Giúp em vs ạ $\frac{3}{2x +3}$ – $\frac{x – 6}{2x^{2}}$ + 6x 04/07/2021 Bởi Caroline Giúp em vs ạ $\frac{3}{2x +3}$ – $\frac{x – 6}{2x^{2}}$ + 6x
Đáp án:Chắc bạn ghi `(x-6)/(2x^2+6x)` thành `(x-6)/(2x^2)+6x`. Điều kiện:`2x+3 ne 0,2x^2+6x ne 0` `<=>x ne 0,x ne -3/2,x ne -3` `3/(2x+3)-(x-6)/(2x^2+6x)` `=3/(2x+3)-(x-6)/(2x(x+3))` `=(3.2x(x+3)-(x-6)(2x+3))/(2x.(x+3)(2x+3))` `=(6x^2+18x-(2x^2+3x-12x-18))/(2x(x+3)(2x+3))` `=(6x^2+18x-(2x^2-9x-18))/(2x(x+3)(2x+3))` `=(4x^2+27x+18)/(2x(x+3)(2x+3))` Bình luận
Điều kiện xác định $\left\{ \begin{array}{l} 2x + 3 \ne 0\\ 2{x^2} + 6x \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ne – \dfrac{3}{2}\\ x \ne 0\\ x \ne – 3 \end{array} \right.$ $\begin{array}{l} \dfrac{3}{{2x + 3}} – \dfrac{{x – 6}}{{2{x^2} + 6x}}\\ = \dfrac{3}{{2x + 3}} – \dfrac{{x – 6}}{{2x\left( {x + 3} \right)}}\\ = \dfrac{{3.2x\left( {x + 3} \right) – \left( {x – 6} \right)\left( {2x + 3} \right)}}{{\left( {2x + 3} \right).2x\left( {x + 3} \right)}}\\ = \dfrac{{6{x^2} + 18x – \left( {2{x^2} – 9x – 18} \right)}}{{\left( {2x + 3} \right).2x}}\\ = \dfrac{{4{x^2} + 27x + 18}}{{\left( {2x + 3} \right)2x.\left( {x + 3} \right)}} \end{array}$ Bình luận
Đáp án:Chắc bạn ghi `(x-6)/(2x^2+6x)` thành `(x-6)/(2x^2)+6x`.
Điều kiện:`2x+3 ne 0,2x^2+6x ne 0`
`<=>x ne 0,x ne -3/2,x ne -3`
`3/(2x+3)-(x-6)/(2x^2+6x)`
`=3/(2x+3)-(x-6)/(2x(x+3))`
`=(3.2x(x+3)-(x-6)(2x+3))/(2x.(x+3)(2x+3))`
`=(6x^2+18x-(2x^2+3x-12x-18))/(2x(x+3)(2x+3))`
`=(6x^2+18x-(2x^2-9x-18))/(2x(x+3)(2x+3))`
`=(4x^2+27x+18)/(2x(x+3)(2x+3))`
Điều kiện xác định $\left\{ \begin{array}{l} 2x + 3 \ne 0\\ 2{x^2} + 6x \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x \ne – \dfrac{3}{2}\\ x \ne 0\\ x \ne – 3 \end{array} \right.$
$\begin{array}{l} \dfrac{3}{{2x + 3}} – \dfrac{{x – 6}}{{2{x^2} + 6x}}\\ = \dfrac{3}{{2x + 3}} – \dfrac{{x – 6}}{{2x\left( {x + 3} \right)}}\\ = \dfrac{{3.2x\left( {x + 3} \right) – \left( {x – 6} \right)\left( {2x + 3} \right)}}{{\left( {2x + 3} \right).2x\left( {x + 3} \right)}}\\ = \dfrac{{6{x^2} + 18x – \left( {2{x^2} – 9x – 18} \right)}}{{\left( {2x + 3} \right).2x}}\\ = \dfrac{{4{x^2} + 27x + 18}}{{\left( {2x + 3} \right)2x.\left( {x + 3} \right)}} \end{array}$