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Nhận xét: \(\left[\sqrt{n^2}\right]=n\); \(\left[\sqrt{a}\right]=n-1\) với (n - 1)2 < a < n2
=> \(\left[\sqrt{1}\right]+\left[\sqrt{2}\right]+\left[\sqrt{3}\right]=1+1+1=1.3\)
\(\left[\sqrt{4}\right]+...+\left[\sqrt{8}\right]=2.5\)
\(\left[\sqrt{9}\right]+...+\sqrt{15}=3.7\)
\(\left[\sqrt{16}\right]+...+\left[\sqrt{24}\right]=4.9\)
Tương tự, nhóm các số có phần nguyên là 5; 6; 7; 8 ;9 ; 10
=> B = 1.3 + 2.5 + 3.7 + 4.9 + 5.11 + 6.13 + 7 .15 + 8.17 + 9.19 + 10.21
B = 825
\(x=\left(1-\dfrac{1}{\sqrt{4}}\right).\left(1-\dfrac{1}{\sqrt{16}}\right).\left(1-\dfrac{1}{\sqrt{36}}\right).\left(1-\dfrac{1}{\sqrt{64}}\right).\left(1-\dfrac{1}{\sqrt{100}}\right)\)
\(x=\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{6}\right).\left(1-\dfrac{1}{8}\right).\left(1-\dfrac{1}{10}\right)\)
\(x=\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.\dfrac{7}{8}.\dfrac{9}{10}\)
\(x=\dfrac{63}{256}\)
và \(y=\sqrt{20+0,25}\)
\(y=\sqrt{20,25}\)
\(y=4,5\)
Do 4,5 > \(\dfrac{63}{256}\)
=> x<y
a)\(\dfrac{3}{4}-\dfrac{5}{2}-\dfrac{3}{5}=\dfrac{15}{20}-\dfrac{50}{20}-\dfrac{12}{20}=-\dfrac{47}{20}\)
b) \(\sqrt{7^2}+\sqrt{\dfrac{25}{16}-\dfrac{3}{2}}=7+\sqrt{\dfrac{1}{16}}=7+\dfrac{1}{4}=\dfrac{29}{4}\)
c) \(\dfrac{1}{2}.\sqrt{100}-\sqrt{\dfrac{1}{16}+\left(\dfrac{1}{3}\right)^0}=\dfrac{1}{2}.10-\sqrt{\dfrac{1}{16}+1}=5-\sqrt{\dfrac{17}{16}}\)
\(\left(\sqrt{100}-1\right).\left(\sqrt{100}-2\right).\left(\sqrt{100}-3\right)...\left(\sqrt{100}-55\right)\)
\(=\left(\sqrt{100}-1\right).\left(\sqrt{100}-2\right).\left(\sqrt{100}-3\right)...\left(\sqrt{100}-10\right)...\left(\sqrt{100}-55\right)\)
\(=\left(\sqrt{100}-1\right).\left(\sqrt{100}-2\right).\left(\sqrt{100}-3\right)...0...\left(\sqrt{100}-55\right)\)
\(=0\)