Ta có: \(M=\dfrac{3\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+4}{\sqrt{x}+1}-\dfrac{9}{x-\sqrt{x}-2}\)

\(=\dfrac{3\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{3x-3-2x+8-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)

Ta có: \(A-1=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-1\)

\(=\dfrac{\sqrt{x}+2-\sqrt{x}-1}{\sqrt{x}+1}\)

\(=\dfrac{1}{\sqrt{x}+1}>0\forall x\) thỏa mãn ĐKXĐ

hay A>1

\(M=\dfrac{3\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+4}{\sqrt{x}+1}-\dfrac{9}{x-\sqrt{x}-2}\\ =\dfrac{3\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{9}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{3\left(x-1\right)-2\left(x-4\right)-9}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\dfrac{x-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{\sqrt{x}+2}{\sqrt{x}+1}=1+\dfrac{1}{\sqrt{x}+1}>1\)

Ta có BĐT phụ \(\dfrac{1+\sqrt{a}}{1-a}\ge4a+1\)

\(\Leftrightarrow-\dfrac{\sqrt{a}\left(2\sqrt{a}-1\right)^2}{\sqrt{a}-1}\ge0\forall\dfrac{1}{4}< a< 0\)

Tương tự cho 3 BĐT còn lại ta cũng có:

\(\dfrac{1+\sqrt{b}}{1-b}\ge4b+1;\dfrac{1+\sqrt{c}}{1-c}\ge4c+1;\dfrac{1+\sqrt{d}}{1-d}\ge4d+1\)

Cộng theo vế 4 BĐT trên ta có:

\(VT\ge4\left(a+b+c+d\right)+4=8=VP\)

Xảy ra khi \(a=b=c=d=\dfrac{1}{4}\)

Ta cần chứng minh :

\(\dfrac{1+\sqrt{a}}{1-a}\ge4a+1\) \(\forall a\in\left(0;\dfrac{1}{4}\right)\)

\(\Leftrightarrow1+\sqrt{a}\ge\left(4a+1\right)\left(1-a\right)\)

\(\Leftrightarrow1+\sqrt{a}\ge4a-4a^2+1-a\)

\(\Leftrightarrow4a^2-4a-1+a+1+\sqrt{a}\ge0\)

\(\Leftrightarrow4a^2-3a+\sqrt{a}\ge0\)

\(\Leftrightarrow\left(4a^2-a\right)-\left(2a-\sqrt{a}\right)\ge0\)

\(\Leftrightarrow\left(2a-\sqrt{a}\right)\left(2a+\sqrt{a}\right)-\left(2a-\sqrt{a}\right)\ge0\)

\(\Leftrightarrow\left(2a-\sqrt{a}\right)\left(2a+\sqrt{a}-1\right)\ge0\)

Ta có: \(2a-\sqrt{a}=\left(\sqrt{2a}-\dfrac{\sqrt{2}}{4}\right)^2-\dfrac{1}{8}\ge0\) \(\forall a\in\left(0;\dfrac{1}{4}\right)\)

\(\left(2a+\sqrt{a}-1\right)=\left(\sqrt{2a}+\dfrac{\sqrt{2}}{4}\right)^2-\dfrac{9}{8}\ge0\)

\(\forall a\in\left(0;\dfrac{1}{4}\right)\)

Vậy: \(\dfrac{1+\sqrt{a}}{1-a}\ge4a+1\) \(\forall a\in\left(0;\dfrac{1}{4}\right)\)

Tương tự: \(\dfrac{1+\sqrt{b}}{1-b}\ge4b+1\forall b\in\left(0;1\right)\)

\(\dfrac{1+\sqrt{c}}{1-c}\ge4c+1\forall c\in\left(0;\dfrac{1}{4}\right)\)

\(\dfrac{1+\sqrt{d}}{1-d}\ge4d+1\forall d\in\left(0;\dfrac{1}{4}\right)\)

Cộng các BĐT vừa chứng minh, ta được:

\(\dfrac{1+\sqrt{a}}{1-a}+\dfrac{1+\sqrt{b}}{1-b}+\dfrac{1+\sqrt{c}}{1-c}+\dfrac{1+\sqrt{d}}{1-d}\ge4\left(a+b+c+d\right)+4=8\)

Vậy: Ta suy ra được điều phải chứng minh

\(1.ĐKXĐ\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)

\(M=\frac{a+1}{\sqrt{a}}+\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

\(\Leftrightarrow M=\frac{a+1}{\sqrt{a}}+\frac{a+\sqrt{a}+1}{\sqrt{a}}-\frac{a-\sqrt{a}+1}{\sqrt{a}}\)

\(\Leftrightarrow M=\frac{a+2\sqrt{a}+1}{\sqrt{a}}\)

2.

có \(M=\frac{a+2\sqrt{a}+1}{\sqrt{a}}\)

\(\Leftrightarrow M=4+\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}}\)

vì \(a>0\) và \(a\ne0\)

\(\Rightarrow\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}}>0\Leftrightarrow4+\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}}>4\)

Vậy M > 4