\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-\frac{64}{34}+\frac{14}{21}=\left(\frac{15}{34}+\frac{9}{34}-\frac{64}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)=\frac{30}{34}+\frac{21}{21}=\frac{15}{17}+1=\frac{32}{17}\)

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 \(2>\sqrt{3}\))

\(=6-3\sqrt{3}\)

Bài 1:

a) \(0,5-\frac{5}{41}+\frac{1}{2}-\frac{36}{41}\)

\(=\frac{1}{2}-\frac{5}{41}+\frac{1}{2}-\frac{36}{41}\)

\(=\left(\frac{1}{2}+\frac{1}{2}\right)-\left(\frac{5}{41}+\frac{36}{41}\right)\)

\(=1-1\)

\(=0.\)

b) \(\left(-\frac{2}{3}+\frac{3}{7}\right):\frac{4}{5}+\left(-\frac{1}{3}+\frac{4}{7}\right):\frac{4}{5}\)

\(=-\frac{2}{3}+\frac{3}{7}:\frac{4}{5}-\frac{1}{3}+\frac{4}{7}:\frac{4}{5}\)

\(=\left[\left(-\frac{2}{3}\right)-\frac{1}{3}\right]+\left(\frac{3}{7}+\frac{4}{7}\right):\frac{4}{5}\)

\(=\left(-1\right)+1:\frac{4}{5}\)

\(=\left(-1\right)+\frac{5}{4}\)

\(=\frac{1}{4}.\)

c) \(\left(-\frac{3}{4}\right).\sqrt{\frac{16}{9}+3.\sqrt{49}}\)

\(=\left(-\frac{3}{4}\right).\sqrt{\frac{16}{9}+3.7}\)

\(=\left(-\frac{3}{4}\right).\sqrt{\frac{16}{9}+21}\)

\(=\left(-\frac{3}{4}\right).\sqrt{\frac{205}{9}}\)

\(=\left(-\frac{3}{4}\right).\frac{\sqrt{205}}{3}\)

\(=-\frac{\sqrt{205}}{4}.\)

d) \(\left(-\frac{1}{3}\right)^2.\frac{4}{11}+1\frac{5}{11}.\left(\frac{1}{3}\right)^2\)

\(=\frac{1}{9}.\frac{4}{11}+\frac{16}{11}.\frac{1}{9}\)

\(=\frac{1}{9}.\left(\frac{4}{11}+\frac{16}{11}\right)\)

\(=\frac{1}{9}.\frac{20}{11}\)

\(=\frac{20}{99}.\)

Chúc bạn học tốt!

Em chỉ làm những bài e biết thôi, thông cảm nhs :D

a/ chịu

b/ \(C=1+7+7^2+.........+7^{50}\)

\(\Leftrightarrow7C=7+7^2+...........+7^{50}+7^{51}\)

\(\Leftrightarrow7C-C=\left(7+7^2+.......+7^{51}\right)-\left(1+7+.....+7^{50}\right)\)

\(\Leftrightarrow6C=7^{51}-1\)

\(\Leftrightarrow C=\dfrac{7^{51}-1}{6}\)

c/ \(A=\dfrac{-1}{4}+\dfrac{7}{3}+\dfrac{3}{4}+\dfrac{9}{2}\)

\(=\left(\dfrac{-1}{4}+\dfrac{3}{4}\right)+\left(\dfrac{7}{3}+\dfrac{9}{2}\right)\)

\(=\dfrac{1}{4}+\dfrac{41}{6}\)

\(=\dfrac{85}{12}\)

d/ Thấy phép tính hơi dài

e/ \(C=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+.........+\dfrac{1}{2015.2016.2017}\)

\(\Leftrightarrow2C=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+.........+\dfrac{2}{2015.2016.2017}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+.......+\dfrac{1}{2015.2016}-\dfrac{1}{2016.2017}\)

\(=\dfrac{1}{1.2}-\dfrac{1}{2016.2017}\)

\(=\dfrac{1}{2}-\dfrac{1}{4066272}\)

\(=\dfrac{2033136}{4066272}\)

\(\Leftrightarrow C=\dfrac{2033136}{4066272}:2\)

\(\Leftrightarrow C=?\)