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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: \(=\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}-2-\sqrt{10}+3\sqrt{7}+2\)
\(=\sqrt{10}-\sqrt{10}+3\sqrt{7}=3\sqrt{7}\)
1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)
2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)
3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2}
\)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)
4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)
5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)
7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)
8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2
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: \(=\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)^2\cdot\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\left(5-2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)\)
\(=5\sqrt{3}-5\sqrt{2}-6\sqrt{2}+4\sqrt{3}=9\sqrt{3}-11\sqrt{2}\)
b: \(=\dfrac{\sqrt{2}}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}}{2-\sqrt{3}+1}\)
\(=\dfrac{\sqrt{2}}{3+\sqrt{3}}+\dfrac{\sqrt{2}}{3-\sqrt{3}}\)
\(=\dfrac{3\sqrt{2}-\sqrt{6}+3\sqrt{2}+\sqrt{6}}{9-3}=\dfrac{6\sqrt{2}}{6}=\sqrt{2}\)
d: \(=2\sqrt{2}-\sqrt{6}-3\sqrt{2}+\sqrt{6}=-\sqrt{2}\)
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
a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)
\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)
c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
\(1.\dfrac{6}{1-\sqrt{3}}-\dfrac{3\sqrt{3}+3}{\sqrt{3}+1}=\dfrac{6}{1-\sqrt{3}}-\dfrac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=\dfrac{6}{1-\sqrt{3}}-3=\dfrac{3+3\sqrt{3}}{1-\sqrt{3}}\) \(2.\dfrac{\sqrt{12}-6}{\sqrt{8}-\sqrt{24}}-\dfrac{3+\sqrt{3}}{\sqrt{3}}+\dfrac{4}{1-\sqrt{7}}=\dfrac{2\sqrt{3}-6}{2\sqrt{2}-2\sqrt{6}}-\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}}=\dfrac{2\sqrt{3}\left(1-\sqrt{3}\right)}{2\sqrt{2}\left(1-\sqrt{3}\right)}-\sqrt{3}-1=\dfrac{\sqrt{3}}{\sqrt{2}}-\sqrt{3}-1=\dfrac{\sqrt{3}-\sqrt{6}-\sqrt{2}}{\sqrt{2}}\) \(3.\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}=\left[\dfrac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\dfrac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right].\left(\sqrt{7}-\sqrt{5}\right)=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=-2\) \(4.\dfrac{\left(\sqrt{2}+1\right)^2-4\sqrt{2}}{\sqrt{2}-1}.\left(\sqrt{2}+1\right)=\dfrac{\left(2-2\sqrt{2}+1\right)\left(\sqrt{2}+1\right)}{\sqrt{2}-1}=\dfrac{\left(\sqrt{2}-1\right)^2\left(\sqrt{2}+1\right)}{\sqrt{2}-1}=1\)
bạn nên tự nghiên cứu rồi giải đi chứ bạn đưa 1 loạt thế thì ai rảnh mà giải, với lại cứ bài gì không biết chưa chịu suy nghĩ đã hỏi rồi thì tiến bộ sao được, đúng không