Tìm \(\lim \frac{{\sqrt {{n^2} + n}  - n}}{{\sqrt {4{n^2} + 3n}  - 2n}}\)

* Hướng dẫn giải

1.  \(\lim \frac{{\sqrt {{n^2} + n}  - n}}{{\sqrt {4{n^2} + 3n}  - 2n}} = \lim \frac{{n\left( {\sqrt {4{n^2} + 3n}  + 2n} \right)}}{{3n\left( {\sqrt {{n^2} + n}  + n} \right)}}\)

\( = \lim \frac{{\sqrt {4{n^2} + 3n}  + 2n}}{{3\left( {\sqrt {{n^2} + n}  + n} \right)}} = \lim \frac{{\sqrt {4 + \frac{3}{n}}  + 2}}{{3\left( {\sqrt {1 + \frac{1}{n}}  + 1} \right)}} = \frac{2}{3}\)

2. \(\left\{ \begin{array}{l}
x + 4 + \sqrt {{x^2} + 8x + 17}  = y + \sqrt {{y^2} + 1} \,\,\left( 1 \right)\\
x + \sqrt y  + \sqrt {y + 21}  + 1 = 2\sqrt {4y - 3x} \,\,\left( 2 \right)
\end{array} \right.\)

Điều kiện: \(y \ge 0\)

 \(\begin{array}{l}
\left( 1 \right) \Leftrightarrow (x - y + 4) + \sqrt {{x^2} + 8x + 17}  - \sqrt {{y^2} + 1}  = 0\\
 \Leftrightarrow \left( {x - y + 4} \right) + \frac{{{{\left( {x + 4} \right)}^2} - {y^2}}}{{\sqrt {{x^2} + 8x + 17}  + \sqrt {{y^2} + 1} }} = 0
\end{array}\)

\(\begin{array}{l}
 \Leftrightarrow \left( {x - y + 4} \right) + \frac{{\left( {x + 4 + y} \right)\left( {x + 4 - y} \right)}}{{\sqrt {{x^2} + 8x + 17}  + \sqrt {{y^2} + 1} }} = 0\\
 \Leftrightarrow \left( {x - y + 4} \right)(1 + \frac{{\left( {x + 4 + y} \right)}}{{\sqrt {{x^2} + 8x + 17}  + \sqrt {{y^2} + 1} }}) = 0\\
 \Leftrightarrow y = x + 4
\end{array}\)

Vì \(1 + \frac{{\left( {x + 4 + y} \right)}}{{\sqrt {{x^2} + 8x + 17}  + \sqrt {{y^2} + 1} }} = \frac{{\sqrt {{{\left( {x + 4} \right)}^2} + 1}  + \left( {x + 4} \right) + \sqrt {{y^2} + 1}  + y}}{{\sqrt {{x^2} + 8x + 17}  = \sqrt {{y^2} + 1} }} > 0,\forall x,y\)

Thay y = x + 4 vào (2) ta được:

\(\begin{array}{l}
\left( 2 \right) \Leftrightarrow x + \sqrt {x + 4}  + \sqrt {x + 25}  + 1 = 2\sqrt {x + 16} \\
 \Leftrightarrow \left( {\sqrt {x + 4}  - 2} \right) + \left( {\sqrt {x + 25}  - 5} \right) + \left( {x + 8 - 2\sqrt {x + 16}  = 0} \right)\\
 \Leftrightarrow x\left( {\frac{1}{{\sqrt {x + 4}  + 2}} + \frac{1}{{\sqrt {x + 25}  + 5}} + \frac{{x + 12}}{{x + 8 + 2\sqrt {x + 16} }}} \right) = 0
\end{array}\)

\(\left[ \begin{array}{l}
x = 0 \Rightarrow y = 4\\
\frac{1}{{\sqrt {x + 4}  + 2}} + \frac{1}{{\sqrt {x + 25}  + 5}} + \frac{{x + 12}}{{x + 8 + 2\sqrt {x + 16} }} = 0
\end{array} \right.\,\,\left( {vn} \right)\)