Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a ) VT = \(\sqrt{\sqrt{6+\sqrt{20}}}=\sqrt{\sqrt{6+\sqrt{4.5}}}=\sqrt{\sqrt{6+2\sqrt{5}}}=\sqrt{\sqrt{\left(1+\sqrt{5}\right)^2}}=\sqrt{1+\sqrt{5}}\)
Có 5 < 6 => \(\sqrt{5}< \sqrt{6}\Rightarrow\sqrt{1+\sqrt{5}}< \sqrt{1+\sqrt{6}}\)
Vậy \(\sqrt{\sqrt{6+\sqrt{20}}}< \sqrt{1+\sqrt{6}}\)
b) VT = \(\sqrt{\sqrt{17+12\sqrt{2}}}=\sqrt{\sqrt{17+2.2\sqrt{2}.3}}=\sqrt{\sqrt{\left(2\sqrt{2}+3\right)^2}=\sqrt{2\sqrt{2}+3}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
=> VT = VP
=> \(\sqrt{\sqrt{17+12\sqrt{2}}}=\sqrt{2}+1\)
c) \(\sqrt{\sqrt{28-16\sqrt{3}}}=\sqrt{\sqrt{16-2.4.2\sqrt{3}+12}}=\sqrt{\sqrt{\left(4-2\sqrt{3}\right)^2}}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
Có -1 > -2 => \(\sqrt{3}-1>\sqrt{3}-2\Rightarrow\sqrt{\sqrt{28-16\sqrt{3}}}>\sqrt{3}-2\)
\(\sqrt{\sqrt{6+\sqrt{20}}}=\sqrt{\sqrt{5+2\sqrt{5}+1}}=\sqrt{\sqrt{\left(\sqrt{5}+1\right)^2}}=\sqrt{\sqrt{5}+1}< \sqrt{\sqrt{6}+1}\)
\(\sqrt{\sqrt{6+\sqrt{20}}}=\sqrt{\sqrt{6+2\sqrt{5}}}=\sqrt{\sqrt{\left(\sqrt{5}+1\right)^2}}=\sqrt{\sqrt{5}+1}\)
Vì \(\sqrt{\sqrt{5}+1}< \sqrt{\sqrt{6}+1}\Rightarrow\sqrt{\sqrt{6+\sqrt{20}}}< \sqrt{1+\sqrt{6}}\)
a) \(\sqrt{2004}-\sqrt{2003}=\frac{1}{\sqrt{2004}+\sqrt{2003}}>\frac{1}{\sqrt{2006}+\sqrt{2005}}=\sqrt{2006}-\sqrt{2005}\)
b) Tương tự.
1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)
2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)
3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2}
\)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)
4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)
5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)
7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)
8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2
a)
\(4\sqrt{7}=\sqrt{4^2.7}=\sqrt{112}\)
\(3\sqrt{13}=\sqrt{3^2.13}=\sqrt{117}\)
\(\sqrt{112}< \sqrt{117}\Rightarrow 4\sqrt{7}< 3\sqrt{13}\)
b) \(3\sqrt{12}=\sqrt{3^2.12}=\sqrt{9.2^2.3}=2\sqrt{27}>2\sqrt{16}\)
c)
\(\frac{1}{4}\sqrt{82}=\sqrt{\frac{82}{16}}=\sqrt{\frac{41}{8}}=\sqrt{5+\frac{1}{8}}\)
\(6\sqrt{\frac{1}{7}}=\sqrt{\frac{36}{7}}=\sqrt{5+\frac{1}{7}}\)
\(\sqrt{5+\frac{1}{8}}< \sqrt{5+\frac{1}{7}}\Rightarrow \frac{1}{4}\sqrt{82}< 6\sqrt{\frac{1}{7}}\)
d)
\(\frac{1}{2}\sqrt{\frac{17}{2}}=\sqrt{\frac{17}{8}}=\sqrt{2+\frac{1}{8}}\)
\(\frac{1}{3}\sqrt{19}=\sqrt{\frac{19}{9}}=\sqrt{2+\frac{1}{9}}\)
\(\sqrt{2+\frac{1}{8}}>\sqrt{2+\frac{1}{9}}\Rightarrow \frac{1}{2}\sqrt{\frac{17}{2}}> \frac{1}{3}\sqrt{19}\)
e)
\(3\sqrt{3}-2\sqrt{2}=\sqrt{27}-\sqrt{8}\)
Mà \(\sqrt{27}>\sqrt{25}; \sqrt{8}< \sqrt{9}\Rightarrow \sqrt{27}-\sqrt{8}> \sqrt{25}-\sqrt{9}=5-3=2\)
Vậy \(3\sqrt{3}-2\sqrt{2}>2\)
f)
\(\sqrt{7}+\sqrt{5}< \sqrt{9}+\sqrt{9}=6\)
\(\sqrt{49}=7\)
\(\Rightarrow \sqrt{7}+\sqrt{5}< 6< 7=\sqrt{49}\)
g)
\(\sqrt{2}< \sqrt{3}; \sqrt{11}< \sqrt{25}=5\)
\(\Rightarrow \sqrt{2}+\sqrt{11}< \sqrt{3}+5\)
h) Lặp lại câu d
i)
\(\sqrt{21}>\sqrt{20}\); \(\sqrt{5}< \sqrt{6}\)
\(\Rightarrow \sqrt{21}-\sqrt{5}> \sqrt{20}-\sqrt{6}\)