16x – 32}}{{\left( {2x + 3} \right)\left( {2x – 3} \right)\left( {2x + 1} \right)}}\\
= \dfrac{{8{x^2} – 24x – 38}}{{\left( {4{x^2} – 9} \right)\left( {2x + 1} \right)}}
\end{array}\)