Lời giải:
\(A=\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)

\(=\sqrt{10+2\sqrt{2}(\sqrt{3}+\sqrt{5})+2\sqrt{15}}=\sqrt{2+(3+5+2\sqrt{15})+2\sqrt{2}(\sqrt{3}+\sqrt{5})}\)

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

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

\(2B=2.\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=2.\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+(\sqrt{4}+\sqrt{6}+\sqrt{8})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=2.\frac{(\sqrt{2}+\sqrt{3}+\sqrt{4})+\sqrt{2}(\sqrt{2}+\sqrt{3}+\sqrt{4})}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=2(1+\sqrt{2})\)

Do đó:

\(A-2B=\sqrt{3}+\sqrt{5}-(2+\sqrt{2})>\sqrt{2}+\sqrt{4}-(2+\sqrt{2})=0\)

\(\Rightarrow A>2B\)

Bạn tham khảo tại link sau:

Câu hỏi của Vi Huỳnh - Toán lớp 9 | Học trực tuyến

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

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

giải ra chưa chỉ mình với

Bạn áp dụng hằng đẳng thức (a+b+c)^2= a^2+b^2+c^2+2(ab+ac+bc)

a/ \(\sqrt{2}+\sqrt{6}\)

b/ Sửa đề:

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

c/ \(1+\sqrt{2}+\sqrt{5}\)

giải rõ ra hộ mình với

a/ \(2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}\)

\(=2\sqrt{3+\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}\)

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

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

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

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

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

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

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

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

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

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

\(=\sqrt{2}+\sqrt{6}\)

b)

\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)

\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)=6-121=-115\)

a)

\(\dfrac{\left(3\sqrt{3}+5\sqrt{2}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)

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

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

\(=\dfrac{9-3\sqrt{6}+5\sqrt{6}-10}{5}=\dfrac{-1+2\sqrt{6}}{5}\)

Lời giải:

a)

\(\frac{2A}{\sqrt{2}}=\frac{4+2\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{4-2\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}=\frac{3+1+2\sqrt{3}}{2+\sqrt{3+1+2\sqrt{3}}}+\frac{3+1-2\sqrt{3}}{2-\sqrt{3+1-2\sqrt{3}}}\)

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

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

$\Rightarrow A=\sqrt{2}$

b)

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

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

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

c)

\(C=\frac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{x^2-4x+4}}=\frac{\sqrt{(x-1)-2\sqrt{x-1}+1}+\sqrt{(x-1)+2\sqrt{x-1}+1}}{\sqrt{(x-2)^2}}\)

\(=\frac{\sqrt{(\sqrt{x-1}-1)^2}+\sqrt{(\sqrt{x-1}+1)^2}}{|x-2|}=\frac{|\sqrt{x-1}-1|+|\sqrt{x-1}+1|}{|x-2|}\)