a: Thay x=9 vào M, ta được:
\(M=\dfrac{3+1}{3-2}=\dfrac{4}{1}=4\)
b: \(N=\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+2}+\dfrac{8}{x-4}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+2}+\dfrac{8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+2\left(\sqrt{x}-2\right)+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)
c: Để N<0 thì \(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}< 0\)
=>\(\sqrt{x}-2< 0\)
=>\(\sqrt{x}< 2\)
=>0<=x<4
mà x nguyên
nên \(x\in\left\{0;1;2;3\right\}\)
d: Để N là số nguyên thì \(\sqrt{x}+2⋮\sqrt{x}-2\)
=>\(\sqrt{x}-2+4⋮\sqrt{x}-2\)
=>\(4⋮\sqrt{x}-2\)
=>\(\sqrt{x}-2\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(\sqrt{x}\in\left\{3;1;4;0;6;-2\right\}\)
=>\(x\in\left\{9;1;16;0;36\right\}\)
e: \(P=\dfrac{M}{N}=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}:\dfrac{\sqrt{x}+2}{\sqrt{x}-2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}+2-1}{\sqrt{x}+2}=1-\dfrac{1}{\sqrt{x}+2}\)
\(\sqrt{x}+2>=2\forall x\) thỏa mãn ĐKXĐ
=>\(\dfrac{1}{\sqrt{x}+2}< =\dfrac{1}{2}\forall x\) thỏa mãn ĐKXĐ
=>\(-\dfrac{1}{\sqrt{x}+2}>=-\dfrac{1}{2}\forall x\) thỏa mãn ĐKXĐ
=>\(P=-\dfrac{1}{\sqrt{x}+2}+1>=-\dfrac{1}{2}+1=\dfrac{1}{2}\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi x=0
f: \(P-1=1-\dfrac{1}{\sqrt{x}+2}-1=-\dfrac{1}{\sqrt{x}+2}< 0\)
=>P<1
