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A=(\(\sqrt{13}\).\(\sqrt{2}\)+5\(\sqrt{2}\))\(\sqrt{19-5\sqrt{13}}\)
=(\(\sqrt{13}\)+5)\(\sqrt{2}\). \(\sqrt{19-5\sqrt{13}}\)
=(\(\sqrt{13}\)+5) \(\sqrt{2\left(19-5\sqrt{13}\right)}\)
= (\(\sqrt{13}\)+5) \(\sqrt{38-2.5\sqrt{13}}\)
=(\(\sqrt{13}\)+5) \(\sqrt{5^2-2.5\sqrt{13}+13}\)
=(\(\sqrt{13}\)+5)\(\sqrt{\left(5-\sqrt{13}\right)^2}\)
=(\(\sqrt{13}\)+5) \(|5-\sqrt{13}|\)
=(5+\(\sqrt{13}\))(5-\(\sqrt{13}\))
= 25-13 = 12
a) Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)\cdot\sqrt{9+2\sqrt{14}}\)
\(=\left(\sqrt{7}-\sqrt{2}\right)\cdot\left(\sqrt{7}+\sqrt{2}\right)\)
=7-2
=5
d) Ta có: \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)
\(=2\sqrt{2}-\sqrt{7}+5\sqrt{7}-\dfrac{2\sqrt{2}\left(3-\sqrt{2}\right)}{3-\sqrt{2}}\)
\(=2\sqrt{2}+4\sqrt{7}-2\sqrt{2}\)
\(=4\sqrt{7}\)
a) Ta có: \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2\cdot\sqrt{20}\cdot3+9}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-2\cdot\sqrt{5}\cdot1+1}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
\(=\sqrt{1}=1\)
b) Ta có: \(\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}\)
\(=\sqrt{6+2\sqrt{5}-\sqrt{20-2\cdot2\sqrt{5}\cdot3+9}}\)
\(=\sqrt{6+2\sqrt{5}-\sqrt{\left(2\sqrt{5}-3\right)^2}}\)
\(=\sqrt{6+2\sqrt{5}-\left(2\sqrt{5}-3\right)}\)
\(=\sqrt{6+3}=3\)
c) Sửa đề: \(\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}\)
Ta có: \(\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}\)
\(=\sqrt{2+\sqrt{5+\sqrt{12-2\cdot2\sqrt{3}\cdot1+1}}}\)
\(=\sqrt{2+\sqrt{5+\sqrt{\left(2\sqrt{3}-1\right)^2}}}\)
\(=\sqrt{2+\sqrt{5+2\sqrt{3}-1}}\)
\(=\sqrt{2+\sqrt{3+2\sqrt{3}\cdot1+1}}\)
\(=\sqrt{2+\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\sqrt{3+\sqrt{3}}\)
d) Ta có: \(\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
\(=\dfrac{\left(6-2\sqrt{5}\right)\sqrt{6+2\sqrt{5}}+\left(6+2\sqrt{5}\right)\sqrt{6-2\sqrt{5}}}{2\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{5}-1\right)^2\cdot\left(\sqrt{5}+1\right)+\left(\sqrt{5}+1\right)^2\cdot\left(\sqrt{5}-1\right)}{2\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\left(\sqrt{5}-1+\sqrt{5}+1\right)}{2\sqrt{2}}\)
\(=\dfrac{4\cdot2\sqrt{5}}{2\sqrt{2}}\)
\(=\dfrac{8\sqrt{5}}{2\sqrt{2}}=\dfrac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right)\left(\sqrt{5}-\sqrt{2}\right)=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-3\)
\(=\left[\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right]\left(\sqrt{5}-\sqrt{2}\right)=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)=-3\)
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
a) \(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)
\(=-10\sqrt{2}+5.2-\left(18-30\sqrt{2}+25\right)\)
\(=-10\sqrt{2}+10-18+30\sqrt{2}-25\)
\(=20\sqrt{2}-33\)
b) câu b đề sai
1.\(5\sqrt{a}+6\sqrt{a.\frac{1}{4}}-\sqrt{a^2.\frac{4}{a}}+\sqrt{5}=5\sqrt{a}+6.\frac{1}{2}\sqrt{a}-2\sqrt{a}\)+\(\sqrt{5}\)
bạn tự làm nốt các câu này và làm tương tự các câu kia nhé!!Nếu khó chỗ nào hãy nhắn tin cho mk!! hihi
Đặt căn2 làm thừa số chung, nhân vào căn19-5căn13. Triệt căn thức, dùng hẳng đẳng thúc là được.
Kq:12