a) Ta sắp xếp theo thứ tự tăng dần như sau:
\(2\sqrt{6};\sqrt{29};4\sqrt{2};3\sqrt{5}\)
b) Ta sắp xếp theo thứ tự tăng dần như sau:
\(\sqrt{38};2\sqrt{14};3\sqrt{7};6\sqrt{2}\)
Hoàng Phong làm bừa
a/
\(2\sqrt{6}=\sqrt{24}< \sqrt{29}< 4\sqrt{2}=\sqrt{32}< 3\sqrt{5}=\sqrt{45}.\)
b/
\(\sqrt{38}< 2\sqrt{14}=\sqrt{56}< 3\sqrt{7}=\sqrt{63}< 6\sqrt{2}=\sqrt{72}\)
a) Ta co :
\(3\sqrt{5}=\sqrt{9.5}=\sqrt{45},2\sqrt{6}=\sqrt{4.6}=\sqrt{24},4\sqrt{2}=\sqrt{32}\)
\(\sqrt{24}< \sqrt{29}< \sqrt{32}< \sqrt{45}\)\(\text{suy ra }\)\(2\sqrt{6}< \sqrt{29}< 4\sqrt{2}< 3\sqrt{5}\)
b) Ta co :
\(6\sqrt{2}=\sqrt{6^2^.2}=\sqrt{72},3\sqrt{7}=\sqrt{9.7}=\sqrt{63},2\sqrt{14}=\sqrt{2^2^.14}=\sqrt{56}\)
\(\sqrt{38}< \sqrt{56}< \sqrt{63}< \sqrt{72}\)\(\text{suy ra}\)\(\sqrt{38}< 2\sqrt{14}< 3\sqrt{7}< 6\sqrt{2}\)
ta có
a)\(2\sqrt{6},\sqrt{29},4\sqrt{2},3\sqrt{5}\)
b)\(\sqrt{38},2\sqrt{14},3\sqrt{7},6\sqrt{2},\)
a) \(2\sqrt{6},\sqrt{29},4\sqrt{2},3\sqrt{5}\) b) \(\sqrt{38},2\sqrt{14},3\sqrt{7},6\sqrt{2}\)
a) 2√6;√29;4√2;3√5
b) √38;2√14;3√7;6√2
a) 2√6<√29<4√2<3√5
b) √38<2√14<3√7<6√2
a. \(2\sqrt{6}\) ; \(\sqrt{29}\) ; \(4\sqrt{2}\) ; \(3\sqrt{5}\)
b. \(\sqrt{38}\) ; \(2\sqrt{14}\) ; \(3\sqrt{7}\) ; \(6\sqrt{2}\)
a) 2 can 6, can 29, 4 can 2, 3 can 5
b) can 38,2 can 14,3 can 7,6 can 2
a) 2căn 6<căn 29<4căn 2<3căn 5
b) căn 38<2căn 14<3căn 7<6căn 2
a) Ta sắp xếp theo thứ tự tăng dần như sau:
b) Ta sắp xếp theo thứ tự tăng dần như sau:
Đưa một thừa số vào trong dấu căn rồi sắp thứ tự các số trong dấu căn
a, Ta có : \(3\sqrt{5}=\sqrt{9.5}=\sqrt{45},2\sqrt{6}=\sqrt{4.6}=\sqrt{24},4\sqrt{2}=32\)
\(\sqrt{24}< \sqrt{29}< \sqrt{32}< \sqrt{45}\)
=> \(2\sqrt{6}< \sqrt{29}< 4\sqrt{2}< 3\sqrt{5}\)
b, Ta có : \(6\sqrt{2}=\sqrt{6^2.2}=\sqrt{72},3\sqrt{7}=\sqrt{9.7}=\sqrt{63,}2\sqrt{14}=\sqrt{2^2.14}=56\)
\(\sqrt{38}< \sqrt{56}< \sqrt{63}< \sqrt{72}\)
=> \(\sqrt{38}< 2\sqrt{14}< 3\sqrt{7}< 6\sqrt{2}\)
A- 2V6 <V29<4V2<3V5
B-V38<2V14<3V7<6V2
a. \(2\sqrt{6};\sqrt{29};4\sqrt{2};3\sqrt{5}\)
b. \(\sqrt{38};2\sqrt{14};3\sqrt{7};6\sqrt{2}\)
a, 2√6; √29; 4√2; 3√5
b, √14; √38; 3√7; 6√2
a) ta có \(3\sqrt{5}=\sqrt{9.5}=\sqrt{45,2}\sqrt{6}=\sqrt{4.6}=\sqrt{24,4}\sqrt{2}=\sqrt{32}\)
\(\sqrt{24}< \sqrt{29}< \sqrt{32}< \sqrt{45}\) suy ra \(2\sqrt{6}< \sqrt{29}< 4\sqrt{2}< 3\sqrt{5}\)
b) ta có \(6\sqrt{2}=\sqrt{6^2.2}=\sqrt{72,3}\sqrt{7}=\sqrt{9.7}=\sqrt{63,}2\sqrt{14}=\sqrt{56}\)
\(\sqrt{38}< \sqrt{56}< \sqrt{63}< \sqrt{72}\) suy ra \(\sqrt{38}< 2\sqrt{14}< 3\sqrt{7}< 6\sqrt{2}\)
a) 3\(\sqrt{5}\) = \(\sqrt{9.5}\) = \(\sqrt{45}\), \(2\sqrt{6}\) = \(\sqrt{4.6}\) = \(\sqrt{24}\), \(4\sqrt{2}\) = \(\sqrt{32}\)
\(\sqrt{24}\) < \(\sqrt{29}\) < \(\sqrt{32}\) < \(\sqrt{45}\) => 2\(\sqrt{6}\)< \(\sqrt{29}\) < \(4\sqrt{2}\) < \(3\sqrt{5}\)
b) 6\(\sqrt{2}\) = \(\sqrt{6^2.2}\) = \(\sqrt{72}\), \(3\sqrt{7}\) = \(\sqrt{9.7}\) = \(\sqrt{63}\), \(2\sqrt{14}\) = \(\sqrt{2^2.14}\) = \(\sqrt{56}\)
\(\sqrt{38}\) < \(\sqrt{56}\) < \(\sqrt{63}\) < \(\sqrt{72}\) => \(\sqrt{38}\) < \(2\sqrt[]{14}\) < \(3\sqrt{7}\) < 6\(\sqrt{2}\)
a, Ta có \(3\sqrt{5}=\sqrt{9.5}=\sqrt{45},2\sqrt{6}=\sqrt{4.6}=\sqrt{24},4\sqrt{2}=\sqrt{32}\)
\(\sqrt{24}< \sqrt{29}< \sqrt{32}< \sqrt{45}\) suy ra \(2\sqrt{6}< \sqrt{29}< 4\sqrt{2}< 3\sqrt{5}\)
a, Ta có 3 căn 5 = căn 9.5 = căn 45, 2căn 6= căn 4.6 = căn 24, 4 căn 2 = căn 32
căn 24< căn 29 < căn 32< căn 45 =>2căn 6<căn 29<4 căn 2<3 căn 5
b, Ta có 6căn 2 = căn 6^2 . 2= căn 72, 3căn 7 = căn 9.7=căn 63, 2 căn 14 = căn 2^2.14=căn 56
Căn 38 < căn 56< căn 63< căn 72=> căn 38<2 căn 14< 3 căn 7< 6căn 2
a) 2\(\sqrt{6}\) < \(\sqrt{29}\) < 4\(\sqrt{2}\) < 3\(\sqrt{5}\)
b) \(\sqrt{38}\) < 2\(\sqrt{14}\) < 3\(\sqrt{7}\) < 6\(\sqrt{2}\)