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a) \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right):\sqrt{3}=7\sqrt{16}+3\sqrt{9}-2\sqrt{4}\)
\(=7.4+3.3-2.2=28+9-4=33\)
b) \(\left(\sqrt{125}+\sqrt{245}-\sqrt{5}\right):\sqrt{5}=\sqrt{25}+\sqrt{49}-1\)
\(=5+7-1=11\)
c) \(\left(\sqrt{\dfrac{1}{7}}-\sqrt{\dfrac{16}{7}}+\sqrt{7}\right):\sqrt{7}=\left(\dfrac{\sqrt{1}}{\sqrt{7}}-\dfrac{\sqrt{16}}{\sqrt{7}}+\sqrt{7}\right):\sqrt{7}\)
\(=\left(\dfrac{1}{\sqrt{7}}-\dfrac{4}{\sqrt{7}}+\sqrt{7}\right):\sqrt{7}=\dfrac{1}{\sqrt{7}.\sqrt{7}}-\dfrac{4}{\sqrt{7}.\sqrt{7}}+1\)
\(=\dfrac{1}{7}-\dfrac{4}{7}+1=\dfrac{1}{7}-\dfrac{4}{7}+\dfrac{7}{7}\Leftrightarrow\dfrac{1-4+7}{7}=\dfrac{4}{7}\)
bạn ghi rõ tại sao từ cái đề mà có ngay phép tính thứ hai cho mình với
1 , \(\left(\sqrt{12}-2\sqrt{75}\right).\sqrt{3}=\sqrt{12.3}-\sqrt{300.3}=6-30=-24\)
2 , \(\sqrt{3}.\left(\sqrt{12}.\sqrt{27}-\sqrt{3}\right)=\sqrt{12.27.3}-\sqrt{3.3}=18\sqrt{3}-3\)
3 , \(\left(7\sqrt{48}+3\sqrt{27}-\sqrt{12}\right):\sqrt{3}=\left(28\sqrt{3}+9\sqrt{3}-2\sqrt{3}\right):\sqrt{3}=35\)
4 , bạn làm tương tự nhé
5 , bạn làm tương tự nhé
6 , bạn làm tương tự nhé
a: \(=\sqrt{7+\sqrt{45}}-\sqrt{7-\sqrt{45}}\)
\(=\dfrac{\sqrt{14+2\sqrt{45}}-\sqrt{14-2\sqrt{45}}}{\sqrt{2}}\)
\(=\dfrac{3+\sqrt{5}-3+\sqrt{5}}{\sqrt{2}}=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)
b: \(=2\cdot\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=\sqrt{\sqrt{3}}\left(2\cdot4\sqrt{5}-2\sqrt{5}-3\cdot2\sqrt{5}\right)\)
\(=\sqrt{\sqrt{3}}\cdot0=0\)
c: \(=\left(2-\sqrt{3}-6+2\sqrt{3}+8+2\sqrt{3}\right)\left(4-3\sqrt{3}\right)\)
\(=\left(4+3\sqrt{3}\right)\left(4-3\sqrt{3}\right)\)
=16-27=-11
a: \(=\left(-\sqrt{5}-\sqrt{7}\right)\cdot\left(\sqrt{7}-\sqrt{5}\right)\)
\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=-2
b: \(=\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}\)
c: \(=\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}-2-\sqrt{10}+3\sqrt{7}+2\)
\(=\sqrt{10}-\sqrt{10}+3\sqrt{7}=3\sqrt{7}\)
a: \(=2\cdot\dfrac{4\sqrt{3}}{5}+3\cdot\dfrac{3\sqrt{3}}{7}-\dfrac{3\sqrt{3}}{2}\)
\(=\dfrac{8\sqrt{3}}{5}+\dfrac{9\sqrt{3}}{7}-\dfrac{3\sqrt{3}}{2}\)
\(=\dfrac{112\sqrt{3}+90\sqrt{3}-105\sqrt{3}}{70}=\dfrac{97\sqrt{3}}{70}\)
b: \(\left(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}\right)^2\)
\(=3-\sqrt{5}+3+\sqrt{5}-2\cdot\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\)
\(=6-2\cdot\sqrt{4}=6-2\cdot2=2\)
c: \(=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}\)
\(=-\dfrac{2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)