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NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 8 2021
\(\sqrt{17-12\sqrt{2}}+\sqrt{8}=\sqrt{\left(3-2\sqrt{2}\right)^2}+2\sqrt{2}=3-2\sqrt{2}+2\sqrt{2}=3\)
\(\sqrt{7+4\sqrt{21-12\sqrt{3}}}=\sqrt{7+4\sqrt{\left(2\sqrt{3}-3\right)^2}}=\sqrt{7+4\left(2\sqrt{3}-3\right)}\)
\(=\sqrt{8\sqrt{3}-5}\) (câu này đề bài bị nhầm dấu, phải là \(\sqrt{7-4\sqrt{21-12\sqrt{3}}}\) mới hợp lý)
\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
25 tháng 8 2021
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}+1\right)\cdot\dfrac{1}{2+\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}+1\right)\cdot\dfrac{1}{\sqrt{6}+2}\)
\(=\dfrac{\sqrt{6}+2}{2}\cdot\dfrac{1}{\sqrt{6}+2}\)
\(=\dfrac{1}{2}\)
25 tháng 10 2021
\(B=\dfrac{\sqrt{13}\left(\sqrt{2}-1\right)}{\sqrt{13}}-\dfrac{\sqrt{2}+1}{2-1}=\sqrt{2}-1-\sqrt{2}-1=-2\)
19 tháng 11 2021
\(C=\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}=\sqrt{16-7}=\sqrt{9}=3\)
5 tháng 10 2021
\(\dfrac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}=\dfrac{\sqrt{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}.\sqrt{3+\sqrt{5}}}{\sqrt{10}+\sqrt{2}}\)
\(=\dfrac{\sqrt{9-5}.\sqrt{3+\sqrt{5}}}{\sqrt{10}+\sqrt[]{2}}=\dfrac{\sqrt{12+4\sqrt{5}}}{\sqrt{10}+\sqrt{2}}\)
\(\dfrac{\sqrt{\left(\sqrt{10}+\sqrt{2}\right)^2}}{\sqrt{10}+\sqrt{2}}=\dfrac{\sqrt{10}+\sqrt{2}}{\sqrt{10}+\sqrt{2}}=1\)
5 tháng 10 2021
\(\dfrac{\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)
\(=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\left(\sqrt{5}+1\right)}\)
\(=\dfrac{2\sqrt{5}+2}{2\left(\sqrt{5}+1\right)}=1\)
27 tháng 9 2021
d: Ta có: \(5\sqrt{20}-3\sqrt{12}+15\cdot\sqrt{\dfrac{1}{5}}-4\sqrt{27}+\sqrt{\sqrt{5^2-4^2}}\)
\(=10\sqrt{5}-6\sqrt{3}+3\sqrt{5}-12\sqrt{3}+\sqrt{3}\)
\(=-17\sqrt{3}+13\sqrt{5}\)
e: Ta có: \(\sqrt{7+4\sqrt{3}}+\sqrt{28-10\sqrt{3}}\)
\(=2+\sqrt{3}+5-\sqrt{3}\)
=7
12 tháng 10 2021
\(=\dfrac{4\left(\sqrt{5}+\sqrt{2}\right)}{3}+\dfrac{3\left(\sqrt{5}+2\right)}{1}-\dfrac{2\left(\sqrt{3}+2\right)}{-1}+\dfrac{\sqrt{3}-1}{6}\\ =\dfrac{8\sqrt{5}+8\sqrt{2}+\sqrt{3}-1}{6}+3\sqrt{5}+6+2\sqrt{3}+4\\ =\dfrac{8\sqrt{5}+8\sqrt{2}+\sqrt{3}-1+18\sqrt{5}+60+12\sqrt{3}}{6}\\ =\dfrac{26\sqrt{5}+8\sqrt{2}+13\sqrt{3}+59}{6}\)
12 tháng 10 2021
\(=\dfrac{5\left(4+\sqrt{11}\right)}{16-11}+\dfrac{3-\sqrt{7}}{9-7}-\dfrac{6\left(\sqrt{7}+2\right)}{7-4}-\dfrac{\sqrt{7}-5}{2}\)
\(=4+\sqrt{11}-2\sqrt{7}-4+\dfrac{3-\sqrt{7}-\sqrt{7}+5}{2}\)
\(=\sqrt{11}-2\sqrt{7}+4-\sqrt{7}=4+\sqrt{11}-3\sqrt{7}\)
12 tháng 10 2021
\(=\dfrac{5\left(4+\sqrt{11}\right)}{5}+\dfrac{3-\sqrt{7}}{2}-\dfrac{6\left(\sqrt{7}+2\right)}{3}-\dfrac{\sqrt{7}-5}{2}\\ =4+\sqrt{11}+\dfrac{3-\sqrt{7}}{2}-\dfrac{\sqrt{7}-5}{2}-2\sqrt{7}-4\\ =\sqrt{11}-2\sqrt{7}+\dfrac{8-2\sqrt{7}}{2}\\ =\sqrt{11}-2\sqrt{7}+4-\sqrt{7}=4+\sqrt{11}-3\sqrt{7}\)
13 tháng 10 2021
c: \(\sqrt{3-\sqrt{5}}\cdot\sqrt{3+\sqrt{5}}=2\)
d: \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\cdot\left(\sqrt{6}+11\right)\)
\(=\left(-3+3\sqrt{6}+4+2\sqrt{6}-12-4\sqrt{6}\right)\cdot\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
=-115
Ta có \(\cot\alpha=\tan\beta\) ; \(\cos^2\alpha+\sin^2\alpha=1\)
Khi đó \(-\frac{\cot58^{\text{o}}+\tan27^{\text{o}}}{\cot63^{\text{o}}+\tan32^{\text{o}}}+1=\frac{-\cot58^{\text{o}}-\tan27^{\text{o}}+\cot63^{\text{o}}+\tan32^{\text{o}}}{\cot63^{\text{o}}+\tan32^{\text{o}}}\)
\(=\frac{\left(\tan32^{\text{o}}-\cot58^{\text{o}}\right)+\left(\cot63^{\text{o}}-\tan27^{\text{o}}\right)}{\cot63^{\text{o}}+\tan32^{\text{o}}}=0\)
=> \(\frac{\cot58^{\text{o}}+\tan27^{\text{o}}}{\cot63^{\text{o}}+\tan32^{\text{o}}}=1\)
=> \(\cos^255^{\text{o}}-\frac{\cot58^{\text{o}}+\tan27^{\text{o}}}{\cot63^{\text{o}}+\tan32^{\text{o}}}=\cos^255^{\text{o}}-1=-\sin^255\)