ĐÁNH GIÁ SAU LIÊN HỢP

$$\begin{array}{l} \sqrt {4x + 1} – \sqrt {3x – 2} = \dfrac{{x + 3}}{5}\\ \Leftrightarrow 5\sqrt {4x + 1} – 5\sqrt {3x – 2} = x + 3\\ \Leftrightarrow 5\left( {\sqrt {4x + 1} – 3} \right) – 5\left( {\sqrt {3x – 2} – 2} \right) – \left( {x – 2} \right) = 0\\ \Leftrightarrow 5.\dfrac{{4\left( {x – 2} \right)}}{{\sqrt {4x + 1} + 3}} – 5.\dfrac{{3\left( {x – 2} \right)}}{{\sqrt {3x – 2} + 2}} – \left( {x – 2} \right) = 0\\ \Leftrightarrow \left( {x – 2} \right)\left[ {\dfrac{{20}}{{\sqrt {4x + 1} + 3}} – \dfrac{{15}}{{\sqrt {3x – 2} + 2}} – 1} \right] = 0\\ \Leftrightarrow \left( {x – 2} \right).A = 0 \end{array}$$

$$\begin{array}{l} A = \dfrac{{20}}{{\sqrt {4x + 1} + 3}} – \dfrac{{15}}{{\sqrt {3x – 2} + 2}} – 1\\ = \dfrac{{20\left( {\sqrt {3x – 2} + 2} \right) – 15\left( {\sqrt {4x + 1} + 3} \right) – \left( {\sqrt {4x + 1} + 3} \right)\left( {\sqrt {3x – 2} + 2} \right)}}{{\left( {\sqrt {4x + 1} + 3} \right)\left( {\sqrt {3x – 2} + 2} \right)}}\\ = \dfrac{{20\sqrt {3x – 2} – 3\sqrt {3x – 2} – 15\sqrt {4x + 1} – 2\sqrt {4x + 1} + 40 – 45 – 6 – \sqrt {\left( {4x + 1} \right)\left( {3x – 2} \right)} }}{{\left( {\sqrt {4x + 1} + 3} \right)\left( {\sqrt {3x – 2} + 2} \right)}}\\ = 17.\dfrac{{\sqrt {3x – 2} – \sqrt {4x + 1} }}{{\left( {\sqrt {4x + 1} + 3} \right)\left( {\sqrt {3x – 2} + 2} \right)}} – \dfrac{{11 + \sqrt {\left( {4x + 1} \right)\left( {3x – 2} \right)} }}{{\left( {\sqrt {4x + 1} + 3} \right)\left( {\sqrt {3x – 2} + 2} \right)}}\\ = 17.\dfrac{{ – x – 3}}{{\left( {\sqrt {3x – 2} – \sqrt {4x + 1} } \right)\left( {\sqrt {4x + 1} + 3} \right)\left( {\sqrt {3x – 2} + 2} \right)}} – \dfrac{{11 + \sqrt {\left( {4x + 1} \right)\left( {3x – 2} \right)} }}{{\left( {\sqrt {4x + 1} + 3} \right)\left( {\sqrt {3x – 2} + 2} \right)}}\\ < 0 \end{array}$$