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Ta có: A = \(\left|\frac{4}{9}-\left(\frac{\sqrt{2}}{2}\right)^2\right|+\left|0,\left(4\right)+\frac{\frac{1}{3}-\frac{2}{5}-\frac{3}{7}}{\frac{2}{3}-\frac{4}{5}-\frac{6}{7}}\right|\)
= \(\left|\frac{4}{7}-\frac{\sqrt{2}^2}{2^2}\right|+\left|0,\left(1\right).4+\frac{\frac{1}{3}-\frac{2}{5}-\frac{3}{7}}{2\left(\frac{1}{3}-\frac{2}{5}-\frac{3}{7}\right)}\right|\)
= \(\left|\frac{4}{7}-\frac{1}{2}\right|+\left|\frac{1}{9}.4+\frac{1}{2}\right|\)
= \(\left|\frac{8-7}{14}\right|+\left|\frac{8+9}{18}\right|\)
= \(\left|\frac{1}{14}\right|+\left|\frac{17}{18}\right|\)
= 1/14 + 17/18 = 64/63
A = \(\left|\frac{4}{9}-\left(\frac{\sqrt{2}}{2}\right)^2\right|+\left|0,\left(4\right)+\frac{\frac{1}{3}-\frac{2}{5}-\frac{3}{7}}{\frac{2}{3}-\frac{4}{5}-\frac{6}{7}}\right|\)
= \(\left|\frac{4}{9}-\left(\frac{\sqrt{2}^2}{2^2}\right)\right|+\left|0,\left(1\right).4+\frac{\frac{1}{3}-\frac{2}{5}-\frac{3}{7}}{2.\left(\frac{1}{3}-\frac{2}{5}-\frac{3}{7}\right)}\right|\)
= \(\left|\frac{4}{9}-\frac{1}{2}\right|+\left|\frac{1}{9}.4+\frac{1}{2}\right|\)
= \(\left|\frac{8-9}{18}\right|+\left|\frac{4}{9}+\frac{1}{2}\right|\)
= \(\left|-\frac{1}{18}\right|+\left|\frac{8+9}{18}\right|\)
= \(\frac{1}{18}+\frac{17}{18}=1\)
\(\frac{3}{4}+\frac{1}{4}:\left(-\frac{2}{3}\right)-\left(-5\right)\)
\(=\frac{3}{4}+\frac{1}{4}.\left(-\frac{3}{2}\right)+5\)
\(=\frac{3}{4}-\frac{3}{8}+5\)
\(=\frac{3}{8}+5=\frac{43}{8}\)
\(12.\left(\frac{2}{5}-\frac{5}{6}\right)^2=12.\left(-\frac{13}{30}\right)^2=12.\frac{169}{900}=\frac{169}{75}\)
\(\left(-2\right)^2+\sqrt{36}-\sqrt{9}+\sqrt{25}=4+6-3+5=12\)
\(\left(9\frac{3}{4}:3.4.2\frac{7}{34}\right):\left(-1\frac{9}{16}\right)=\left(\frac{39}{4}:3.4.\frac{75}{34}\right):\left(-\frac{25}{16}\right)=\frac{975}{34}.\left(-\frac{16}{25}\right)=-\frac{312}{17}\)
\(\frac{\sqrt{3^2}+\sqrt{39^2}}{\sqrt{91^2}-\sqrt{\left(-7\right)^2}}=\frac{3+39}{91-7}=\frac{42}{84}=\frac{1}{2}\)
\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-\frac{64}{34}+\frac{14}{21}=\left(\frac{15}{34}+\frac{9}{34}-\frac{64}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)=\frac{30}{34}+\frac{21}{21}=\frac{15}{17}+1=\frac{32}{17}\)
a) Ta có: \(A=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\left(\sqrt{9}-\sqrt{4}\right)\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)
\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\)(Vì \(\sqrt{5}>\sqrt{3}\))
\(=5-3-\sqrt{5}\)
\(=2-\sqrt{5}\)
b) Ta có: \(B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)\)
\(=\left(\frac{\sqrt{3}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}+\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}+\sqrt{6}\right)\)
\(=\sqrt{3}+\sqrt{3}+\sqrt{6}\)
\(=2\sqrt{3}+\sqrt{6}\)
c) Ta có: \(C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}}+\sqrt{3}\right):\sqrt{3}\)
\(=2\sqrt{3}+\sqrt{4-2\cdot2\cdot\sqrt{3}+3}+\sqrt{\frac{1}{3}:3}-\sqrt{\frac{4}{3}:3}+\sqrt{3:3}\)
\(=2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\frac{1}{9}}-\sqrt{\frac{4}{9}}+\sqrt{1}\)
\(=2\sqrt{3}+\left|2-\sqrt{3}\right|+\frac{1}{3}-\frac{2}{3}+1\)
\(=2\sqrt{3}+2-\sqrt{3}+\frac{2}{3}\)(Vì \(2>\sqrt{3}\))
\(=\sqrt{3}+\frac{8}{3}\)
d) Ta có: \(D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}\)
\(=\left(\frac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\right)\cdot\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\)
\(=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\cdot\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\frac{60}{20}\cdot\left|2-\sqrt{3}\right|\)
\(=3\cdot\left(2-\sqrt{3}\right)\)(Vì \(2>\sqrt{3}\))
\(=6-3\sqrt{3}\)
Ta có từ n3 + 1 đến (n + 1)3 - 1 có
(n + 1)3 - 1 - n3 - 1 + 1 = 3n2 + 3n số có phần nguyên bằng n
Áp dụng vào cái ban đầu ta có
\(=\frac{3.1^2+3.1}{1}+\frac{3.2^2+3.2}{2}+...+\frac{3.2011^2+3.2011}{2011}\)
= 3.1 + 3 + 3.2 + 3 + ...+ 3.2011 + 3
= 3.2011 + 3(1 + 2 +...+ 2011)
= 6075231
