\(\sqrt{11-4\sqrt{7}}-\sqrt{29-4\sqrt{7}}\)

\(=\sqrt{7-2.\sqrt{7}.2+4}-\sqrt{28-2.2\sqrt{7}.1+1}\)

\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}.2+2^2}-\sqrt{\left(2\sqrt{7}\right)^2-2.2\sqrt{7}.1+1^2}\)

\(=\sqrt{\left(\sqrt{7}-2\right)^2}-\sqrt{\left(2\sqrt{7}-1\right)^2}\)

\(=\left|\sqrt{7}-2\right|-\left|2\sqrt{7}-1\right|\)

\(=\sqrt{7}-2-2\sqrt{7}+1\)

\(=-\sqrt{7}-1\)

Bài 2. 

ĐKXĐ của biểu thức đã cho là: 

\(\hept{\begin{cases}x\ge0,\sqrt{x}\ne0\\\sqrt{x}-1\ne0\\\sqrt{x}-2\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>0\\x\ne1,x\ne2\end{cases}}\).

\(A=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right)\div\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)

\(=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left[\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right]\)

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

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

\(A>\frac{1}{6}\Rightarrow\frac{\sqrt{x}-2}{\sqrt{x}}>\frac{1}{6}\Leftrightarrow6\left(\sqrt{x}-2\right)>\sqrt{x}\)

\(\Leftrightarrow5\sqrt{x}>12\Leftrightarrow x>\frac{144}{25}\).

ĐK: \(-\sqrt{3}\le x\le\sqrt{3}\).

\(\left(x-2\right)\sqrt{3-x^2}=x^2-x-2\)

\(\Leftrightarrow\left(x-2\right)\sqrt{3-x^2}=\left(x-2\right)\left(x+1\right)\)

\(\Leftrightarrow\left(x-2\right)\left(\sqrt{3-x^2}-x-1\right)=0\)

\(\Leftrightarrow\sqrt{3-x^2}=x+1\)(vì \(-\sqrt{3}\le x\le\sqrt{3}\))

\(\Rightarrow3-x^2=\left(x+1\right)^2\)

\(\Leftrightarrow2x^2+2x-2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1+\sqrt{5}}{2}\left(tm\right)\\x=\frac{-1-\sqrt{5}}{2}\left(l\right)\end{cases}}\)

\(n^{2}={\underbrace{999\dots 9}_{\text{50 chữ số 9}}}^{2}=\left(10^{50}-1\right)^{2}=10^{100}-2\cdot 10^{50}+1=\left(10^{50}-2\right)\cdot 10^{50}+1=\underbrace{999\dots 9}_{\text{49 chữ số 9}}8\cdot10^{50}+1=\underbrace{999\dots 9}_{\text{49 chữ số 9}}8\underbrace{000\dots 0}_{\text{49 chữ số 0}}1\)

Giả sử đồ thị hàm số đã cho luôn đi qua điểm cố định \(\left(x_0,y_0\right)\)với mọi \(m\).

\(y_0=\left(3m^2+1\right)x_0+m^2-4,\forall m\)

\(\Leftrightarrow m^2\left(3x_0+1\right)+x_0-y_0-4=0,\forall m\)

\(\Leftrightarrow\hept{\begin{cases}3x_0+1=0\\x_0-y_0-4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x_0=-\frac{1}{3}\\y_0=-\frac{13}{3}\end{cases}}\)

Vậy điểm cố định mà đồ thị hàm số đã cho luôn đi qua có tọa độ là \(\left(-\frac{1}{3},-\frac{13}{3}\right)\).