\(\sqrt{10+\sqrt{19}}+\sqrt{10-\sqrt{19}}\)

\(=\sqrt{10^2-\left(\sqrt{19}\right)^2}\)

\(=\sqrt{100-19}\)

= \(\sqrt{81}\)

\(=9\)

\(\sqrt{10+\sqrt{19}}+\sqrt{10-\sqrt{19}}\)

=\(\sqrt{10^2-\left(\sqrt{19}\right)^2}\)

=\(\sqrt{100-19}\)

=\(\sqrt{81}\)

= 9 (đpcm)

\(3T=\left(\sqrt{x^2-6x+19}-\sqrt{x^2-6x+10}\right)\left(\sqrt{x^2-6x+19}+\sqrt{x^2-6x+10}\right)\)

\(=x^2-6x+19-\left(x^2-6x+10\right)=9\)

\(\Rightarrow T=3\)

Bài 2: 

\(A=\sqrt{26}+\sqrt{10}>\sqrt{25}+\sqrt{9}=5+3=8\)

\(B=\sqrt{64}=8\)

Do đó: A>B

1.Ta có:

\(A=\)\(\sqrt{13}+\sqrt{20}=\sqrt{13}+2\sqrt{5}\)

\(B=\)\(\sqrt{24}+\sqrt{19}=\sqrt{19}+2\sqrt{6}\)

So sánh ta thấy:

\(\sqrt{13}<\sqrt{19}\) ; \(2\sqrt{5}<2\sqrt{6}\)

Vậy A < B

a)

\((4+\sqrt{15})(\sqrt{10}-\sqrt{6})\sqrt{4-\sqrt{15}}=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{3+5-2\sqrt{3.5}}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)

\(=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})^2=(4+\sqrt{15})(8-2\sqrt{15})=2(4+\sqrt{15})(4-\sqrt{15})\)

\(=2(4^2-15)=2\)

b)

\(\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}=\sqrt{(8+2\sqrt{15})+2+2(\sqrt{6}+\sqrt{10})}\)

\(=\sqrt{(\sqrt{5}+\sqrt{3})^2+2\sqrt{2}(\sqrt{3}+\sqrt{5})+2}\)

\(=\sqrt{(\sqrt{5}+\sqrt{3}+\sqrt{2})^2}=\sqrt{5}+\sqrt{3}+\sqrt{2}\)

c)

\((\sqrt{5+2\sqrt{9\sqrt{5}-19}}-\sqrt{7-\sqrt{5}}):(2\sqrt{\sqrt{5}-2})\)

\(=(\sqrt{(5+2\sqrt{9\sqrt{5}-19})(\sqrt{5}+2)}-\sqrt{(7-\sqrt{5})(\sqrt{5}+2)}):(2\sqrt{(\sqrt{5}-2)(\sqrt{5}+2)})\)

\(=[\sqrt{10+5\sqrt{5}+2\sqrt{(9\sqrt{5}-19)(9+4\sqrt{5})}}-\sqrt{9+5\sqrt{5}}]:2\)

\(=[\sqrt{10+5\sqrt{5}+2\sqrt{9+5\sqrt{5}}}-\sqrt{9+5\sqrt{5}}]:2\)

\(=[\sqrt{(9+5\sqrt{5})+2\sqrt{9+5\sqrt{5}}+1}-\sqrt{9+5\sqrt{5}}]:2\)

\(=[\sqrt{(\sqrt{9+5\sqrt{5}}+1)^2}-\sqrt{9+5\sqrt{5}}]:2\)

\(=[\sqrt{9+5\sqrt{5}}+1-\sqrt{9+5\sqrt{5}}]:2=\frac{1}{2}\)

d)

\((\sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}})^2=18+2\sqrt{(9+\sqrt{5})(9-\sqrt{5})}=18+4\sqrt{19}\)

\(\Rightarrow \sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}}=\sqrt{18+4\sqrt{19}}\)

Do đó:
\(\frac{\sqrt{9+\sqrt{5}}+\sqrt{9-\sqrt{5}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{3-2\sqrt{2}}=\frac{\sqrt{18+4\sqrt{19}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{2+1-2\sqrt{2.1}}\)

\(=\frac{\sqrt{2}.\sqrt{9+2\sqrt{19}}}{\sqrt{9+2\sqrt{19}}}-\sqrt{(\sqrt{2}-1)^2}=\sqrt{2}-(\sqrt{2}-1)=1\)

b, t = \(\sqrt{3- \sqrt{5}}\)(3 +\(\sqrt{5}\)).(\(\sqrt{10}\)-\(\sqrt{2}\))

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

t = (\(\sqrt{5}\) -1).(\(\sqrt{5}\) -1).(3 +\(\sqrt{5}\))

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

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

t = 15 + \(5\sqrt{5}\) \(-6\sqrt{5}\)-10+1+\(\sqrt{5}\)

t = 6

bạn kiểm tra lại biểu thức A đi bạn

\(19+6\sqrt{10}=10+2.3\sqrt{10}+9=\left(\sqrt{10}+3\right)^2\)

=> \(A=\sqrt[10]{\frac{19+6\sqrt{10}}{2}}\cdot\sqrt[5]{3\sqrt{2}-2\sqrt{5}}\)

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

= \(\sqrt[5]{\frac{\sqrt{10}+3}{\sqrt{2}}}.\sqrt[5]{\sqrt{2}\left(3-\sqrt{10}\right)}\)

= \(\sqrt[5]{\frac{\sqrt{10}+3}{\sqrt{2}}.\sqrt{2}\left(3-\sqrt{10}\right)}\)

\(=\sqrt[5]{3^2-10}=-1\)

Đặt \(a=\sqrt{x^2-6x+19},a\ge0\) ; \(b=\sqrt{x^2-6x+10},b\ge0\)

\(\Rightarrow\begin{cases}a-b=3\\a^2-b^2=9\end{cases}\)  \(\Rightarrow A=a+b=3\)