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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
\(\sqrt{3}+2+\sqrt{7-4\sqrt{3}}=\sqrt{3}+2+\sqrt{4-2.2\sqrt{3}+3}\)
=\(\sqrt{3}+2+\sqrt{\left(2-\sqrt{3}\right)^2}=\sqrt{3}+2+2-\sqrt{3}=4\)
=>ĐPCM
Ap dung \(\frac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Ta co \(P< 2\left(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2007}}-\frac{1}{\sqrt{2008}}\right)\)
=> \(P< 2\left(1-\frac{1}{\sqrt{2008}}\right)< 2.1=2\)
Suy ra P khong phai so nguyen to
a, A=15√x−11x+2√x−3+3√x−21−√x−2√x+3√x+3A=15x−11x+2x−3+3x−21−x−2x+3x+3
=15√x−11x−√x+3√x−3−3√x−2√x−1−2√x+3√x+3=15x−11x−x+3x−3−3x−2x−1−2x+3x+3
=15√x−11√x(√x−1)+3(√x−1)−3√x−2√x−1−2√x+3√x+3=15x−11x(x−1)+3(x−1)−3x−2x−1−2x+3x+3
=15√x−11(√x−1)(√x+3)−3√x−2√x−1−2√x+3√x+3=15x−11(x−1)(x+3)−3x−2x−1−2x+3x+3
=15√x−11−(3√x−2)(√x+3)−(2√x+3)(√x−1)(√x−1)(√x+3)=15x−11−(3x−2)(x+3)−(2x+3)(x−1)(x−1)(x+3)
=15√x−11−(3x+9√x−2√x−6)−(2x−2√x+3√x−3)(√x−1)(√x+3)=15x−11−(3x+9x−2x−6)−(2x−2x+3x−3)(x−1)(x+3)
=15√x−11−3x−9√x+2√x+6−2x+2√x−3√x+3(√x−1)(√x+3)=15x−11−3x−9x+2x+6−2x+2x−3x+3(x−1)(x+3)
=7√x−5x−8(√x−1)(√x+3)
x02 = 8 - ( \(2\sqrt{2+\sqrt{3}}\)+ \(2\sqrt{6-3\sqrt{3}}\)) (1)
Ta có ( \(2\sqrt{2+\sqrt{3}}\)+ \(2\sqrt{6-3\sqrt{3}}\))2 = 32
Do đó x02 = 8 - \(\sqrt{32}\)(2)
PT <=> (x2 - 8)2 - 32 = 0 (3)
Thế (2) vào (3) thì đúng
Vậy x0 là nghiệm của PT
\(\sqrt{\sqrt{3}+2+\sqrt{7-4\sqrt{3}}}=\sqrt{\sqrt{3}+2+2-\sqrt{3}}=\sqrt{4}=2\)LÀ MỘT SỐ NGUYÊN
\(\sqrt{\sqrt{3}+2+\left|2\right|-\sqrt{3}}\)
<=>4 là số nguyên => t là số nguyên