a, \(\sqrt{25.16}=5.4=20\)

Chắc vậy á.

ukm đúng rùi đóa

Ta có công thức tổng quát là \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}\left(n+1-n\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

Vậy \(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{400\sqrt{399}+399\sqrt{400}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{399}}-\dfrac{1}{\sqrt{400}}=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{400}}=1-\dfrac{1}{20}=\dfrac{19}{20}\)

a)   = \(\frac{\sqrt{2}.\sqrt{4+\sqrt{15}}}{\sqrt{2}}+\frac{\sqrt{2}.\sqrt{4-\sqrt{15}}}{\sqrt{2}}\)
      = \(\frac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}+\frac{\sqrt{8-2\sqrt{15}}}{\sqrt{2}}\)
      =  \(\frac{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{2}}\)
      = \(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\)

      =   \(\frac{2\sqrt{5}}{\sqrt{2}}\)
      =   \(\sqrt{10}\)
b) = \(5\sqrt{3}+4\sqrt{3}-10\sqrt{3}\)

    = \(-\sqrt{3}\)
c) = \(\sqrt{\left(\sqrt{2}-\sqrt{1}\right)^2}+\frac{1\left(\sqrt{2}+1\right)}{2-1}\)
    = \(\sqrt{2}-\sqrt{1}+\frac{1\left(\sqrt{2}+1\right)}{1}\)
    = \(\sqrt{2}-\sqrt{1}+\sqrt{2}+\sqrt{1}\)
    = \(2\sqrt{2}\)
Chúc bạn học tốt ^^

a,

\(\sqrt{4-2\sqrt{3}}-\sqrt{3}\\ =\sqrt{3-2\cdot1\cdot\sqrt{3}+1}-\sqrt{3}\\ =\sqrt{\left(\sqrt{3}\right)^2-2\cdot1\cdot\sqrt{3}+1^2}-\sqrt{3}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\\ =\sqrt{3}-1-\sqrt{3}\\ =-1\)

b,

\(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\\ =\sqrt{9+2\cdot3\cdot\sqrt{2}+2}-3+\sqrt{2}\\ =\sqrt{3^2+2\cdot3\cdot\sqrt{2}+\left(\sqrt{2}\right)^2}-3+\sqrt{2}\\ =\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}\\ =3+\sqrt{2}-3+\sqrt{2}\\ =2\sqrt{2}\)

c,

\(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}\\ =\sqrt{5+2\cdot\sqrt{2\cdot5}+2}-\sqrt{5-2\cdot\sqrt{2\cdot5}+2}\\ =\sqrt{\left(\sqrt{5}\right)^2+2\cdot\sqrt{2}\cdot\sqrt{5}+\left(\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}\right)^2-2\cdot\sqrt{2}\cdot\sqrt{5}+\left(\sqrt{2}\right)^2}\\ =\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\\ =\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}\\ =2\sqrt{2}\)

d,

\(\left(20\sqrt{300}-15\sqrt{675}+5\sqrt{75}\right):\sqrt{15}\\ =\left(20\cdot\sqrt{20}\cdot\sqrt{15}-15\cdot\sqrt{45}\cdot\sqrt{15}+5\cdot\sqrt{5}\cdot\sqrt{15}\right):\sqrt{15}\\ =\left(20\cdot2\cdot\sqrt{5}\cdot\sqrt{15}-15\cdot3\cdot\sqrt{5}\cdot\sqrt{15}+5\cdot\sqrt{5}\cdot\sqrt{15}\right):\sqrt{15}\\ =\sqrt{15}\cdot\left(20\cdot2\cdot\sqrt{5}-15\cdot3\cdot\sqrt{5}+5\cdot\sqrt{5}\right):\sqrt{15}\\ =20\cdot2\cdot\sqrt{5}-15\cdot3\cdot\sqrt{5}+5\cdot\sqrt{5}\\ =40\sqrt{5}-45\sqrt{5}+5\sqrt{5}\\ =0\)

Ta có:

\(\frac{1}{n\sqrt{n+1}+\left(n+1\right)\sqrt{n}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)

\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}.\text{ Vì thế, }A=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-...-\frac{1}{\sqrt{401}}< 1.\)

chịu rồi

a: \(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)

\(=\left(2\sqrt{6}-4\sqrt{3}+\dfrac{9}{2}\sqrt{2}\right)\cdot5\sqrt{6}\)

\(=60-20\sqrt{18}+\dfrac{45}{2}\sqrt{12}\)

\(=60-60\sqrt{2}+45\sqrt{3}\)

b: \(=\dfrac{\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{2}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}\)

\(=\dfrac{2\sqrt{5}+3}{3}\cdot\dfrac{1}{3+2\sqrt{2}}=\dfrac{2\sqrt{5}+3}{9+6\sqrt{2}}\)

Bằng 66

\(2\sqrt{3}\). (\(3\sqrt{3}+8\sqrt{3}-5\))+ \(10\sqrt{3}\)

= \(2\sqrt{3}.3\sqrt{3}+8\sqrt{3}+10\sqrt{3}-5\)

=\(18+2\sqrt{3}-5\)

=\(13+21\sqrt{3}\)

91,35898385

\(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-2\sqrt{25}+\sqrt{300}\right)\)

\(=2\sqrt{3}.\left(\sqrt{9.3}+2\sqrt{16.3}-10+\sqrt{100.3}\right)\)

\(=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-10+10\sqrt{3}\right)\)

\(=2\sqrt{3}.\left(21\sqrt{3}-10\right)\)

\(=126-20\sqrt{3}\)

\(a.6\sqrt{3}-2\sqrt{12}+5\sqrt{300}-7\sqrt{243}=6\sqrt{3}-4\sqrt{3}+50\sqrt{3}-63\sqrt{3}=\left(6-4+50-63\right)\sqrt{3}=-11\sqrt{3}\)

\(b.\sqrt{28}+3\sqrt{63}-6\sqrt{175}-\dfrac{1}{5}\sqrt{252}=2\sqrt{7}+9\sqrt{7}-30\sqrt{7}-\dfrac{6}{5}\sqrt{7}=\left(2+9-30-\dfrac{6}{5}\right)\sqrt{7}=-20,2\sqrt{7}\)\(c.5\sqrt{44}-2\sqrt{275}-3\sqrt{176}=10\sqrt{11}-10\sqrt{11}-12\sqrt{11}=-12\sqrt{11}\)

\(d.2\sqrt{75}-\sqrt{12}+2\sqrt{147}-7\sqrt{103}=10\sqrt{3}-2\sqrt{3}+14\sqrt{3}-7\sqrt{103}=22\sqrt{3}-7\sqrt{103}\)

\(a.6\sqrt{3}-2\sqrt{12}+5\sqrt{300}-7\sqrt{243}=6\sqrt{3}-4\sqrt{3}+50\sqrt{3}-63\sqrt{3}=-11\sqrt{3}\)

\(b.\sqrt{28}+3\sqrt{63}-6\sqrt{175}-\dfrac{1}{5}\sqrt{252}=2\sqrt{7}+9\sqrt{7}-30\sqrt{7}-\dfrac{6}{5}\sqrt{7}=-\dfrac{101}{5}\sqrt{7}\)

\(c.5\sqrt{44}-2\sqrt{275}-3\sqrt{176}=20\sqrt{11}-10\sqrt{11}-12\sqrt{11}=-2\sqrt{11}\)

\(d.2\sqrt{75}-\sqrt{12}+2\sqrt{147}-7\sqrt{103}=10\sqrt{3}-2\sqrt{3}+14\sqrt{3}-7\sqrt{103}=22\sqrt{3}-7\sqrt{103}\)