Toán 2x-4/ (√x^2-3x-10) >1 thành lời: 2x-4 phần căn x mũ 2 -3x-10 lớn hơn 1 30/08/2021 By Skylar 2x-4/ (√x^2-3x-10) >1 thành lời: 2x-4 phần căn x mũ 2 -3x-10 lớn hơn 1
Đáp án: $\begin{array}{l}Dkxd:{x^2} – 3x – 10 > 0\\ \Rightarrow \left( {x – 5} \right)\left( {x + 2} \right) > 0\\ \Rightarrow \left[ \begin{array}{l}x > 5\\x < – 2\end{array} \right.\left( 1 \right)\\\dfrac{{2x – 4}}{{\sqrt {{x^2} – 3x – 10} }} > 1\\ \Rightarrow \dfrac{{2x – 4 – \sqrt {{x^2} – 3x – 10} }}{{\sqrt {{x^2} – 3x – 10} }} > 0\\ \Rightarrow 2x – 3 – \sqrt {{x^2} – 3x – 10} > 0\\ \Rightarrow 2x – 3 > \sqrt {{x^2} – 3x – 10} > 0\\ \Rightarrow \left\{ \begin{array}{l}2x – 3 > 0\\{\left( {2x – 3} \right)^2} > {x^2} – 3x – 10\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}x > \dfrac{3}{2}\\4{x^2} – 12x + 9 > {x^2} – 3x – 10\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}x > \dfrac{3}{2}\\3{x^2} – 9x + 19 > 0\left( {tm} \right)\end{array} \right.\left( 2 \right)\\\left( 1 \right);\left( 2 \right) \Leftrightarrow x > 5\\Vậy\,x > 5\end{array}$ Trả lời
Đáp án:
$\begin{array}{l}
Dkxd:{x^2} – 3x – 10 > 0\\
\Rightarrow \left( {x – 5} \right)\left( {x + 2} \right) > 0\\
\Rightarrow \left[ \begin{array}{l}
x > 5\\
x < – 2
\end{array} \right.\left( 1 \right)\\
\dfrac{{2x – 4}}{{\sqrt {{x^2} – 3x – 10} }} > 1\\
\Rightarrow \dfrac{{2x – 4 – \sqrt {{x^2} – 3x – 10} }}{{\sqrt {{x^2} – 3x – 10} }} > 0\\
\Rightarrow 2x – 3 – \sqrt {{x^2} – 3x – 10} > 0\\
\Rightarrow 2x – 3 > \sqrt {{x^2} – 3x – 10} > 0\\
\Rightarrow \left\{ \begin{array}{l}
2x – 3 > 0\\
{\left( {2x – 3} \right)^2} > {x^2} – 3x – 10
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x > \dfrac{3}{2}\\
4{x^2} – 12x + 9 > {x^2} – 3x – 10
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x > \dfrac{3}{2}\\
3{x^2} – 9x + 19 > 0\left( {tm} \right)
\end{array} \right.\left( 2 \right)\\
\left( 1 \right);\left( 2 \right) \Leftrightarrow x > 5\\
Vậy\,x > 5
\end{array}$