\(\frac{-20+32\sqrt{7}}{9}\)

các bn trình bày bài giải cho mk nha :D

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+2\sqrt{12}}}}}\)

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-2\sqrt{75}}}}\)

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)

\(C=\sqrt{4+5}\)

\(C=3\)

* \(\frac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\frac{6\sqrt{2}-4}{3-\sqrt{2}}\)\(=\frac{\left(\sqrt{8}-\sqrt{7}\right)}{\left(\sqrt{8}+\sqrt{7}\right)\left(\sqrt{8}-\sqrt{7}\right)}+\sqrt{25.7}-\frac{2\sqrt{2}\left(3-\sqrt{2}\right)}{3-\sqrt{2}}\)

\(=\sqrt{8}-\sqrt{7}+5\sqrt{7}-2\sqrt{2}=2\sqrt{2}+4\sqrt{7}-2\sqrt{2}=4\sqrt{7}\)

** \(\frac{\sqrt{6-\sqrt{11}}}{\sqrt{22}-\sqrt{2}}+\frac{6}{\sqrt{2}}-\frac{3}{\sqrt{2}+1}\)\(=\frac{\sqrt{2}\sqrt{6-\sqrt{11}}}{\sqrt{2}\left(\sqrt{22}-\sqrt{2}\right)}+\frac{6\sqrt{2}}{2}-\frac{3\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}\)

\(=\frac{\sqrt{12-2\sqrt{11}}}{2\sqrt{11}-2}+3\sqrt{2}-\frac{3\sqrt{2}-3}{1}\)\(=\frac{\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}+1^2}}{2\left(\sqrt{11}-1\right)}+3\sqrt{2}-3\sqrt{2}+3\)

\(=\frac{\sqrt{11}-1}{2\left(\sqrt{11}-1\right)}+3=\frac{1}{2}+3=\frac{7}{2}\).

Cảm ơn bạn :)

1)

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

\(=\frac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{4+2.2.\sqrt{2}+2}}+\frac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{4-2.2.\sqrt{2}+2}}\)

\(=\frac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{\left(2+\sqrt{2}\right)^2}}+\frac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{\left(2-\sqrt{2}\right)^2}}\)

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

\(=\frac{2.\left(3+2\sqrt{2}\right)}{2.\left(1+\sqrt{2}\right)}+\frac{2.\left(3-2\sqrt{2}\right)}{2.\left(\sqrt{2}-1\right)}\)

\(=\frac{3+2\sqrt{2}}{\sqrt{2}+1}+\frac{3-2\sqrt{2}}{\sqrt{2}-1}\)

\(=\frac{\left(3+2\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\frac{\left(3-2\sqrt{2}\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}\)

\(=1+\sqrt{2}+\sqrt{2}-1=2\sqrt{2}\)

đặt \(\sqrt[3]{2}\)=a \(\Rightarrow\)a³=2, ta có:

x=\(\frac{1}{a+a^2+a^3}\)=\(\frac{a-1}{a\cdot\left(a^3-1\right)}\)=\(\frac{a-1}{a}\)

y=\(\frac{6}{a^4-a^3+a^2}\)=\(\frac{6\cdot\left(a+1\right)}{a^2\left(a^3+1\right)}\)=\(\frac{2\left(a+1\right)}{a^2}\)=\(\sqrt[3]{2}\cdot\left(a+1\right)\)

THeo cách đặt thì tính được x,y. Sau đó thay vào B thì tính được bạn nhé

\(\frac{\sqrt{7}-5}{2}-\frac{6-2\sqrt{7}}{4}+\frac{6}{\sqrt{7}-2}-\frac{5}{4+\sqrt{7}}\)

\(=\frac{\sqrt{7}-5}{2}-\frac{6+2\sqrt{7}}{4}+\frac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}\right)^2-2^2}-\frac{5\left(4-\sqrt{7}\right)}{4^2-\left(\sqrt{7}\right)^2}\)

\(=\frac{\sqrt{7}-5}{2}-\frac{6+2\sqrt{7}}{4}+\frac{6\sqrt{7}+12}{3}-\frac{20-5\sqrt{7}}{8}\)

\(=\frac{12\left(\sqrt{7}-5\right)}{24}-\frac{6\left(6+2\sqrt{7}\right)}{24}+\frac{8\left(6\sqrt{7}+12\right)}{24}-\frac{3\left(20-5\sqrt{7}\right)}{24}\)

\(=\frac{12\sqrt{7}-60-36-12\sqrt{7}+48\sqrt{7}+96-60+15\sqrt{7}}{24}\)

\(=\frac{-60+63\sqrt{7}}{24}\)