Bài 25 :
1. M = \(\left(\dfrac{x+3\sqrt{x}}{x-25}+\dfrac{1}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\) ( ĐKXĐ : x ≥ 0 ; x ≠ 25 )
= \(\dfrac{x+3\sqrt{x}+\sqrt{x}-5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
= \(\dfrac{x+4\sqrt{x}-5}{\sqrt{x}+5}.\dfrac{1}{\sqrt{x}+2}\)
= \(\dfrac{\left(\sqrt{x}+5\right).\left(\sqrt{x}-1\right)}{\sqrt{x}+5}.\dfrac{1}{\sqrt{x+2}}\)
= \(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
2. Ta có : \(\dfrac{1}{\sqrt{3}-1}-\dfrac{1}{\sqrt{3}+1}\) = \(\dfrac{\sqrt{3}+1-\sqrt{3}+1}{2}=1\)
Khi thay x = 1, ta được :
\(\dfrac{\sqrt{1}-1}{\sqrt{1}+2}=\dfrac{1-1}{1+2}=0\)
3. Vì M = 0 nên không có GTNN
M=\(\left(\dfrac{x+3\sqrt{x}}{x-25}+\dfrac{1}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\left(x\ge0,x\ne25\right)\)
=\(\left(\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}+\dfrac{\sqrt{x}-5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right).\dfrac{\sqrt{x}-5}{\sqrt{x}+2}\)
\(=\dfrac{x+3\sqrt{x}+\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}=\dfrac{x+4\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
2.
x=\(\dfrac{1}{\sqrt{3}-1}-\dfrac{1}{\sqrt{3}+1}=\dfrac{\sqrt{3}+1-\sqrt{3}+1}{2}=1\)
thay x=1 vào M=>M=\(\dfrac{1-1}{\sqrt{1}+2}=0\)
3.\(M=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=1+\dfrac{-3}{\sqrt{x}+2}\)
M=\(1-\dfrac{3}{\sqrt{x}+2}\)
có \(\sqrt{x}\ge0\left(\forall x\ge0\right)=>\sqrt{x}+2\ge2\left(\forall x\ge0\right)\)
=>M\(\ge1-\dfrac{3}{2}=-\dfrac{1}{2}\)
dấu'=' xảy ra <=> x=0(TM)
vậy Min M=\(-\dfrac{1}{2}\)<=>x=0
