a: \(\frac{3}{4+\sqrt{9+4\sqrt5}}\)

\(=\frac{3}{4+\sqrt{\left(\sqrt5+2\right)^2}}\)

\(=\frac{3}{4+\sqrt5+2}=\frac{3}{6+\sqrt5}=\frac{3\left(6-\sqrt5\right)}{36-5}=\frac{3\left(6-\sqrt5\right)}{31}\)

b: \(\frac{\sqrt3}{\sqrt2+\sqrt{5+2\sqrt6}}\)

\(=\frac{\sqrt3}{\sqrt2+\sqrt{\left(\sqrt3+\sqrt2\right)^2}}=\frac{\sqrt3}{\sqrt2+\sqrt3+\sqrt2}\)

\(=\frac{\sqrt3}{2\sqrt2+\sqrt3}=\frac{\sqrt3\left(2\sqrt2-\sqrt3\right)}{8-3}=\frac{2\sqrt6-3}{5}\)

c: \(\frac{3}{\sqrt5+\sqrt7-\sqrt2}=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)}{\left(\sqrt5+\sqrt7\right)^2-2}\)

\(=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)}{10+2\sqrt{35}}=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)\left(\sqrt{35}-5\right)}{2\left(\sqrt{35}+5\right)\left(\sqrt{35}-5\right)}\)

\(=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)\left(\sqrt{35}-5\right)}{2\left(35-25\right)}=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)\left(\sqrt{35}-5\right)}{20}\)

\(E=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\sqrt{\left(4+\sqrt{15}\right)^2}.\sqrt{\left(\sqrt{10}-\sqrt{6}\right)^2}.\frac{4^2-15}{\sqrt{4+\sqrt{15}}}\)

\(=\sqrt{4+\sqrt{15}}.\sqrt{10+6-2\sqrt{10}.\sqrt{6}}\)

\(=\sqrt{4+\sqrt{15}}.\sqrt{16-2\sqrt{60}}\)

\(=\sqrt{4+\sqrt{15}}.\sqrt{4\left(4-\sqrt{15}\right)}\)

\(=2\sqrt{\left(4+\sqrt{15}\right).\left(4-\sqrt{15}\right)}\)

\(=2\sqrt{16-15}=2\)

 Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề hơn nhé. 

\(\sqrt{3}-\frac{5}{2}>\sqrt{3}-4\text{ vì }-\frac{5}{2}>-4\)

\(\Rightarrow2.\left(\sqrt{3}-\frac{5}{2}\right)>\sqrt{3}-4\)

\(\Rightarrow2.\sqrt{3}-5>\sqrt{3}-4\)

b) vì \(\sqrt{5}-\sqrt{12}< 0\), ta có: 

 \(5\sqrt{5}-2\sqrt{3}=4\sqrt{5}+\sqrt{5}-\sqrt{12}< 4\sqrt{5}< 4\sqrt{5}+6\) 

Vậy \(5\sqrt{5}-2\sqrt{3}< 6+4\sqrt{5}\)

a) \(\left(\sqrt{\dfrac{9}{20}}-\sqrt{\dfrac{1}{2}}\right).\sqrt{2}=\sqrt{\dfrac{9}{20}.2}-\sqrt{\dfrac{1}{2}.2}=\sqrt{\dfrac{9}{10}}-1=\dfrac{3}{\sqrt{10}}-1\)

\(=\dfrac{3\sqrt{10}}{10}-1\)

b) \(\left(\sqrt{12}+\sqrt{27}-\sqrt{3}\right)\sqrt{3}=\sqrt{12.3}+\sqrt{27.3}-\sqrt{3.3}\)

\(=\sqrt{36}+\sqrt{81}-\sqrt{9}=6+9-3=12\)

c) \(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right)\sqrt{6}=\sqrt{\dfrac{8}{3}.6}-\sqrt{24.6}+\sqrt{\dfrac{50}{3}.6}\)

\(=\sqrt{16}-\sqrt{144}+\sqrt{100}=4-12+10=2\)

Đặt \(\frac{a}{\sqrt3}=\frac{b}{\sqrt2}=\frac{2c}{\sqrt6-\sqrt2}=k\)

=>\(a=k\cdot\sqrt3;b=k\cdot\sqrt2;2c=k\left(\sqrt6-\sqrt2\right)\)

=>\(a=k\cdot\sqrt3;b=k\cdot\sqrt2;c=\frac{k\left(\sqrt6-\sqrt2\right)}{2}\)

Xét ΔABC có \(cosA=\frac{b^2+c^2-a^2}{2\cdot b\cdot c}\)

\(=\left\lbrack2k^2+k^2\cdot\frac{\left(\sqrt6-\sqrt2\right)^2}{4}-3k^2\right\rbrack:\left\lbrack2\cdot k\cdot\sqrt2\cdot\frac{k\left(\sqrt6-\sqrt2\right)}{2}\right\rbrack\)

\(=\left\lbrack2k^2+k^2\cdot\frac{8-4\sqrt3}{4}-3k^2\right\rbrack:\left\lbrack k^2\cdot\left(\sqrt{12}-2\right)\right\rbrack\)

\(=\frac{\left(2+2-\sqrt3-3\right)}{\sqrt{12}-2}=\frac{1-\sqrt3}{2\left(\sqrt3-1\right)}=-\frac12\)

=>\(\hat{BAC}=120^0\)

Xét ΔABC có \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)

=>\(\frac{k\cdot\sqrt3}{\sin120}=\frac{k\cdot\sqrt2}{\sin B}=\frac{\frac{k\left(\sqrt6-\sqrt2\right)}{2}}{\sin C}\)

=>\(\frac{\sqrt2}{\sin B}=\frac{\frac{\left(\sqrt6-\sqrt2\right)}{2}}{\sin C}=\sqrt3:\frac{\sqrt3}{2}=2\)

=>\(\sin B=\frac{\sqrt2}{2};\sin C=\frac{\sqrt6-\sqrt2}{4}\)

=>\(\hat{B}=45^0;\hat{C}=15^0\)

Xét ΔABC có \(\frac{BC}{\sin A}=2R\)

=>\(2R=2\sqrt3:\sin120=2\sqrt3:\frac{\sqrt3}{2}=2\sqrt3\cdot\frac{2}{\sqrt3}=4\)

=>R=2