`5\sqrt3+5\sqrt48-10\sqrt7+\sqrt4. \sqrt3`

`=5\sqrt3 + 5\sqrt(4^2 .3)-10\sqrt7 + 2\sqrt3`

`=5\sqrt3+20\sqrt3 -10\sqrt7 +2\sqrt3`

`=27\sqrt3-10\sqrt7`

\(r=\left(p-a\right)tan\frac{A}{2}=\left(p-b\right)tan\frac{B}{2}=\left(p-c\right)tan\frac{C}{2}\)

\(\Rightarrow\frac{5\sqrt{3}}{3}=\frac{b+c-a}{2}.tan30^0\Leftrightarrow\frac{5\sqrt{3}}{3}=\frac{b+c-10}{2}.\frac{\sqrt{3}}{3}\)

\(\Rightarrow b+c=20\) \(\Rightarrow p=\frac{a+b+c}{2}=15\)

Mặt khác \(R=\frac{a}{2sinA}=\frac{10}{\sqrt{3}}=\frac{10\sqrt{3}}{3}\)

\(S=\frac{abc}{4R}\) ; \(S=pr\Rightarrow\frac{abc}{4R}=pr\)

\(\Leftrightarrow\frac{10.bc}{4.\frac{10}{\sqrt{3}}}=p.\frac{5\sqrt{3}}{3}\Leftrightarrow bc=\frac{20p}{3}=100\)

\(\Rightarrow\left\{{}\begin{matrix}b+c=20\\bc=100\end{matrix}\right.\) theo Viet đảo b;c là nghiệm:

\(x^2-20x+100=0\Leftrightarrow x=10\Rightarrow b=c=10\)

=

\(\left(3+\frac{\sqrt{5}}{\sqrt{10}}+\sqrt{3}+\sqrt{5}\right)-\left(3-\frac{\sqrt{5}}{\sqrt{10}}+\sqrt{3}-\sqrt{5}\right)=\sqrt{34.64911064}\)

Phép tính:

\(2\times\sqrt{15}-2\times\sqrt{10}+\sqrt{6}=1421411372\)

\(2\times\sqrt{15}-2\times\sqrt{10}+\sqrt{3}+\sqrt{6}=5602951922\)

P/s: Em ko biết đúng hay sai đâu mới lớp 4 thôi à