\(a,b,M=\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x}\left(x\ge0;x\ne0;x\ne1\right)\\ M=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{x}{\sqrt{x}+1}\\ M=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\cdot\dfrac{x}{\sqrt{x}+1}\\ M=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}\cdot\dfrac{x}{\sqrt{x}+1}=\sqrt{x}\left(\sqrt{x}-1\right)\)

\(c,M=\sqrt{x}\left(\sqrt{x}-1\right)=x-\sqrt{x}\\ =x-\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu \("="\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\)

\(M=\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x}\)

ĐKXĐ: \(x>0;x\ne1\)

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

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

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

\(=\sqrt{x}-1\)

Kết quả rút gọn: \(P=\frac{\sqrt{x}+2}{\sqrt{x}-1}\)

\(M=\frac{x+12}{\sqrt{x}-1}.\frac{\sqrt{x}-1}{\sqrt{x}+2}=\frac{x+12}{\sqrt{x}+2}\)

\(M=\frac{x-4+16}{\sqrt{x}+2}=\sqrt{x}-2+\frac{16}{\sqrt{x}+2}=\left(\sqrt{x}+2+\frac{16}{\sqrt{x}+2}\right)-4\)

Âp dụng BĐT AM-GM cho 2 số không âm ta có: 

\(M\ge2\sqrt{\left(\sqrt{x}+2\right).\frac{16}{\sqrt{x}+2}}-4=2.4-4=4\)

Vậy min M =4. Dấu bằng xảy ra \(\Leftrightarrow\left(\sqrt{x}+2\right)^2=16\Leftrightarrow\sqrt{x}+2=4\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)

\(P=\left(\frac{3}{x-1}+\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\) \(ĐKXĐ:x\ne1\)

\(P=\left(\frac{3}{x-1}+\frac{\sqrt{x}-1}{x-1}\right):\frac{1}{\sqrt{x}+1}\)

\(P=\frac{\sqrt{x}+2}{x-1}.\left(\sqrt{x}+1\right)\)

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

b) theo câu a) \(P=\frac{\sqrt{x}+2}{\sqrt{x}-1}\) với \(ĐKXĐ:x\ne1\)

theo bài ra \(P=\frac{5}{4}\)thì \(\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{5}{4}\)

\(\Leftrightarrow\left(\sqrt{x}+2\right).4=\left(\sqrt{x}-1\right).5\)

\(\Leftrightarrow4\sqrt{x}+8=5\sqrt{x}-5\)

\(\Leftrightarrow-\sqrt{x}+13=0\)

\(\Leftrightarrow-\sqrt{x}=-13\)

\(\Leftrightarrow\sqrt{x}=13\)

\(\Leftrightarrow x=169\)

vậy \(x=169\)khi \(P=\frac{5}{4}\)

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