Rút Gọn
a,\(\sqrt{75}-\sqrt{5\frac{1}{3}}+\frac{9}{2}\sqrt{2\frac{2}{3}}+2\sqrt{27}\)
b,\(\sqrt{48}+\sqrt{5\frac{1}{3}}+2\sqrt{75}-5\sqrt{1\frac{1}{3}}\)
c,\(\left(\sqrt{12}+2\sqrt{27}\right)\frac{\sqrt{3}}{2}-\sqrt{150}\)
d,\(\left(\sqrt{18}+\sqrt{0,5}-3\sqrt{\frac{1}{3}}\right)-\left(\sqrt{\frac{1}{8}-\sqrt{75}}\right)\)
e,\(6\sqrt{\frac{8}{9}}-5\sqrt{\frac{32}{25}}+14\sqrt{\frac{18}{49}}\)
f,\(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\)
g,\(\left(2\sqrt{\frac{16}{3}}-3\sqrt{\frac{1}{27}}-6\sqrt{\frac{4}{75}}\right)\sqrt{3}\)
h,\(\left(6\sqrt{\frac{8}{9}}-5\sqrt{\frac{32}{25}}+14\sqrt{\frac{18}{49}}\right)\sqrt{\frac{1}{2}}\)
i,\(\frac{1}{2}\sqrt{48}-2\sqrt{75}-\frac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\frac{1}{3}}\)
j,\(\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+7\right):\sqrt{7}\)
B1:
\(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}\)
\(=\left|\sqrt{2}-\sqrt{3}\right|+3\sqrt{2}\)
\(=\sqrt{3}-\sqrt{2}+3\sqrt{2}\)
\(=\sqrt{3}+2\sqrt{2}\)
\(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=\sqrt{4-4\sqrt{3}+3}+\left|1+\sqrt{3}\right|\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+1+\sqrt{3}\)
\(=2-\sqrt{3}+1+\sqrt{3}\)
\(=3\)
B2:
đk: \(x\ge-2\)
Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\frac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\)
\(\Leftrightarrow\sqrt{x+2}=3\)
\(\Leftrightarrow x+2=9\)
\(\Rightarrow x=7\)
Vậy x = 7