Bài 1 :

a) \(Cos30^o=Cos\left(2.15^o\right)=2cos^215^o-1\)

\(\Rightarrow cos^215^o=\dfrac{cos30^o+1}{2}\)

\(\Rightarrow cos^215^o=\dfrac{\dfrac{\sqrt[]{3}}{2}+1}{2}\)

\(\Rightarrow cos^215^o=\dfrac{\sqrt[]{3}+2}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{\sqrt[]{3}+2}}{2}\)

\(\Rightarrow cos15^o=\dfrac{2\sqrt[]{\sqrt[]{3}+2}}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{4\sqrt[]{3}+8}}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{6+2.2\sqrt[]{2}\sqrt[]{6}+2}}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{\left(\sqrt[]{6}+\sqrt[]{2}\right)^2}}{4}\)

\(\Rightarrow cos15^o=\dfrac{\sqrt[]{6}+\sqrt[]{2}^{ }}{4}\left(dpcm\right)\)

a)

 Dựng tam giác ABC vuông tại A với \(\widehat{C}=15^o\). Trên đoạn thẳng AC lấy điểm D sao cho \(\widehat{CBD}=15^o\). Không mất tính tổng quát, ta chuẩn hóa \(AB=1\). \(\Rightarrow\left\{{}\begin{matrix}BD=\dfrac{AB}{cos60^o}=2\\AD=AB.tan60^o=\sqrt{3}\end{matrix}\right.\)

 Dễ thấy tam giác DBC cân tại D \(\Rightarrow BD=CD=2\) \(\Rightarrow AC=AD+DC=2+\sqrt{3}\)

  \(\Rightarrow tanC=\dfrac{AB}{AC}=\dfrac{1}{2+\sqrt{3}}=2-\sqrt{3}\) 

\(\Rightarrow\dfrac{sinC}{cosC}=2-\sqrt{3}\)

\(\Rightarrow sinC=\left(2-\sqrt{3}\right)cosC\)

Mà \(sin^2C+cos^2C=1\)

\(\Rightarrow\left(7-4\sqrt{3}\right)cos^2C+cos^2C=1\)

\(\Leftrightarrow\left(8-4\sqrt{3}\right)cos^2C=1\)

\(\Leftrightarrow cos^2C=\dfrac{1}{8-4\sqrt{3}}=\dfrac{2+\sqrt{3}}{4}\)

\(\Leftrightarrow cosC=\sqrt{\dfrac{2+\sqrt{3}}{4}}\) \(=\dfrac{\sqrt{2+\sqrt{3}}}{2}=\dfrac{\sqrt{8+4\sqrt{3}}}{4}\) \(=\dfrac{\sqrt{6}+\sqrt{2}}{4}\) 

\(\Rightarrow cos15^o=\dfrac{\sqrt{6}+\sqrt{2}}{4}\)

a) \(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)

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

\(=\dfrac{\sqrt{\left(3-\sqrt{5}\right)^2}}{\sqrt{3^2-\left(\sqrt{5}\right)^2}}\)

\(=\dfrac{\left|3-\sqrt{5}\right|}{\sqrt{9-5}}\)

\(=\dfrac{3-\sqrt{5}}{2}\)

b) \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)

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

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

\(=\dfrac{\left|2-\sqrt{3}\right|}{\sqrt{4-3}}\)

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

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

a: \(=\sqrt{\dfrac{\left(3-\sqrt{5}\right)\left(3-\sqrt{5}\right)}{4}}=\dfrac{3-\sqrt{5}}{2}\)

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

d: \(=\left(-3+3\sqrt{6}+4+2\sqrt{6}-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

=(căn 6-11)(căn 6+11)

=6-121=-115

\(a,\dfrac{7}{\sqrt{12}}=\dfrac{7\sqrt{3}}{\sqrt{12}\cdot\sqrt{3}}\)

\(=\dfrac{7\sqrt{3}}{\sqrt{36}}=\dfrac{7\sqrt{3}}{6}\)

\(b,\dfrac{3}{2\sqrt{3}}=\dfrac{3\sqrt{3}}{2\sqrt{3}\cdot\sqrt{3}}\)

\(=\dfrac{3\sqrt{3}}{2\cdot3}=\dfrac{3\sqrt{3}}{6}=\dfrac{\sqrt{3}}{2}\)

\(c,\dfrac{1}{5\sqrt{12}}=\dfrac{\sqrt{3}}{5\cdot2\sqrt{3}\cdot\sqrt{3}}\)

\(=\dfrac{\sqrt{3}}{10\cdot3}=\dfrac{\sqrt{3}}{30}\)

\(d,\dfrac{2\sqrt{3}+3}{4\sqrt{3}}=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{4\sqrt{3}}\)

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

a) \(\dfrac{7}{\sqrt[]{12}}=\dfrac{7}{2\sqrt[]{3}}=\dfrac{7\sqrt[]{3}}{2\sqrt[]{3}.\sqrt[]{3}}=\dfrac{7\sqrt[]{3}}{6}\)

b) \(\dfrac{3}{2\sqrt[]{3}}=\dfrac{\sqrt[]{3}.\sqrt[]{3}}{2\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{2}\)

c) \(\dfrac{1}{5\sqrt[]{12}}=\dfrac{1}{10\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{10\sqrt[]{3}.\sqrt[]{3}}=\dfrac{\sqrt[]{3}}{30}\)

d) \(\dfrac{2\sqrt[]{3}+3}{4\sqrt[]{3}}=\dfrac{\sqrt[]{3}\left(2\sqrt[]{3}+3\right)}{4\sqrt[]{3}.\sqrt[]{3}}=\dfrac{3\left(2+\sqrt[]{3}\right)}{12}=\dfrac{2+\sqrt[]{3}}{4}\)

Trước hết, ta cần tính giá trị của a và b trong G và H:
$$G^2 = \frac{1}{a+b} \Rightarrow a+b = \frac{1}{G^2}$$
$$H^2 = 4a - 4\sqrt{ab} + 4b = 4(\sqrt{a} - \sqrt{b})^2 \Rightarrow \sqrt{a} - \sqrt{b} = \frac{H}{2}$$
Từ đó, suy ra được:
$$\sqrt{a} + \sqrt{b} = \frac{1}{G}\sqrt{\frac{1}{G^2} + 4}$$
$$\Rightarrow 2\sqrt{a} = \frac{1}{G}\sqrt{\frac{1}{G^2} + 4} + H$$
$$\Rightarrow a = \left(\frac{1}{G}\sqrt{\frac{1}{G^2} + 4} + H\right)^2/4$$
$$\Rightarrow b = \left(\frac{1}{G}\sqrt{\frac{1}{G^2} + 4} - H\right)^2/4$$

Tiếp theo, ta tính giá trị của F:
$$F = 6\sqrt{3} + \sqrt{2} = 6\sqrt{3} + \sqrt{2}\frac{\sqrt{6}+\sqrt{2}}{2} = 6\sqrt{3} + 3\sqrt{2} + 3\sqrt{6}$$

Cuối cùng, ta tính giá trị của K:
$$K = 2xy\left(2\sqrt{x} + 3\sqrt{y}\right) = 2\sqrt{xy}(4\sqrt{x} + 6\sqrt{y})$$

Vậy, ta đã tính được giá trị của F, G, H và K.

a) Ta có: \(\dfrac{1}{\sqrt{5}-\sqrt{3}-\sqrt{2}}\)

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

\(=\dfrac{\sqrt{5}+\sqrt{3}+\sqrt{2}}{5-5-2\sqrt{6}}\)

\(=\dfrac{-\sqrt{5}-\sqrt{3}-\sqrt{2}}{2\sqrt{6}}\)

\(=\dfrac{-\sqrt{6}\left(\sqrt{5}+\sqrt{3}+\sqrt{2}\right)}{12}\)

b) Ta có: \(\dfrac{2}{-1-\sqrt{2}+\sqrt{3}}\)

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

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

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

\(\dfrac{-2}{\sqrt{3}-1}=\dfrac{-2\left(\sqrt{3}+1\right)}{2}=-\sqrt{3}-1\\ \dfrac{\sqrt{5}}{\sqrt{7}-3}=\dfrac{-\sqrt{5}\left(\sqrt{7}+3\right)}{2}\\ \dfrac{3\sqrt{3}-2}{1-2\sqrt{3}}=\dfrac{\left(3\sqrt{3}-2\right)\left(1+2\sqrt{3}\right)}{-11}=\dfrac{\sqrt{3}-16}{11}\\ \dfrac{14}{\sqrt{10}+\sqrt{3}}=\dfrac{14\left(\sqrt{10}-\sqrt{3}\right)}{7}=2\sqrt{10}-2\sqrt{3}\)

Lời giải:

a.

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

b.

\(=\frac{2[(\sqrt{3}-(\sqrt{2}-1)]}{[(\sqrt{3}+(\sqrt{2}-1)][\sqrt{3}-(\sqrt{2}-1)]}=\frac{2(\sqrt{3}-\sqrt{2}+1)}{3-(\sqrt{2}-1)^2}=\frac{2(\sqrt{3}-\sqrt{2}+1)}{2\sqrt{2}}\)

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

c.

\(=\frac{5(\sqrt[3]{2^2}-3\sqrt[3]{2}+3^2)}{(\sqrt[3]{2})^3+3^3}=\frac{5(\sqrt[3]{4}+3\sqrt[3]{2}+9)}{29}\)