Đáp án:
1) \(\dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\sqrt x }}\)
2) x=4
3) đpcm
Giải thích các bước giải:
\(\begin{array}{l}
1)A = \dfrac{{x - 1}}{{\sqrt x }}:\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) + 1 - \sqrt x }}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x - 1}}{{\sqrt x }}.\dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{x - 1 + 1 - \sqrt x }}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{\sqrt x }}.\dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{x - \sqrt x }}\\
= \dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{\sqrt x }}.\dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x \left( {\sqrt x - 1} \right)}}\\
= \dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\sqrt x }}\\
2)A.\sqrt x = 9\\
\to \dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\sqrt x }}.\sqrt x = 9\\
\to {\left( {\sqrt x + 1} \right)^2} = 9\\
\to \sqrt x + 1 = 3\\
\to \sqrt x = 2\\
\to x = 4\\
3)A > 4\\
\to \dfrac{{{{\left( {\sqrt x + 1} \right)}^2}}}{{\sqrt x }} > 4\\
\to \dfrac{{x + 2\sqrt x + 1 - 4\sqrt x }}{{\sqrt x }} > 0\\
\to x - 2\sqrt x + 1 > 0\left( {do:\sqrt x > 0\forall x > 0} \right)\\
\to {\left( {\sqrt x - 1} \right)^2} > 0\\
\to \sqrt x - 1 \ne 0\\
\to x \ne 1;x > 0
\end{array}\)
⇒ Điều phải chứng minh
