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1/\(\sqrt{\frac{4}{5}}\)+\(\sqrt{\frac{1}{2}}\)
=\(\sqrt{\frac{4.5}{5.5}}\)+\(\sqrt{\frac{1.2}{2.2}}\)
= \(5.2.\sqrt{5}\)+\(2\sqrt{2}\)
=\(10\sqrt{5}+2\sqrt{2}\)
2.
\(\sqrt{\frac{1}{12}}\)+\(\sqrt{\frac{1}{3}}\)
=\(\sqrt{\frac{1.12}{12.12}}\)+\(\sqrt{\frac{1.3}{3.3}}\)
=\(12.2\sqrt{3}\)+\(3\sqrt{3}\)
=\(\sqrt{3}\left(24+3\right)\)
=\(27\sqrt{3}\)
a,
\(\frac{\sqrt{6}\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)}+\sqrt{\frac{\left(2-\sqrt{2}\right)^2}{\left(2+\sqrt{2}\right).\left(2-\sqrt{2}\right)}}\)
=\(\sqrt{2}+\frac{2-\sqrt{2}}{\sqrt{2}}\)
=\(\sqrt{2}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}}\)
=\(\sqrt{2}+\sqrt{2}-1\)
=\(2\sqrt{2}-1\)
còn tiếp
b=,\(\frac{6\sqrt{3}}{3}-\frac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}-\frac{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{2}-\sqrt{3}}\)
=\(6-1+\sqrt{3}-\sqrt{6}\)
=\(5+\sqrt{3}+\sqrt{6}\)
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) \(\frac{3}{2+\sqrt{3}}+\frac{13}{4-\sqrt{3}}+\frac{6}{\sqrt{3}}\)
\(=\frac{3\left(2-\sqrt{3}\right)}{2^2-3}+\frac{13\left(4+\sqrt{3}\right)}{4^2-3}+\frac{6\sqrt{3}}{3}\)
\(=3\left(2-\sqrt{3}\right)+\left(4+\sqrt{3}\right)+2\sqrt{3}\)
\(=3.2+4=6+4=10\)
b) \(=\left[\frac{\left(\sqrt{14}-\sqrt{7}\right)\left(\sqrt{2}+1\right)}{2-1}+\frac{\left(\sqrt{15}-\sqrt{5}\right)\left(\sqrt{3}+1\right)}{3-1}\right]:\frac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\) (nhân bung mấy cái trong ngoặc vuông ra, rút gọn)
c) Gợi ý: \(28-10\sqrt{3}=5^2-2.5.\sqrt{3}+\sqrt{3}=\left(5-\sqrt{3}\right)^2\)
d) \(=\frac{3\left(3-2\sqrt{3}\right)}{3^2-\left(2\sqrt{3}\right)^2}+\frac{3\left(3+2\sqrt{3}\right)}{3^2-\left(2\sqrt{3}\right)^2}=-6\)
e) Tự làm.
1/\(\sqrt{8-2\sqrt{15}}-\sqrt{21-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(2\sqrt{5}-1\right)^2}\)
Bạn tự làm tiếp
2/ \(\frac{4}{\sqrt{7-4\sqrt{3}}}-\frac{4}{7-4\sqrt{3}}=\frac{4}{\sqrt{\left(2-\sqrt{3}\right)^2}}-\frac{4}{\left(2-\sqrt{3}\right)^2}=\frac{4}{2-\sqrt{3}}-\frac{4}{\left(2-\sqrt{3}\right)^2}\)
\(=\frac{8-4\sqrt{3}-4}{\left(2-\sqrt{3}\right)^2}=\frac{4-4\sqrt{3}}{\left(2-\sqrt{3}\right)^2}\) đến đây ko rút gọn được nữa, nghi bạn chép sai đề.
Tử số của phân số thứ hai là 4 hay 1 vậy?
3/ \(\frac{\sqrt{8+2\sqrt{15}}-\sqrt{4-2\sqrt{3}}}{\sqrt{6-2\sqrt{5}}}=\frac{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{\left(\sqrt{5}-1\right)^2}}=\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{3+\sqrt{5}}{2}\)
4/ \(\frac{10}{\sqrt{\left(\sqrt{5}-2\right)^2}}-\frac{12}{\sqrt{\left(3+\sqrt{5}\right)^2}}+\frac{20}{\sqrt{\left(\sqrt{5}-1\right)^2}}=\frac{10}{\sqrt{5}-2}-\frac{12}{3+\sqrt{5}}+\frac{20}{\sqrt{5}-1}\)
\(=\frac{10\left(\sqrt{5}+2\right)}{1}-\frac{12\left(3-\sqrt{5}\right)}{4}+\frac{20\left(\sqrt{5}+1\right)}{4}=16+18\sqrt{5}\)
\(\frac{10}{\sqrt{5}-2.\sqrt{5}.2+4}-\frac{12}{\sqrt{\sqrt{5}+2.\sqrt{5}.3+9}}+\frac{20}{\sqrt{5-2.\sqrt{5}.1+1}}=\frac{10}{\left(\sqrt{5}-2\right)^2}-\frac{12}{\sqrt{\left(\sqrt{5}+3\right)^2}}+\frac{20}{\sqrt{\left(\sqrt{5}-1\right)^2}}=\frac{10}{\sqrt{5}-2}-\frac{12}{\sqrt{5}+3}+\frac{20}{\sqrt{5}-1}=\frac{10\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right).\left(\sqrt{5}+2\right)}-\frac{12.\left(\sqrt{5}-3\right)}{\left(\sqrt{5}+3\right).\sqrt{5}-3\left(\right)}+\frac{20.\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right).\left(\sqrt{5}+1\right)}=\frac{10\sqrt{5}-20}{5-4}-\frac{12\sqrt{5}-36}{5-9}+\frac{20\sqrt{5}+20}{5-1}\\=\frac{40\sqrt{5}-80+12\sqrt{5}+36+20\sqrt{5}+20}{4}=\\ 18\sqrt{5}-6\)
1.
\(\sqrt{\frac{2+\sqrt{3}}{2}}\\ =\frac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}\\ =\frac{\sqrt{4+2\sqrt{3}}}{2}\\ =\frac{\sqrt{\left(1+\sqrt{3}\right)^2}}{2}\\ =\frac{1+\sqrt{3}}{2}\)
2.
\(\sqrt{\frac{14+5\sqrt{3}}{2}}\\ =\frac{\sqrt{14+5\sqrt{3}}}{\sqrt{2}}\\ =\frac{\sqrt{28+10\sqrt{3}}}{2}\\ =\frac{\sqrt{\left(5+\sqrt{3}\right)^2}}{2}\\ =\frac{5+\sqrt{3}}{2}\)