Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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:
\(\sqrt{38};2\sqrt{14};3\sqrt{7};6\sqrt{2}\)
a. \(3\sqrt{5}=\sqrt{45}\) ; \(2\sqrt{6}=\sqrt{24}\) ; \(4\sqrt{2}=\sqrt{32}\)
Vì 24 < 29 < 32 < 45 nên \(\sqrt{24}< \sqrt{29}< \sqrt{32}< \sqrt{45}\)
Hay \(2\sqrt{6}< \sqrt{29}< 4\sqrt{2}< 3\sqrt{5}\)
b. \(6\sqrt{2}=\sqrt{72}\) ; \(3\sqrt{7}=\sqrt{63}\) ; \(2\sqrt{14}=\sqrt{56}\)
Vì 38 < 56 < 63 < 72 nên \(\sqrt{38}< \sqrt{56}< \sqrt{63}< \sqrt{72}\)
Hay \(\sqrt{38}< 2\sqrt{14}< 3\sqrt{7}< 6\sqrt{2}\)
a)
\(3\sqrt{5}=\sqrt{9.5}=\sqrt{45}\)
\(2\sqrt{6}=\sqrt{4.6}=\sqrt{24}\)
\(4\sqrt{2}=\sqrt{16.2}=\sqrt{32}\)
Do 24 < 29 < 32 < 45 => \(\sqrt{24}< \sqrt{29}< \sqrt{32}< \sqrt{45}\)
=> \(2\sqrt{6}< \sqrt{29}< 4\sqrt{2}< 3\sqrt{5}\)
b)
\(5\sqrt{2}=\sqrt{25.2}=\sqrt{50}\\ 3\sqrt{8}=\sqrt{9.8}=\sqrt{72}\\ 2\sqrt{15}=\sqrt{4.15}=\sqrt{60}\)
Do 39 < 50 < 60 < 72 nên \(\sqrt{39}< \sqrt{50}< \sqrt{60}< \sqrt{72}\)
=> \(\sqrt{39}< 5\sqrt{2}< 2\sqrt{15}< 3\sqrt{8}\)
a: 3căn5=căn 45
2căn 6=căn 24
căn 29=căn 29
4căn2=căn 32
=>2căn6<căn29<4căn2<3căn5
b: 5căn 2=căn 50
căn 39=căn 39
3căn 8=căn 72
2căn 15=căn60
=>căn 39<5căn2<2căn15<3căn8
+ Ta có:
2√6−√5=2(√6+√5)(√6−√5)(√6+√5)26−5=2(6+5)(6−5)(6+5)
=2(√6+√5)(√6)2−(√5)2=2(√6+√5)6−5=2(6+5)(6)2−(5)2=2(6+5)6−5
=2(√6+√5)1=2(√6+√5)=2(6+5)1=2(6+5).
+ Ta có:
3√10+√7=3(√10−√7)(√10+√7)(√10−√7)310+7=3(10−7)(10+7)(10−7)
=3(√10−√7)(√10)2−(√7)2=3(10−7)(10)2−(7)2=3(√10−√7)10−7=3(10−7)10−7
=3(√10−√7)3=√10−√7=3(10−7)3=10−7.
+ Ta có:
1√x−√y=1.(√x+√y)(√x−√y)(√x+√y)1x−y=1.(x+y)(x−y)(x+y)
=√x+√y(√x)2−(√y)2=√x+√yx−y=x+y(x)2−(y)2=x+yx−y
+ Ta có:
2ab√a−√b=2ab(√a+√b)(√a−√b)(√a+√b)2aba−b=2ab(a+b)(a−b)(a+b)
=2ab(√a+√b)(√a)2−(√b)2=2ab(√a+√b)a−b=2ab(a+b)(a)2−(b)2=2ab(a+b)a−b.
\(\frac{2}{\sqrt{6}-\sqrt{5}}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{6-5}=2\left(\sqrt{6}+\sqrt{5}\right)\)
\(\frac{3}{\sqrt{10}+\sqrt{7}}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}-\sqrt{7}\right)\left(\sqrt{10}+\sqrt{7}\right)}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\sqrt{10}-\sqrt{7}\)
\(\frac{1}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}+\sqrt{y}}{x-y}\)
\(\frac{2ab}{\sqrt{a}-\sqrt{b}}=\frac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\)
a,\(\left(5+4\sqrt{2}\right)\left(3+2\sqrt{1+\sqrt{2}}\right)\left(3-2\sqrt{1+\sqrt{2}}\right)\)
=\(\left(5+4\sqrt{2}\right)\left(9-4\left(1+\sqrt{2}\right)\right)\)
=\(\left(5+4\sqrt{2}\right)\left(9-4-4\sqrt{2}\right)\)
=\(\left(5+4\sqrt{2}\right)\left(5-4\sqrt{2}\right)=25-\left(4\sqrt{2}\right)^2\)
=-7
b, \(\sqrt{\frac{9}{4}-\sqrt{2}}=\sqrt{\frac{9-4\sqrt{2}}{4}}=\frac{\sqrt{9-4\sqrt{2}}}{2}=\frac{\sqrt{9-2\sqrt{8}}}{2}=\frac{\sqrt{\left(\sqrt{8}-1\right)^2}}{2}=\frac{\left|\sqrt{8}-1\right|}{2}=\frac{\sqrt{8}-1}{2}\)
So sánh:
1) \(2\sqrt{27}\) và \(\sqrt{147}\)
+ \(2\sqrt{27}\) = \(6\sqrt{3}\)
+ \(\sqrt{147}\) = \(7\sqrt{3}\)
⇒ \(6\sqrt{3}\) < \(7\sqrt{3}\)
Vậy: \(2\sqrt{27}\)< \(\sqrt{147}\)
2) \(2\sqrt{15}\) và \(\sqrt{59}\)
+ \(2\sqrt{15}\) = \(\sqrt{60}\)
⇒ \(\sqrt{60}\) > \(\sqrt{59}\)
Vậy: \(2\sqrt{15}\) > \(\sqrt{59}\)
3) \(2\sqrt{2}-1\) và 2
\(giống\left(-1\right)\left\{{}\begin{matrix}3-1\\2\sqrt{2}-1\end{matrix}\right.\)
So sánh: 3 và \(2\sqrt{2}\)
+ 3 = \(\sqrt{9}\)
+ \(2\sqrt{2}=\sqrt{8}\)
⇒ \(\sqrt{8}\) < \(\sqrt{9}\)
⇒ \(\sqrt{8}\) -1 < \(\sqrt{9}\) -1
⇒ \(2\sqrt{2}\) - 1 < 3 - 1
Vậy: \(2\sqrt{2}-1< 2\)
4) \(\frac{\sqrt{3}}{2}\) và 1
+ 1 = \(\frac{2}{2}\)
⇒ \(\frac{\sqrt{3}}{2}\) < \(\frac{2}{2}\)
Vậy: \(\frac{\sqrt{3}}{2}\) < 1
5) \(\frac{-\sqrt{10}}{2}\) và \(-2\sqrt{5}\)
+ \(-2\sqrt{5}\) = \(\frac{-4\sqrt{5}}{2}\) = \(\frac{-\sqrt{80}}{2}\)
⇒ \(\frac{-\sqrt{10}}{2}\) > \(\frac{-\sqrt{80}}{2}\)
Vậy: \(\frac{-\sqrt{10}}{2}\) > \(-2\sqrt{5}\)
a) 2 \sqrt{6}, \sqrt{29}, 4 \sqrt{2}, 3 \sqrt{5} ;26,29,42,35;
b) \sqrt{38}, 2 \sqrt{14}, 3 \sqrt{7}, 6 \sqrt{2}38,214,37,62
a) \(2\sqrt{6}< \sqrt{29}< 4\sqrt{2}< 3\sqrt{5}\)
b) \(\sqrt{38}< 2\sqrt{14}< 3\sqrt{7}< 6\sqrt{2}\)