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B1.
\(A=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{12+4\sqrt{3}+1}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{2+\sqrt{3}}}{\sqrt{2}\left(\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{3+2\sqrt{3}+1}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{3}+1}\)
\(=\frac{\sqrt{3}+1}{\sqrt{3}+1}=1\)
Vậy ...
Ta có: M= \(\frac{1+2x}{1+\sqrt{1+2x}}+\frac{1-2x}{1-\sqrt{1-2x}}\)= \(\frac{\left(1+2x\right)\left(1-\sqrt{1+2x}\right)+\left(1-2x\right)\left(1+\sqrt{1+2x}\right)}{1-\left(1-2x\right)}\)=\(\frac{1-\sqrt{1+2x}+2x-2x\sqrt{1+2x}+1+\sqrt{1+2x}-2x-2x\sqrt{1+2x}}{2x}\)
=\(\frac{2}{2x}=\frac{1}{x}\)
Với x=\(\frac{\sqrt{3}}{4}\)=> M=\(\frac{4}{\sqrt{3}}\)
Ta có \(\sqrt{\left(1+2x\right)^2}\)= 1 + 2x (1)
+ \(\sqrt{\left(1-2x\right)^2}\)= 1 - 2x (2)
(1) +(2) = 2
Có \(\sqrt{1+2x}.\sqrt{1-2x}\)= \(\sqrt{1-4x^2}=\frac{1}{2}\) (3)
Từ (1),(2),(3) \(\Rightarrow\)\(\left(\sqrt{1+2x}+\sqrt{1-2x}\right)^2\)= 3 \(\Rightarrow\)\(\sqrt{1+2x}+\sqrt{1-2x}\)=\(\sqrt{3}\) (4)
\(\left(\sqrt{1+2x}-\sqrt{1-2x}\right)^2\)= 1 \(\Rightarrow\) \(\sqrt{1+2x}-\sqrt{1-2x}\)= 1 (5)
Có M= \(\frac{\left(1+2x\right).\left(1-\sqrt{1-2x}\right)+\left(1-2x\right).\left(1+\sqrt{1+2x}\right)}{\left(1+\sqrt{1+2x}\right).\left(1-\sqrt{1-2x}\right)}\)
Xét TS= \(1-\sqrt{1-2x}+2x-2x.\sqrt{1-2x}+1+\sqrt{1+2x}-2x-2x.\sqrt{1+2x}\)
= 2+ \(\sqrt{1+2x}-\sqrt{1-2x}\)- 2x\(\left(\sqrt{1+2x}+\sqrt{1-2x}\right)\)
Thay (4), (5) và x vào TS ta có TS= \(2+1-2.\frac{\sqrt{3}}{4}.\sqrt{3}=\frac{3}{2}\) (6)
Xét MS=\(1-\sqrt{1-2x}+\sqrt{1+2x}-\sqrt{1-4x^2}\)
Thay (5) và x vào MS ta có MS= \(1+1-\frac{1}{2}\)=\(\frac{3}{2}\) (7)
Từ (6),(7) ta có giá trị của M= 1
Ta có: \(Q=\frac{1-2x}{1-\sqrt{1-2x}}+\frac{1-2x}{1-\sqrt{1-2x}}\)
\(=\frac{1-2x}{1-\sqrt{1^2-2.x.1+\left(\sqrt{x}\right)^2}}+\frac{1-2x}{1-\sqrt{1-2.x.1+\left(\sqrt{x}\right)^2}}\)
\(=\frac{1-2x}{1-\sqrt{\left(1-\sqrt{x}\right)^2}}+\frac{1-2x}{1-\sqrt{\left(1-\sqrt{x}\right)^2}}\)
\(=\frac{1-2x}{1-\left|1-\sqrt{x}\right|}+\frac{1-2x}{1-\left|1-\sqrt{x}\right|}\)
\(=\frac{1-2x}{\sqrt{x}}+\frac{1-2x}{\sqrt{x}}=\left(\frac{1-2x}{\sqrt{x}}\right)^2\)
Thế \(x=\frac{\sqrt{3}}{4}\) ta được: \(Q=\left(\frac{1-2.\frac{\sqrt{3}}{4}}{\sqrt{\frac{\sqrt{3}}{4}}}\right)^2=0,04145188433\)