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a) Ta có:
5√15+12√20+√5515+1220+5
=√52.15+√(12)2.20+√5=√25.15+√14.20+√5=√255+√204+√5=√5+√5+√5=(1+1+1)√5=3√5=52.15+(12)2.20+5=25.15+14.20+5=255+204+5=5+5+5=(1+1+1)5=35
b) Ta có:
√12+√4,5+√12,512+4,5+12,5
=√12+√92+√252=√12+√9.12+√25.12=√12+√32.12+√52.12=√12+3√12+5√12=(1+3+5).√12=9√12=91√2=9.√22=9√22=12+92+252=12+9.12+25.12=12+32.12+52.12=12+312+512=(1+3+5).12=912=912=9.22=922
c) Ta có:
√20−√45+3√18+√72=√4.5−√9.5+3√9.2+√36.2=√22.5−√32.5+3√32.2+√62.2=2√5−3√5+3.3√2+6√2=2√5−3√5+9√2+6√2=(2√5−3√5)+(9√2+6√2)=(2−3)√5+(9+6)√2=−√5+15√2=15√2−√520−45+318+72=4.5−9.5+39.2+36.2=22.5−32.5+332.2+62.2=25−35+3.32+62=25−35+92+62=(25−35)+(92+62)=(2−3)5+(9+6)2=−5+152=152−5
d) Ta có:
0,1√200+2√0,08+0,4.√50=0,1√100.2+2√0,04.2+0,4√25.2=0,1√102.2+2√0,22.2+0,4√52.2=0,1.10√2+2.0,2√2+0,4.5√2=1√2+0,4√2+2√2=(1+0,4+2)√2=3,4√2
a)
\(A=\sqrt{26+15\sqrt{3}}=\sqrt{\frac{52+30\sqrt{3}}{2}}=\sqrt{\frac{27+25+2\sqrt{27.25}}{2}}\)
\(=\sqrt{\frac{(\sqrt{27}+\sqrt{25})^2}{2}}=\frac{\sqrt{27}+\sqrt{25}}{\sqrt{2}}=\frac{3\sqrt{3}+5}{\sqrt{2}}=\frac{3\sqrt{6}+5\sqrt{2}}{2}\)
b)
\(B\sqrt{2}=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)
\(=\sqrt{7+1+2\sqrt{7}}-\sqrt{7+1-2\sqrt{7}}-2\)
\(=\sqrt{(\sqrt{7}+1)^2}-\sqrt{(\sqrt{7}-1)^2}-2=\sqrt{7}+1-(\sqrt{7}-1)-2=0\)
\(\Rightarrow B=0\)
c)
\(C=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{3+5-2\sqrt{3.5}}-\sqrt{3+5+2\sqrt{3.5}}\)
\(=\sqrt{(\sqrt{5}-\sqrt{3})^2}-\sqrt{(\sqrt{5}+\sqrt{3})^2}=(\sqrt{5}-\sqrt{3})-(\sqrt{5}+\sqrt{3})=-2\sqrt{3}\)
d)
\(D=(\sqrt{6}-2)(5+2\sqrt{6})\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})(2+3+2\sqrt{2.3})\sqrt{2+3-2\sqrt{2.3}}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})^2\sqrt{(\sqrt{3}-\sqrt{2})^2}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})^2(\sqrt{3}+\sqrt{2})^2=\sqrt{2}[(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})]^2\)
\(=\sqrt{2}.1^2=\sqrt{2}\)
e)
\(E=(\sqrt{10}-\sqrt{2})\sqrt{3+\sqrt{5}}=(\sqrt{5}-1).\sqrt{2}.\sqrt{3+\sqrt{5}}\)
\(=(\sqrt{5}-1)\sqrt{6+2\sqrt{5}}=(\sqrt{5}-1)\sqrt{5+1+2\sqrt{5.1}}\)
\(=(\sqrt{5}-1)\sqrt{(\sqrt{5}+1)^2}=(\sqrt{5}-1)(\sqrt{5}+1)=4\)
f)
\(F=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20+9-2\sqrt{20.9}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{(\sqrt{20}-3)^2}}}=\sqrt{\sqrt{5}-\sqrt{3-(\sqrt{20}-3)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5+1-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{(\sqrt{5}-1)^2}}=\sqrt{\sqrt{5}-(\sqrt{5}-1)}=\sqrt{1}=1\)
Câu 1
a: \(=\dfrac{\sqrt{2}}{4}\)
b: \(=\sqrt{\dfrac{6}{100}}=\dfrac{\sqrt{6}}{10}\)
d: \(=\dfrac{7}{3\sqrt{3}}=\dfrac{7\sqrt{3}}{9}\)
Câu 2:
a: \(=\sqrt{\dfrac{1}{\sqrt{3}}}=\sqrt{\dfrac{\sqrt{3}}{3}}\)
b: \(=\sqrt{\dfrac{3\sqrt{3}+9}{6}}\)
c: \(=\dfrac{\sqrt{4}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}=2\)
d: \(=\dfrac{5\sqrt{2}}{6}\)
Em thử nhá, ko chắc đâu
1) \(\frac{2}{\sqrt{20}}=\frac{2\sqrt{20}}{20}\) 2) \(\frac{4}{\sqrt{8}}=\frac{4\sqrt{8}}{8}\)
3) \(\frac{2+\sqrt{3}}{\sqrt{2}}=\frac{2\sqrt{2}+\sqrt{6}}{2}\) 4) \(\frac{1}{\sqrt{6}-2}=\frac{\sqrt{6}+2}{6-4}=\frac{\sqrt{6}+2}{2}\)
5) \(\frac{1}{\sqrt{2}-\sqrt{3}}=\frac{\sqrt{2}+\sqrt{3}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}=-\left(\sqrt{2}+\sqrt{3}\right)\)
6) \(\frac{9a-b}{3\sqrt{a}-\sqrt{b}}=\frac{\left(9a-b\right)\left(3\sqrt{a}+b\right)}{\left(3\sqrt{a}-\sqrt{b}\right)\left(3\sqrt{a}+\sqrt{b}\right)}=\left(3\sqrt{a}+b\right)\)
7) + 8) em chưa nghĩ ra
ong tth :v
\(\frac{2}{\sqrt{20}}=\frac{\sqrt{4}}{\sqrt{4}.\sqrt{5}}=\frac{1}{\sqrt{5}}\)
\(\frac{4}{\sqrt{8}}=\frac{\sqrt{16}}{\sqrt{8}}=\sqrt{2}\)
\(\frac{2+\sqrt{3}}{\sqrt{2}}=\sqrt{2}+\frac{\sqrt{3}}{\sqrt{2}}=\sqrt{2}+\sqrt{1,5}\)
\(\frac{1}{\sqrt{6}-2}=\frac{\sqrt{6}+2}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}=\frac{\sqrt{6}+2}{2}\)
\(\frac{1}{\sqrt{2}-\sqrt{3}}=\frac{\sqrt{3}+\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{3}\right)}=\frac{\sqrt{3}+\sqrt{2}}{-1}=-\sqrt{3}-\sqrt{2}\)
7: chưa
8: chưa
9:\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(\sqrt{2}+\sqrt{3}+2\right)+\left(2+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=1+\sqrt{2}\)