\(A=\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}=\sqrt{\left(4+\sqrt{8}\right)^2}-\sqrt{\left(4-\sqrt{8}\right)^2}=\left|4+\sqrt{8}\right|-\left|4-\sqrt{8}\right|=4+\sqrt{8}-4+\sqrt{8}=4\sqrt{2}\)

\(A=\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}\)

\(=\sqrt{8+2.4.2\sqrt{2}+16}-\sqrt{16-2.4.2\sqrt{2}+8}\)

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

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

\(=4\sqrt{2}\)

\(\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}=\sqrt{8\left(3+2\sqrt{2}\right)}-\sqrt{8\left(3-2\sqrt{2}\right)}\)

\(=\sqrt{8}.\left[\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}\right]=\sqrt{8}.\left(\sqrt{2}+1-\sqrt{2}+1\right)=2\sqrt{8}=4\sqrt{2}\)

\(8\sqrt{2}\left(\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}\right)\)

\(=8\sqrt{2}\left(\sqrt{16+2.4.\sqrt{8}+8}-\sqrt{16-2.4\sqrt{8}+8}\right)\)

\(=8\sqrt{2}\left(\sqrt{\left(4+\sqrt{8}\right)^2}-\sqrt{\left(4-\sqrt{8}\right)^2}\right)\)

\(=8\sqrt{2}\left(4+\sqrt{8}-4+\sqrt{8}\right)\)

\(=8\sqrt{2}.2\sqrt{8}\)

= 64

ta có\(8\sqrt{2}\cdot\left(\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}\right)=8\sqrt{2}\cdot\left(\sqrt{\left(4+\sqrt{8}\right)^2}-\sqrt{\left(4-\sqrt{8}\right)^2}\right)=8\sqrt{2}\cdot\left(4+\sqrt{8}-4+\sqrt{8}\right)=8\sqrt{2}\cdot2\sqrt{8}=64\)vây..................

Ta có: \(x=\sqrt{97-56\sqrt{3}}+\sqrt{52+16\sqrt{3}}\)

\(=\sqrt{49-2\cdot7\cdot4\sqrt{3}+48}+\sqrt{48+2\cdot4\sqrt{3}\cdot2+4}\)

\(=\sqrt{\left(7-4\sqrt{3}\right)^2}+\sqrt{\left(4\sqrt{3}+2\right)^2}\)

\(=\left|7-4\sqrt{3}\right|+\left|4\sqrt{3}+2\right|\)

\(=7-4\sqrt{3}+4\sqrt{3}+2\)

\(=9\)

Làm luôn phần y :D

y = \(\sqrt{33+20\sqrt{2}}+\sqrt{24-16\sqrt{2}}\)

y = \(\sqrt{33+2.10\sqrt{2}}+\sqrt{24-2.8\sqrt{2}}\)

y = \(\sqrt{33+2.5.2\sqrt{2}}+\sqrt{24-2.4.2\sqrt{2}}\)

y = \(\sqrt{25+2.5.\sqrt{8}+8}+\sqrt{16-2.4.\sqrt{8}+8}\)

y = \(\sqrt{\left(5+\sqrt{8}\right)^2}+\sqrt{\left(4-\sqrt{8}\right)^2}\)

y = |5 + \(\sqrt{8}\)| + |4 - \(\sqrt{8}\)| 

y = 5 + \(\sqrt{8}\) + 4 - \(\sqrt{8}\)   (Vì 4 > \(\sqrt{8}\) nên 4 - \(\sqrt{8}\) > 0)

y = 9

Vậy y = 9

Chúc bn học tốt!

ai lam ho voi

A=\(\sqrt{\left(4+\sqrt{8}\right)^2}\)\(-\sqrt{\left(4-\sqrt{8}\right)^2}\)=\(4+\sqrt{8}\)\(-\left(4-\sqrt{8}\right)\)=\(2\sqrt{8}\)

Giờ mình chỉ giải đc câu a thôi để hồi nao mình rảnh giải típ cho

ok ban

\(\sqrt{24-16\sqrt{2}}+\sqrt{12-8\sqrt{2}}=\dfrac{\sqrt{32-2.4.4\sqrt{2}+16}+\sqrt{12-2.4.2\sqrt{2}+16}}{\sqrt{2}}=\dfrac{4\sqrt{2}-4+4-2\sqrt{2}}{\sqrt{2}}=\dfrac{2\sqrt{2}}{\sqrt{2}}=1\)

\(1,\\ a,=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}=3+\sqrt{7}-\sqrt{7}+1=4\\ b,K=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\sqrt{\left(6-2\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-4\right)^2}=6-2\sqrt{6}+2\sqrt{6}-4=2\\ e,=\sqrt{\left(2-\sqrt{2}\right)^2}-\left(\sqrt{6}-\sqrt{2}\right)=2-\sqrt{2}-\sqrt{6}+\sqrt{2}=2-\sqrt{6}\)

\(2,\\ a,A=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\\ A=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\\ b,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{1}{\sqrt{3}+1-3}=\dfrac{1}{\sqrt{3}+2}=2-\sqrt{3}\)

a,ĐK: x\(\ge\)1

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

⇔\(\sqrt{\left(\sqrt{x-1}-1\right)^2}\)=\(\sqrt{2}\)

⇔\(\left|\sqrt{x-1}-1\right|\)=\(\sqrt{2}\)

TH1:\(\sqrt{x-1}\)-1≥0⇒\(\left|\sqrt{x-1}-1\right|\)=\(\sqrt{x-1}\)-1   bn tự giải ra nha

TH2:\(\sqrt{x-1}\)-1<0⇒\(\left|\sqrt{x-1}-1\right|\)=1-\(\sqrt{x-1}\)    bn tự lm nha