\(A=\frac{cos\alpha+\sin\alpha}{\cos\alpha-\sin\alpha}+3=\frac{1+\frac{\sin\alpha}{\cos\alpha}}{1-\frac{\sin\alpha}{\cos\alpha}}+3=\frac{1+\tan\alpha}{1-\tan\alpha}+3=\frac{1+0,5}{1-0,5}+3=3+3=6\)

c) \(\left(\sqrt{10}+\sqrt{2}\right)\left(\sqrt{3+\sqrt{5}}\right)\left(6-2\sqrt{5}\right)\)

= \(\left(\sqrt{5}+1\right)\sqrt{2}\left(\sqrt{3+\sqrt{5}}\right)\left(5-2\sqrt{5}+1\right)\)

= \(\left(\sqrt{5}+1\right)\left(\sqrt{6+2\sqrt{5}}\right)\left(\sqrt{5}-1\right)^2\)

= \(\left(\sqrt{5}+1\right)\left(\sqrt{5+2\sqrt{5}+1}\right)\left(\sqrt{5}-1\right)^2\)

= \(\left(\sqrt{5}+1\right)\left(\sqrt{\left(\sqrt{5}+1\right)^2}\right)\left(\sqrt{5}-1\right)^2\)

= \(\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)^2\)

= \(4.4=16\)

d) \(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{2}-\sqrt{5}\)

= \(\sqrt{1+2+5+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{2}-\sqrt{5}\)

= \(\sqrt{\left(\sqrt{2}+\sqrt{1}+\sqrt{5}\right)^2}-\sqrt{2}-\sqrt{5}\)

= \(\sqrt{2}+\sqrt{5}+1-\sqrt{2}-\sqrt{5}\)

= \(1\)

a) \(\sqrt{\dfrac{2-\sqrt{3}}{2}}+\dfrac{1-\sqrt{3}}{2}\)

= \(\sqrt{\dfrac{4-2\sqrt{3}}{4}}+\dfrac{1-\sqrt{3}}{2}\)

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

= \(\dfrac{\sqrt{3}-1+1-\sqrt{3}}{2}\)

= 0

b) \(\sqrt{41+6\sqrt{6}-12\sqrt{10}-4\sqrt{15}}+2\sqrt{5}-\sqrt{3}\)

= \(\sqrt{18+20+3+2\sqrt{54}-2\sqrt{360}-2\sqrt{60}}+2\sqrt{5}-\sqrt{3}\)

= \(\sqrt{\left(\sqrt{18}-\sqrt{20}+\sqrt{3}\right)^2}+2\sqrt{5}-\sqrt{3}\)

= \(\sqrt{18}-2\sqrt{5}+\sqrt{3}+2\sqrt{5}-\sqrt{3}\)

= \(\sqrt{18}\)

a) \(\sqrt{12}-4\sqrt{75}-3\sqrt{27}+5\sqrt{48}\)

= \(2\sqrt{3}-20\sqrt{3}-9\sqrt{3}+20\sqrt{3}\)

= \(2\sqrt{3}-9\sqrt{3}\)

= \(-7\sqrt{3}\)

b)\(\sqrt{\left(1-2\sqrt{7}\right)^2}+\sqrt{8+2\sqrt{7}}\)

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

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

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

= \(3\sqrt{7}\)

c) \(\dfrac{1}{1-\sqrt{3}}-\dfrac{1}{1+\sqrt{3}}\)

= \(\dfrac{1+\sqrt{3}}{\left(1-\sqrt{3}\right)\left(1+\sqrt{3}\right)}-\dfrac{1-\sqrt{3}}{\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)}\)

= \(\dfrac{1+\sqrt{3}-1+\sqrt{3}}{1-3}\)

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

Ta có:

\(\dfrac{x}{2014}+\dfrac{x+1}{2015}+\dfrac{x+2}{2016}+\dfrac{x+3}{2017}+\dfrac{x+4}{2018}=5\)

\(\Leftrightarrow\left(\dfrac{x}{2014}-1\right)+\left(\dfrac{x+1}{2015}-1\right)+\left(\dfrac{x+2}{2016}-1\right)+\left(\dfrac{x+3}{2017}-1\right)+\left(\dfrac{x+4}{2018}-1\right)=0\)\(\Leftrightarrow\dfrac{x-2014}{2014}+\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}+\dfrac{x-2014}{2017}+\dfrac{x-2014}{2018}=0\)\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)=0\) (1)

Mà \(\dfrac{1}{2014}+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}>0\) (2)

Từ (1) và (2) => \(x-2014=0\) \(\Leftrightarrow x=2014\)

1/ D = \(x^2+8x+20\)

D = \(x^2+8x+16+4\)

D = \(\left(x+4\right)^2+4\ge4\)

=> biểu thức luôn dương

2/ E = \(4x^2-2x+17\)

E = \(4\left(x^2-\dfrac{1}{2}x+\dfrac{17}{4}\right)\)

E = \(4\left(x^2-2.\dfrac{1}{4}x+\dfrac{1}{16}+\dfrac{67}{16}\right)\)

E = \(4\left(x-\dfrac{1}{4}\right)^2+\dfrac{67}{4}\ge\dfrac{67}{4}\)

=> biểu thức luôn dương

3/ F = \(9x^2+12x+8\)

F = \(9\left(x^2+\dfrac{4}{3}x+\dfrac{8}{9}\right)\)

F = \(9\left(x^2+2.\dfrac{4}{6}x+\dfrac{16}{36}+\dfrac{4}{9}\right)\)

F = \(9\left(x+\dfrac{4}{6}\right)^2+4\ge4\)

=> biểu thức luôn dương

Ta có:

\(\dfrac{30}{\sqrt{6}+1}+\dfrac{2}{\sqrt{6}-2}+\dfrac{6}{3-\sqrt{6}}\)

= \(\dfrac{30\left(\sqrt{6}-1\right)}{6-1}+\dfrac{2\left(\sqrt{6}+2\right)}{6-4}-\dfrac{6\left(3+\sqrt{6}\right)}{9-6}\)

= \(6\left(\sqrt{6}-1\right)+\left(\sqrt{6}+2\right)-2\left(3+\sqrt{6}\right)\)

= \(6\sqrt{6}-6+\sqrt{6}+2-6-2\sqrt{6}\)

= \(5\sqrt{6}-10\)

= \(5\left(\sqrt{6}-2\right)\)