a) Đặt \(A=\frac{\sqrt{x}+5}{\sqrt{x}+1}\)(1)

\(\left(1\right)\Leftrightarrow\frac{\sqrt{x}+1+4}{\sqrt{x}+1}=1+\frac{4}{\sqrt{x}+1}\)

\(A\inℤ\Leftrightarrow\frac{4}{\sqrt{x}+1}\inℤ\Leftrightarrow4⋮\left(\sqrt{x}+1\right)\)

\(\Leftrightarrow\sqrt{x}+1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Mà \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+1\ge1\)nên \(\sqrt{x}+1\in\left\{1;2;4\right\}\)

\(TH1:\sqrt{x}+1=1\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)

\(TH2:\sqrt{x}+1=2\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)

\(TH2:\sqrt{x}+1=4\Leftrightarrow\sqrt{x}=3\Leftrightarrow x=9\)

Vậy \(x\in\left\{0;1;9\right\}\)thì \(\frac{\sqrt{x}+5}{\sqrt{x}+1}\)đạt giá trị nguyên

b) \(\frac{2}{\sqrt{x}+2}\inℤ\Leftrightarrow2⋮\left(\sqrt{x}+2\right)\)

\(\Leftrightarrow\sqrt{x}+2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Vì \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\)nên \(\sqrt{x}+2=2\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)

Vậy x = 0 thì \(\frac{2}{\sqrt{x}+2}\inℤ\)

ĐK \(x\ge-\frac{1}{2}\)

Đặt như trên... (\(a\ge\sqrt{\frac{1}{2}};b\ge0\)) ta có hệ:

\(\hept{\begin{cases}2a^2b=a+b^3\\2a^2-b^2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(b^2+1\right)b=a+b^3\\2a^2=b^2+1\end{cases}}\)

Xét pt trình đầu của hệ \(\Leftrightarrow a=b\). Thay b bởi a ở pt dưới ta được:

\(2a^2-a^2-1=0\Leftrightarrow\orbr{\begin{cases}a=1\left(TM\right)\\a=-\frac{1}{2}\left(KTM\right)\end{cases}}\). Với a = 1 thì ta có:

\(\sqrt{1+x}=1\Leftrightarrow x=0\) (TM)

Vậy...

\(x^2-2\left(m-1\right)x+m-3=0\)

\(\Delta'=b'^2-ac=\left[-\left(m-1\right)^2\right]-1.\left(m-3\right)=m^2-2m+1-m+3=m^2-3m+4\)

\(=m^2-2.m.\frac{3}{2}+\frac{9}{4}+\frac{7}{4}=\left(m-\frac{3}{2}\right)^2+\frac{7}{4}\)

Vì\(\left(m-\frac{3}{2}\right)^2\ge0\forall m\)

\(\Rightarrow\left(m-\frac{3}{2}\right)^2+\frac{7}{4}\ge\frac{7}{4}>0\forall m\)

\(\Rightarrow\Delta'>0\forall m\)

Vậy...

\(1+\left(\frac{a+2\sqrt{a}-1}{1-a}-\frac{2a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}\right)\cdot\frac{a-\sqrt{a}}{2\sqrt{a}-1}\)

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

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

\(=1+\left(\frac{\left(1-\sqrt{a}\right)}{\left(1+\sqrt{a}\right)}-\frac{\sqrt{a}}{\left(1-\sqrt{a}\right)}\right)\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)

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

\(=1+\frac{1-2\sqrt{a}+a-\sqrt{a}-a}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)

\(=1+\frac{1-2\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{2\sqrt{a}-1}\)

\(=1+\frac{1-2\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\cdot\frac{\sqrt{a}\left(1-\sqrt{a}\right)}{1-2\sqrt{a}}\)

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

\(=\frac{1+\sqrt{a}+\sqrt{a}}{1+\sqrt{a}}\)

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