cho 3 đường thẳng (d) y = x+2 ; (d') y = $\frac{3}{2}$ \left(1+\frac{2+\sqrt{2}}{1+\sqrt{2}}\right)$$ \left(1-\frac{2-\sqrt{2}}{1-\sqrt{2}}\right)$$ \frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\frac{1}{\sqrt{6}}$$ \frac{6}{1+\sqrt{7}}+\frac{1}{\sqrt{7}}$$ \frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{1+\sqrt{2}}-\frac{1}{2-\sqrt{3}}$$ \left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right).\left(\sqrt{5}-\sqrt{2}\right)$$ \frac{2}{\sqrt{5}+1}-\sqrt{\frac{2}{3-\sqrt{5}}}$$ \frac{1}{4}$$ A=\frac{2x-\sqrt{x}+8}{2\sqrt{x}-1}$$ \frac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}$$ \frac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}$$ \frac{x+y+2xy}{1-xy}$$ \frac{2}{2+\sqrt{3}}$$ P=\frac{\sqrt{x}-2}{\sqrt{x}-3}+\frac{\sqrt{x}+1}{\sqrt{x}+3}+\frac{x-4\sqrt{x}-9}{9-x}$$ Q=\frac{\sqrt{x}+5}{3-\sqrt{x}}$$ x\ge 0;x\ne 9$$ M=P:Q$$ \left|M\right|<\frac{1}{2}$$ P=\frac{a+b+c}{2}$$ \le $$ \frac{1}{8}abc$$ \frac{1}{P-a}+\frac{1}{P-b}+\frac{1}{P-c}\ge 2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$ \frac{1}{{\left(P-a\right)}^2}+\frac{1}{{\left(P-b\right)}^2}+\frac{1}{{\left(P-c\right)}^2}\ge \frac{P}{\left(P-a\right)\left(P-b\right)\left(P-c\right)}$$ \left(1+\frac{2+\sqrt{2}}{1+\sqrt{2}}\right)$$ \left(1-\frac{2-\sqrt{2}}{1-\sqrt{2}}\right)$$ \frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\frac{1}{\sqrt{6}}$$ \frac{6}{1+\sqrt{7}}+\frac{1}{\sqrt{7}}$$ \frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{1+\sqrt{2}}-\frac{1}{2-\sqrt{3}}$$ \left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right).\left(\sqrt{5}-\sqrt{2}\right)$$ \frac{2}{\sqrt{5}+1}-\sqrt{\frac{2}{3-\sqrt{5}}}$$ \frac{}{}$$ \sqrt{\frac{9}{{\left(x-2\right)}^2}}$$ \sqrt{\frac{a}{6}}$$ \sqrt{\frac{2a}{3}}$$ \sqrt{\frac{3a}{2}}$$ \sqrt{9a^2}$$ \sqrt{4a^2}$$ \sqrt{{\left(1-a\right)}^2}$$ \sqrt{16a^2}$$ \left|x-2\right|$$ \frac{\sqrt{x^2}}{x}$$ \frac{x^2+2\sqrt{3}.x+3}{x^2-3}$$ \left|x-\sqrt{{\left(x-1\right)}^2}\right|$$ -$$ \frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge \frac{1}{2}$$ \$\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{6}$$ \left(x\sqrt{\frac{6}{x}}+\sqrt{\frac{2x}{3}}+\sqrt{6x}\right):\sqrt{6x}=2\frac{1}{3}$$ \frac{3}{ab+bc+ac}+\frac{2}{a^2+b^2+c^2}>14$$