Toán Lớp 10: tìm đk của bpt : 2x-3/2x+3>x+1
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Toán Lớp 10: tìm đk của bpt : 2x-3/2x+3>x+1
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\dfrac{{2x – 3}}{{2x + 3}} > x + 1\\
\Leftrightarrow \dfrac{{2x – 3}}{{2x + 3}} – x – 1 > 0\\
\Leftrightarrow \dfrac{{2x – 3 – \left( {x + 1} \right)\left( {2x + 3} \right)}}{{2x + 3}} > 0\\
\Leftrightarrow \dfrac{{2x – 3 – 2{x^2} – 3x – 2x – 3}}{{2x + 3}} > 0\\
\Leftrightarrow \dfrac{{ – 2{x^2} – 3x – 6}}{{2x + 3}} > 0\\
\Leftrightarrow \dfrac{{{x^2} + \dfrac{3}{2}x + 3}}{{2x + 3}} < 0\\
\Leftrightarrow \dfrac{{{x^2} + 2.x.\dfrac{3}{4} + \dfrac{9}{{16}} + \dfrac{{39}}{{16}}}}{{2x + 3}} < 0\\
\Leftrightarrow \dfrac{{{{\left( {x + \dfrac{3}{4}} \right)}^2} + \dfrac{{39}}{{16}}}}{{2x + 3}} < 0\\
\Leftrightarrow 2x + 3 < 0\left( {do:{{\left( {x + \dfrac{3}{4}} \right)}^2} + \dfrac{{39}}{{16}} > 0} \right)\\
\Leftrightarrow 2x < – 3\\
\Leftrightarrow x < \dfrac{{ – 3}}{2}\\
Vậy\,x < – \dfrac{3}{2}
\end{array}$