a) = \(\frac{7}{2}\)

b) = \(\frac{643}{64}\)

c) = 0

a) \(9,2\cdot2\frac{1}{2}-\left(2\cdot0,125-1\frac{5}{12}\right):\frac{1}{4}\)

\(=23-\left(0,25-\frac{17}{12}\right):\frac{1}{4}\)

\(=23-\left(-\frac{7}{6}\right):\frac{1}{4}\)

\(=23-\left(-\frac{14}{3}\right)\)

\(=\frac{83}{3}\)

b) \(\frac{\sqrt{3^2}-\sqrt{39^2}}{\sqrt{59-10-\sqrt{91^2}}}\)

\(=\frac{3-39}{\sqrt{59-10-91}}\)

\(=\frac{-36}{\sqrt{-42}}\)

Vì -42 < 0 \(\Leftrightarrow\)Biểu thức không tồn tại

c) \(\frac{5}{18}-1,456:\frac{7}{25}+4,5\cdot\frac{4}{5}\)

\(=\frac{5}{18}-\frac{26}{5}+\frac{18}{5}\)

\(=\frac{5}{18}-\frac{8}{5}\)

\(=\frac{-119}{90}\)

a, \(\frac{8^{15^{ }}.3^{16}}{4^{23^{ }}.9^8}=\frac{2^{45}.3^{16}}{2^{46}.3^{16}}=\frac{2^{45}}{2^{46}}=\frac{1}{2}\)

b, \(\sqrt{121}-4.\sqrt{9}+\sqrt{36}=11-4.3+6=11-12+6=5\)

c,

\(\frac{2^5}{5^2}+5\frac{1}{2}.\left(4,5-2,5\right)+\frac{2^3}{-4}+\left(-2016\right)^0\)

\(\frac{4}{25}+\frac{11}{2}.2+\frac{8}{-4}+1=\frac{4}{25}+11+\left(-2\right)+1=\frac{4}{25}+10\)

= \(\frac{254}{25}\)

C = \(25.\left(\frac{-1}{3}\right)^3\) \(+\frac{1}{5}\) \(-2.\left(\frac{-1}{2}\right)^2\) \(-\frac{1}{2}\)

C = \(25.\left(\frac{-1}{27}\right)+\frac{1}{5}\) \(-2.\frac{1}{4}\) \(-\frac{1}{2}\)

C = \(\frac{-25}{27}\) \(+\frac{1}{5}\) \(-\frac{1}{2}\) \(-\frac{1}{2}\)

C = \(\frac{-25}{27}\) \(+\frac{1}{5}\) \(-1\)

C = \(\frac{-125}{135}\) \(+\frac{27}{135}\) \(-\frac{135}{135}\)

C = \(\frac{-233}{135}\)

D =  \(-8.\left(\frac{3}{4}-\frac{1}{4}\right):\left(\frac{9}{4}-\frac{7}{6}\right)\)

D = \(-8.\frac{1}{2}\) \(.\frac{12}{13}\)

D = \(-4.\frac{12}{13}\)

D = \(\frac{-48}{13}\)

E = \(5\sqrt{16}\) \(-4\sqrt{9}\) \(+\sqrt{25}\) \(-0,3\sqrt{400}\)

E = \(5.4-4.3+5-0,3.20\)

E = \(20-12+5-6\)

E = \(8+\left(-1\right)\)

E = \(7\)

F = \(\left(\frac{-3}{2}\right)\) \(+\left|\frac{-5}{6}\right|\) \(-1\frac{1}{2}\) \(:6\)

F = \(\left(\frac{-3}{2}\right)\) \(+\frac{5}{6}\) \(-\frac{3}{2}\) \(.\frac{1}{6}\)

F = \(\left(\frac{-3}{2}\right)\) \(+\frac{5}{6}\) \(-\frac{1}{4}\) 

F = \(\left(\frac{-18}{12}\right)\) \(+\frac{10}{12}\) \(-\frac{3}{12}\)

F = \(\frac{-11}{12}\)

 Chúc cậu hk tốt ~ 

a) Ta có: \(A=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)\)

\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\left(\sqrt{9}-\sqrt{4}\right)\)

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

\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)

\(=\left(\sqrt{5}-\sqrt{3}\right)\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)(Vì \(\sqrt{5}>\sqrt{3}\))

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

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

b) Ta có: \(B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)\)

\(=\left(\frac{\sqrt{3}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}+\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}+\sqrt{6}\right)\)

\(=\sqrt{3}+\sqrt{3}+\sqrt{6}\)

\(=2\sqrt{3}+\sqrt{6}\)

c) Ta có: \(C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}}+\sqrt{3}\right):\sqrt{3}\)

\(=2\sqrt{3}+\sqrt{4-2\cdot2\cdot\sqrt{3}+3}+\sqrt{\frac{1}{3}:3}-\sqrt{\frac{4}{3}:3}+\sqrt{3:3}\)

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

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

\(=2\sqrt{3}+2-\sqrt{3}+\frac{2}{3}\)(Vì \(2>\sqrt{3}\))

\(=\sqrt{3}+\frac{8}{3}\)

d) Ta có: \(D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}\)

\(=\left(\frac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\right)\cdot\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\)

\(=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\cdot\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(=\frac{60}{20}\cdot\left|2-\sqrt{3}\right|\)

\(=3\cdot\left(2-\sqrt{3}\right)\)(Vì \(2>\sqrt{3}\))

\(=6-3\sqrt{3}\)