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Bài 2
\(P=\frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{12+2\sqrt{12}+1}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{12}-1}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{4-\sqrt{12}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{\left(\sqrt{3}-1\right)^2}}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{6}+\sqrt{2}}\)
\(=\frac{\sqrt{2}\cdot\sqrt{2}\cdot\sqrt{2+\sqrt{3}}}{\sqrt{2}\left(\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{2}\cdot\sqrt{4+2\sqrt{3}}}{\sqrt{2}\left(\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{3+2\sqrt{3}+1}}{\left(\sqrt{3}+1\right)}\)
=\(\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\left(\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{3}+1}{\left(\sqrt{3}+1\right)}=1\)
Vậy P là một số nguyên
NX \(A=\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}\)
\(A^2=1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}=\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}\)
\(=\frac{a^2\left(a^2+2a+1+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}\)=\(\frac{a^4+2a^3+2a^2+\left(a+1\right)^2}{a^2\left(a+1\right)^2}\)
\(=\frac{a^4+2a^2\left(a+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}=\frac{\left(a^2+a+1\right)^2}{a^2\left(a+1\right)^2}\)=\(\left[\frac{a^2+a+1}{a\left(a+1\right)}\right]^2\)suy ra A=\(\frac{a^2+a+1}{a\left(a+1\right)}\)
=\(\frac{a\left(a+1\right)+1}{a\left(a+1\right)}=1+\frac{1}{a\left(a+1\right)}=1+\frac{1}{a}-\frac{1}{a+1}\)
ap dung vao bai ta co =\(\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2012}-\frac{1}{2013}\right)\)
=\(2011+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)= \(2011+\frac{1}{2}-\frac{1}{2013}=2011,499503\)
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
\(P\sqrt{2}=\sqrt{2x-1+14\sqrt{2x-1}+49}+\sqrt{2x-1+6\sqrt{2x-1}+9}\)
\(=\sqrt{\left(\sqrt{2x-1}+7\right)^2}+\sqrt{\left(\sqrt{2x-1}+3\right)^2}\)
\(=\left|\sqrt{2x-1}+7\right|+\left|\sqrt{2x-1}+3\right|\)
\(=2\sqrt{2x-1}+10\)
Chỉ tính được đến đây, chắc bạn ghi nhầm đề, muốn ra số cụ thể thì trước \(7\sqrt{2x-1}\) hoặc \(3\sqrt{2x-1}\) phải là dấu "-" chứ ko thể là dấu "+"