Thực hiện phép tính: (2 x+1/2x-1-2x-1/2x+1):4x/10x-5 02/08/2021 Bởi Josie Thực hiện phép tính: (2 x+1/2x-1-2x-1/2x+1):4x/10x-5
Đáp án: \(\dfrac{{5\left( {6x – 1} \right)}}{{4x\left( {2x + 1} \right)}}\) Giải thích các bước giải: \(\begin{array}{l}\left( {\dfrac{{2x}}{{2x – 1}} – \dfrac{{2x – 1}}{{2x + 1}}} \right):\dfrac{{4x}}{{10x – 5}}\\ = \dfrac{{2x\left( {2x + 1} \right) – \left( {2x – 1} \right)\left( {2x – 1} \right)}}{{\left( {2x – 1} \right)\left( {2x + 1} \right)}}:\dfrac{{4x}}{{5\left( {2x – 1} \right)}}\\ = \dfrac{{4{x^2} + 2x – 4{x^2} + 4x – 1}}{{\left( {2x – 1} \right)\left( {2x + 1} \right)}}.\dfrac{{5\left( {2x – 1} \right)}}{{4x}}\\ = \dfrac{{6x – 1}}{{\left( {2x – 1} \right)\left( {2x + 1} \right)}}.\dfrac{{5\left( {2x – 1} \right)}}{{4x}}\\ = \dfrac{{\left( {6x – 1} \right).5\left( {2x – 1} \right)}}{{\left( {2x – 1} \right)\left( {2x + 1} \right).4x}} = \dfrac{{5\left( {6x – 1} \right)}}{{4x\left( {2x + 1} \right)}}\end{array}\) Bình luận
Đáp án: Giải thích các bước giải: \(\left( {\dfrac{{2x}}{{2x – 1}} – \dfrac{{2x – 1}}{{2x + 1}}} \right):\dfrac{{4x}}{{10x – 5}}\) \(= \dfrac{{2x\left( {2x + 1} \right) – \left( {2x – 1} \right)\left( {2x – 1} \right)}}{{\left( {2x – 1} \right)\left( {2x + 1} \right)}}:\dfrac{{4x}}{{5\left( {2x – 1} \right)}}\) \( = \dfrac{{4{x^2} + 2x – 4{x^2} + 4x – 1}}{{\left( {2x – 1} \right)\left( {2x + 1} \right)}}.\dfrac{{5\left( {2x – 1} \right)}}{{4x}}\) \(= \dfrac{{6x – 1}}{{\left( {2x – 1} \right)\left( {2x + 1} \right)}}.\dfrac{{5\left( {2x – 1} \right)}}{{4x}}\) \(= \dfrac{{\left( {6x – 1} \right).5\left( {2x – 1} \right)}}{{\left( {2x – 1} \right)\left( {2x + 1} \right).4x}}\) \(= \dfrac{{5\left( {6x – 1} \right)}}{{4x\left( {2x + 1} \right)}}\) Bình luận
Đáp án:
\(\dfrac{{5\left( {6x – 1} \right)}}{{4x\left( {2x + 1} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
\left( {\dfrac{{2x}}{{2x – 1}} – \dfrac{{2x – 1}}{{2x + 1}}} \right):\dfrac{{4x}}{{10x – 5}}\\
= \dfrac{{2x\left( {2x + 1} \right) – \left( {2x – 1} \right)\left( {2x – 1} \right)}}{{\left( {2x – 1} \right)\left( {2x + 1} \right)}}:\dfrac{{4x}}{{5\left( {2x – 1} \right)}}\\
= \dfrac{{4{x^2} + 2x – 4{x^2} + 4x – 1}}{{\left( {2x – 1} \right)\left( {2x + 1} \right)}}.\dfrac{{5\left( {2x – 1} \right)}}{{4x}}\\
= \dfrac{{6x – 1}}{{\left( {2x – 1} \right)\left( {2x + 1} \right)}}.\dfrac{{5\left( {2x – 1} \right)}}{{4x}}\\
= \dfrac{{\left( {6x – 1} \right).5\left( {2x – 1} \right)}}{{\left( {2x – 1} \right)\left( {2x + 1} \right).4x}} = \dfrac{{5\left( {6x – 1} \right)}}{{4x\left( {2x + 1} \right)}}
\end{array}\)
Đáp án:
Giải thích các bước giải:
\(\left( {\dfrac{{2x}}{{2x – 1}} – \dfrac{{2x – 1}}{{2x + 1}}} \right):\dfrac{{4x}}{{10x – 5}}\)
\(= \dfrac{{2x\left( {2x + 1} \right) – \left( {2x – 1} \right)\left( {2x – 1} \right)}}{{\left( {2x – 1} \right)\left( {2x + 1} \right)}}:\dfrac{{4x}}{{5\left( {2x – 1} \right)}}\)
\( = \dfrac{{4{x^2} + 2x – 4{x^2} + 4x – 1}}{{\left( {2x – 1} \right)\left( {2x + 1} \right)}}.\dfrac{{5\left( {2x – 1} \right)}}{{4x}}\)
\(= \dfrac{{6x – 1}}{{\left( {2x – 1} \right)\left( {2x + 1} \right)}}.\dfrac{{5\left( {2x – 1} \right)}}{{4x}}\)
\(= \dfrac{{\left( {6x – 1} \right).5\left( {2x – 1} \right)}}{{\left( {2x – 1} \right)\left( {2x + 1} \right).4x}}\)
\(= \dfrac{{5\left( {6x – 1} \right)}}{{4x\left( {2x + 1} \right)}}\)