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19 tháng 3 2020

a) \(9,2\cdot2\frac{1}{2}-\left(2\cdot0,125-1\frac{5}{12}\right):\frac{1}{4}\)

\(=23-\left(0,25-\frac{17}{12}\right):\frac{1}{4}\)

\(=23-\left(-\frac{7}{6}\right):\frac{1}{4}\)

\(=23-\left(-\frac{14}{3}\right)\)

\(=\frac{83}{3}\)

b) \(\frac{\sqrt{3^2}-\sqrt{39^2}}{\sqrt{59-10-\sqrt{91^2}}}\)

\(=\frac{3-39}{\sqrt{59-10-91}}\)

\(=\frac{-36}{\sqrt{-42}}\)

Vì -42 < 0 \(\Leftrightarrow\)Biểu thức không tồn tại

c) \(\frac{5}{18}-1,456:\frac{7}{25}+4,5\cdot\frac{4}{5}\)

\(=\frac{5}{18}-\frac{26}{5}+\frac{18}{5}\)

\(=\frac{5}{18}-\frac{8}{5}\)

\(=\frac{-119}{90}\)

19 tháng 3 2020

Thanks bạn nhá!

28 tháng 7 2016

\(\left(\sqrt{4,5}-\frac{1}{2}.\sqrt{72}+5\sqrt{\frac{1}{2}}\right).\left(42\sqrt{\frac{25}{6}}-10\sqrt{\frac{3}{2}}-12\sqrt{\frac{98}{3}}\right)\)

=\(\left(\frac{3\sqrt{2}}{2}-3\sqrt{2}+\frac{5\sqrt{2}}{2}\right).\left(35\sqrt{6}-5\sqrt{6}-28\sqrt{6}\right)\)

=\(\left(\frac{3\sqrt{2}-6\sqrt{2}+5\sqrt{2}}{2}\right).2\sqrt{6}\)

=\(2\sqrt{2}.\sqrt{6}=4\sqrt{3}\)

\(a,\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)

\(=\frac{-5}{9}.\frac{-1}{10}\)

\(=\frac{1}{18}\)

\(b,2^8:2^5+3^3.2-12\)

\(=2^3+9.2-12\)

\(=8+18-12\)

\(=26-12\)

\(=14\)

Câu c,d em chưa học nên không biết làm ạ, mong mọi người thông cảm!!!

Sửa lại câu b

\(=2^3+27.2-12\)

\(=8+54-12\)

\(=62-12\)

\(=50\)

21 tháng 7 2019

a) \(\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}\)

\(\frac{\left(10+2\sqrt{10}\right)\left(1-\sqrt{5}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(1-\sqrt{5}\right)}+\frac{8\left(\sqrt{5}+\sqrt{2}\right)}{\left(1-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)

\(\frac{\left(10+2\sqrt{10}\right)\left(1-\sqrt{5}\right)+8\left(\sqrt{5}+\sqrt{2}\right)}{\left(1-\sqrt{5}\right)\left(\sqrt{5}+\sqrt{2}\right)}\)

\(\frac{10-2\sqrt{5}+2\sqrt{10}-2\sqrt{2}}{\sqrt{5}+\sqrt{2}-5-\sqrt{10}}\)

\(\frac{2\left(5-\sqrt{5}+\sqrt{10}-\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}-5-\sqrt{10}}\)

= -2

b); c); d) làm tương tự

13 tháng 7 2016

a) Kết quả rút gọn xấu (+dài) nữa. (có thể đề sai)

b) 

\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)

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

\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=-\left(7-5\right)=-2\)

c) \(\frac{\sqrt{5-2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}=\frac{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)

\(=\frac{\sqrt{3}-\sqrt{2}+\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}=\frac{\left(\sqrt{5}-\sqrt{2}\right)^2}{3}\)

14 tháng 7 2016

a) \(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}}=\left[\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right].\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{6}}{2}-2\sqrt{6}\right).\frac{1}{\sqrt{6}}=\frac{1}{2}-2=-\frac{3}{2}\)

10 tháng 2 2020

\(a,\left(\frac{1}{\sqrt{625}}+\frac{1}{5}+1\right):\left(\frac{1}{25}-\frac{1}{\sqrt{25}}-1\right)\)

\(=\left(\frac{1}{25}+\frac{1}{5}+1\right):\left(\frac{1}{25}-\frac{1}{5}-1\right)\)

\(=\frac{31}{25}:\left(-\frac{29}{25}\right)\)

\(=\frac{31}{25}.\frac{-25}{29}\)

\(=-\frac{31}{29}\)

\(b,\left[18\frac{1}{6}-\left(0,06:7\frac{1}{2}+3\frac{2}{5}.0,38\right)\right]:\left(19-2\frac{2}{3}.4\frac{3}{4}\right)\)

\(=\left[\frac{109}{6}-\left(\frac{3}{50}:\frac{15}{2}+\frac{17}{5}.\frac{19}{50}\right)\right]:\left(19-\frac{8}{3}.\frac{19}{4}\right)\)

\(=\left(\frac{109}{6}-\frac{13}{10}\right):\frac{19}{3}\)

\(=\frac{253}{15}.\frac{3}{19}\)

\(=\frac{253}{95}\)

Số to :v

17 tháng 9 2019

\(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}=4\)

\(\Leftrightarrow\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+...+\sqrt{2}-\sqrt{1}=4\)

\(\Leftrightarrow\sqrt{25}-\sqrt{1}=4\Leftrightarrow5-1=4\)(đúng)

Vậy \(\frac{1}{\sqrt{25}+\sqrt{24}}+\frac{1}{\sqrt{24}+\sqrt{23}}+...+\frac{1}{\sqrt{2}+\sqrt{1}}=4\)(đpcm)

17 tháng 9 2019

\(M=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{11-6\sqrt{2}}}}\)

\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{2-6\sqrt{2}+9}}}\)

\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+\sqrt{\left(3-\sqrt{2}\right)^2}}}\)

\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{3+\sqrt{2}+3-\sqrt{2}}}\)

\(=\left(2\sqrt{2}\right)\sqrt{2+4\sqrt{6}}\)

\(=\sqrt{16+32\sqrt{6}}\)

a: \(=\dfrac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\dfrac{1}{\sqrt{6}}\)

\(=\dfrac{2\left(2\sqrt{2}-\sqrt{3}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{1}{\sqrt{6}}\)

\(=\dfrac{-\sqrt{6}}{3}-\dfrac{1}{\sqrt{6}}=\dfrac{-\sqrt{6}}{2}\)

b: \(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\sqrt{5}+2}\)

\(=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\sqrt{5}+2}=\dfrac{2\sqrt{5}+2}{2\sqrt{5}+2}=1\)

d: \(=-\left(\sqrt{5}+\sqrt{2}\right)\cdot\left(\sqrt{5}-\sqrt{2}\right)=-3\)