\(\frac{\sqrt{x^4+y^4}+\sqrt{x^4-y^4}}{\sqrt{x^4+y^4}-\sqrt{x^4-y^4}}=\frac{\left(\sqrt{x^4+y^4}+\sqrt{x^4-y^4}\right)^2}{\left(x^4+y^4\right)-\left(x^4-y^4\right)}\)

\(=\frac{x^4+y^4+x^4-y^4+2\sqrt{x^8-y^8}}{2y^4}=\frac{x^4}{y^4}+\sqrt{\frac{x^8-y^8}{y^8}}=\frac{x^4}{y^4}+\sqrt{\frac{x^8}{y^8}-1}\)

ĐK:x\(\ge0,x\ne16\)

\(\frac{x\sqrt{x}-2x+28}{x-3\sqrt{x}-4}-\frac{\sqrt{x}-4}{\sqrt{x}+1}-\frac{\sqrt{x}+8}{\sqrt{x}-4}=\frac{x\sqrt{x}-2x+28}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}-\frac{\left(\sqrt{x}-4\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}-\frac{\left(\sqrt{x}+8\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\frac{x\sqrt{x}-2x+28}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}-\frac{x-8\sqrt{x}+16}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}-\frac{x+9\sqrt{x}+8}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\frac{x\sqrt{x}-2x+28-x+8\sqrt{x}-16-x-9\sqrt{x}-8}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\frac{x\sqrt{x}-4x-\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\frac{x\left(\sqrt{x}-4\right)-\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\frac{\left(\sqrt{x}-4\right)\left(x-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\sqrt{x}-1\)

Tử số của phân số đầu phải là \(\sqrt{x}+2\) chứ không phải \(\sqrt{x+2}\), vì cái \(\sqrt{x}+2\) nó mới logic để rút gọn: )

\(Q=\left(\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}^3+8}-\dfrac{x-\sqrt{x}}{\sqrt{x}^3+8}\right)\left(\dfrac{5x-10\sqrt{x}+20}{5\sqrt{x}+4}\right)\\ =\left(\dfrac{x+4\sqrt{x}+4-x+\sqrt{x}}{\sqrt{x}^3+8}\right)\left(\dfrac{5x-10\sqrt{x}+20}{5\sqrt{x}+4}\right)\\ =\dfrac{\left(5\sqrt{x}+4\right).5.\left(x-2\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)\left(5\sqrt{x}+4\right)}\\ =\dfrac{5}{\sqrt{x}+2}\)

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

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

- Với \(x\ge8\Rightarrow\sqrt{x-4}-2\ge0\)

\(\Rightarrow A=\frac{\sqrt{x-4}+2+\sqrt{x-4}-2}{\frac{x-4}{x}}=\frac{2x\sqrt{x-4}}{x-4}=\frac{2x}{\sqrt{x-4}}\)

- Với \(4< x\le8\)

\(\Rightarrow A=\frac{\sqrt{x-4}+2+2-\sqrt{x-4}}{\frac{x-4}{x}}=\frac{4x}{x-4}\)

a) \(\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)

\(=\sqrt{4-4\sqrt{3}+3}-\sqrt{4+4\sqrt{3}+3}\)

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

\(=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\)

\(=2-\sqrt{3}-2-\sqrt{3}\)

\(=-2\sqrt{3}\)