49 + 20 căn 6 =  25 + 2.5.(2 căn 6) +24 =  (5 + 2 căn 6)²

tương tự vs 49 - 20 căn 6 = (5 - 2 căn 6)² =) căn ( 49 - 20 căn 6 ) = 5 - 2 căn 6

7 - 4 căn 3 = 4 - 4 căn 3 + 3 = (2 - căn 3)²  =) căn ( 7 - 4 căn 3 ) = 2 - căn 3

tự giải nhé

a) -25 b) -64

\(\dfrac{1}{\sqrt{49+20\sqrt{6}}}-\dfrac{1}{\sqrt{49-20\sqrt{6}}}+\dfrac{1}{\sqrt{7-4\sqrt{3}}}\)

\(=\dfrac{1}{\sqrt{5^2+2\cdot2\sqrt{6}\cdot5+\left(2\sqrt{6}\right)^2}}-\dfrac{1}{\sqrt{5^2-2\cdot2\sqrt{6}\cdot5+\left(2\sqrt{6}\right)^2}}+\dfrac{1}{\sqrt{2^2-2\cdot2\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}}\)

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

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

\(=\dfrac{5-2\sqrt{6}}{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}-\dfrac{5+2\sqrt{6}}{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}+\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)

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

\(=-4\sqrt{6}+2+\sqrt{3}\)

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

\(=5-2\sqrt{6}-5-2\sqrt{6}+2+\sqrt{3}\)

\(=2-4\sqrt{6}+\sqrt{3}\)

\(=\dfrac{1}{\sqrt{\sqrt{25}^2+2.\sqrt{25}.\sqrt{24}+\sqrt{24}^2}}-\dfrac{1}{\sqrt{\sqrt{25}^2-2.\sqrt{25}.\sqrt{24}+\sqrt{24}^2}}+\dfrac{1}{\sqrt{\sqrt{4}^2-2\sqrt{4}\sqrt{3}+\sqrt{3}^2}}\\ =\dfrac{1}{\sqrt{\left(\sqrt{25}+\sqrt{24}\right)^2}}-\dfrac{1}{\sqrt{\left(\sqrt{25}-\sqrt{24}\right)^2}}+\dfrac{1}{\sqrt{\left(\sqrt{4}-\sqrt{3}\right)^2}}\\ =\dfrac{1}{5+\sqrt{24}}-\dfrac{1}{5-\sqrt{24}}+\dfrac{1}{2-\sqrt{3}}\)

\(=\dfrac{5-\sqrt{24}}{25-24}-\dfrac{5+\sqrt{24}}{25-24}+\dfrac{2+\sqrt{3}}{4-3}\\ =5-\sqrt{24}-5-\sqrt{24}+2+\sqrt{3}\\ =2-4\sqrt{6}+\sqrt{3}\)

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

\(49+20\sqrt{6}=25+2.5.2\sqrt{6}+24=\left(5+2\sqrt{6}\right)^2=\left(3+2.\sqrt{3}\sqrt{2}+2\right)^2=\left(\sqrt{3}+\sqrt{2}\right)^4\)

\(\Leftrightarrow\sqrt[4]{49+20\sqrt{6}}=\sqrt{3}+\sqrt{2}\)

tuiwng tự \(\Leftrightarrow\sqrt[4]{49-20\sqrt{6}}=\sqrt{3}-\sqrt{2}\)

=> Cộng lại  = > dpcm

Ta có \(\sqrt[4]{49+20\sqrt{6}}=\sqrt[4]{25+10\sqrt{24}+24}=\sqrt[4]{\left(5+2\sqrt{6}\right)^2}\)

                               \(=\sqrt[4]{\left(\sqrt{3}+\sqrt{2}\right)^4}=\sqrt{3}+\sqrt{2}\)

Tương tự : \(\sqrt[4]{49-20\sqrt{6}}=\sqrt{3}-\sqrt{2}\) ( Do \(\sqrt{3}>\sqrt{2}\) )

Suy ra \(\sqrt[4]{49+20\sqrt{6}}+\sqrt[4]{49-20\sqrt{6}}=2\sqrt{3}\)

a, [1 - 2 ]+[ 3 - 4] +[5 - 6.]..+ [49 - 50] có 25 số hạng

=-1+[-1]+[-1]+...+[-1]

=-1.25

=-25

vậy b=-25

Bạn có thể làm nốt câu b ko ạ ?

a: =-2/49-5/3+2/49=-5/3

b: \(=\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}\)

\(=\dfrac{5\cdot2^{30}\cdot3^{18}-3^{20}\cdot2^{27}\cdot2^2}{5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{5\cdot2^{30}\cdot3^{18}-3^{20}\cdot2^{29}}{5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)

\(=\dfrac{2^{29}\cdot3^{18}\left(2\cdot3-3^2\right)}{2^{28}\cdot3^{18}\left(5\cdot3-7\cdot2\right)}=2\cdot\dfrac{6-9}{15-14}=2\cdot\left(-3\right)=-6\)

a) \(\sqrt[4]{49+20\sqrt{6}}+\sqrt[4]{49-20\sqrt{6}}=\sqrt[4]{25+2\sqrt{600}+24}+\sqrt[4]{25-2\sqrt{600}+24}\\ =\sqrt[4]{\left(\sqrt{25}+\sqrt{24}\right)^2}+\sqrt[4]{\left(\sqrt{25}-\sqrt{24}\right)^2}=\sqrt{\sqrt{25}+\sqrt{24}}+\sqrt{\sqrt{25}-\sqrt{24}}\\ =\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{6}+2}+\sqrt{3-2\sqrt{6}+2}\\ =\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}\\ =2\sqrt{3}\)