$ĐKXĐ:x > 0$
$a)P=1+\dfrac {x^{2}+\sqrt {x}}{x-\sqrt {x}+1}-\dfrac {2x+\sqrt {x}}{\sqrt {x}}=1+\dfrac {\sqrt {x}\left( x\sqrt {x}+1\right) }{x-\sqrt {x}+1}-\dfrac {\sqrt {x}\left( 2\sqrt {x}+1\right) }{\sqrt {x}}$
$\Leftrightarrow P=1+\dfrac {\sqrt {x}\left( x-\sqrt {x}+1\right) \left( \sqrt {x}+1\right) }{x-\sqrt {x}+1}-\left( 2\sqrt {x}+1\right) =1+\sqrt {x}\left( \sqrt {x}+1\right) -2\sqrt {x}-1$
$\Leftrightarrow P=x+\sqrt {x}-2\sqrt {x}=x-\sqrt {x}$
$b)$Ta có : $x=4+2\sqrt {3}=\left( \sqrt {3}+1\right) ^{2}\Leftrightarrow x=\sqrt {3}+1$ $($do $\sqrt {3}+1 > 0$$)$
Khi đó : $P=x-\sqrt {x}=4+2\sqrt {3}-\left( \sqrt {3}+1\right) =4+2\sqrt {3}-\sqrt {3}-1=3+\sqrt {3}$
$c)$Ta có : $P=6\Leftrightarrow x-\sqrt {x}=6\Leftrightarrow x-\sqrt {x}-6=0\Leftrightarrow \left( \sqrt {x}-3\right) \left( \sqrt {x}+2\right) =0$
$\Leftrightarrow \left( \sqrt {x}-3\right) \left( \sqrt {x}+2\right) =0\Leftrightarrow \sqrt {x}-3=0$ $($do $\sqrt {x}+2> 2 > 0$$)$$\Leftrightarrow \sqrt {x}=3\Leftrightarrow x=9$ $($thoả$)$
$d)$Ta có : $P < 0\Leftrightarrow x-\sqrt {x} < 0\Leftrightarrow \sqrt {x}\left( \sqrt {x}-1\right) < 0\Leftrightarrow \sqrt {x}-1 < 0$ $($do $\sqrt {x} > 0$$)$$\Leftrightarrow \sqrt {x} < 1\Leftrightarrow x < 1$
Kết hợp điều kiện đề bài, ta có : $P < 0\Leftrightarrow 0 < x < 1$
$e)$$P=x-\sqrt {x}=\left( \sqrt {x}-\dfrac {1}{2}\right) ^{2}-\dfrac {1}{4}\geq -\dfrac {1}{4}$
$ĐTXR \Leftrightarrow \sqrt {x}-\dfrac {1}{2}=0\Leftrightarrow \sqrt {x}=\dfrac {1}{2}\Leftrightarrow x=\dfrac {1}{4}$ $($thoả$)$
Vậy $Min_{P}=-\dfrac {1}{4}\Leftrightarrow x=\dfrac {1}{4}$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
