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1) \(2\sqrt{2}=\sqrt{8}< \sqrt{9}=3\)
\(\Rightarrow\)\(6+2\sqrt{2}< 6+3=9\)
2) \(4\sqrt{5}=\sqrt{80}>\sqrt{49}=7\)
\(\Rightarrow\)\(9+4\sqrt{5}>9+7=16\)
3) \(2=\sqrt{4}>\sqrt{3}\)
\(\Rightarrow\)\(2-1>\sqrt{3}-1\)
hay \(1>\sqrt{3}-1\)
4) \(9-4\sqrt{5}< 16\)
5) \(\sqrt{2}>\sqrt{1}=1\)
\(\Rightarrow\)\(\sqrt{2}+1>2\)
\(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}=\left(2\sqrt{5}+3\right)-\left(2\sqrt{5}-3\right)=6\)
\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)-\left(2\sqrt{5}-\sqrt{3}\right)=-\sqrt{5}\)
\(\sqrt{8-12\sqrt{5}}+\sqrt{48+6\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)+\left(3\sqrt{5}+\sqrt{3}\right)=4\sqrt{5}\)
\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\left(5-2\sqrt{6}\right)+\left(5+2\sqrt{6}\right)=10\)
\(\sqrt{15-6\sqrt{15}}+\sqrt{33-12\sqrt{6}}\) đề này sai ạ
\(\sqrt{16-6\sqrt{7}}+\sqrt{64-24\sqrt{7}}=\left(3-\sqrt{7}\right)+\left(6-2\sqrt{7}\right)=9-3\sqrt{7}\)
\(\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}=\left(3-\sqrt{5}\right)+\left(3+\sqrt{5}\right)=6\)
\(\sqrt{1-6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\left(2\sqrt{2}+5\right)+\left(2\sqrt{2}-5\right)=4\sqrt{2}\)
\(\sqrt{46-6\sqrt{5}}+\sqrt{29-12\sqrt{5}}=\left(3\sqrt{5}-1\right)+\left(2\sqrt{5}-3\right)=5\sqrt{5}-4\)
#Học tốt ạ
a)(\(\sqrt{2006}-\sqrt{2005}\)).(\(\sqrt{2006}+\sqrt{2005}\))
=\(\sqrt{2006}^2-\sqrt{2005}^2\)
=2006-2005
=1
\(1)\) Ta có :
\(\left(\sqrt{3\sqrt{2}}\right)^4=\left[\left(\sqrt{3\sqrt{2}}\right)^2\right]^2=\left(3\sqrt{2}\right)^2=9.2=18\)
\(\left(\sqrt{2\sqrt{3}}\right)^4=\left[\left(\sqrt{2\sqrt{3}}\right)^2\right]^2=\left(2\sqrt{3}\right)^2=4.3=12\)
Vì \(18>12\) nên \(\left(\sqrt{3\sqrt{2}}\right)^4>\left(\sqrt{2\sqrt{3}}\right)^4\)
\(\Rightarrow\)\(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
Vậy \(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
Chúc bạn học tốt ~
1. Đặt A =\(\sqrt{\frac{129}{16}+\sqrt{2}}\)
\(\sqrt{16}\)A = \(\sqrt{129+16\sqrt{2}}\)
4A = \(\sqrt{\left(8\sqrt{2}+1\right)^2}\)
4A = \(8\sqrt{2}+1\)
⇒ A = \(\frac{\text{}8\sqrt{2}+1}{4}\)= \(2\sqrt{2}\) + \(\frac{1}{4}\)
2. Đặt B = \(\sqrt{\frac{289+4\sqrt{72}}{16}}\)
\(\sqrt{16}\)B = \(\sqrt{289+24\sqrt{2}}\)
4B = \(\sqrt{\left(12\sqrt{2}+1\right)^2}\)
4B = \(12\sqrt{2}+1\)
⇒ B = \(\frac{12\sqrt{2}+1}{4}\)= \(3\sqrt{2}+\frac{1}{4}\)
3. \(\sqrt{2-\sqrt{3}}\). \(\left(\sqrt{6}+\sqrt{2}\right)\)
= \(\sqrt{2-\sqrt{3}}\). \(\sqrt{2}.\left(\sqrt{3}+1\right)\)
= \(\sqrt{4-2\sqrt{3}}\) . \(\left(\sqrt{3}+1\right)\)
= \(\sqrt{\left(\sqrt{3}-1\right)^2}\) . \(\left(\sqrt{3}+1\right)\)
= \(\left(\sqrt{3}-1\right)\). \(\left(\sqrt{3}+1\right)\)
= \(\left(\sqrt{3}\right)^2\) - 12
= 3 - 1
= 2
4. \(\left(\sqrt{21}+7\right)\). \(\sqrt{10-2\sqrt{21}}\)
= \(\left(\sqrt{21}+7\right)\) . \(\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
= \(\sqrt{7}\left(\sqrt{3}+\sqrt{7}\right)\) . \(\left(\sqrt{7}-\sqrt{3}\right)\)
= \(\sqrt{7}\) \(\left[\left(\sqrt{7}\right)^2-\left(\sqrt{3}\right)^2\right]\)
= \(\sqrt{7}\) . (7 - 3)
= 4\(\sqrt{7}\)
5. \(2.\left(\sqrt{10}-\sqrt{2}\right)\). \(\sqrt{4+\sqrt{6-2\sqrt{5}}}\)
= \(2.\left(\sqrt{10}-\sqrt{2}\right)\) . \(\sqrt{4+\sqrt{5}-1}\)
= \(2.\left(\sqrt{10}-\sqrt{2}\right)\) . \(\sqrt{3+\sqrt{5}}\)
= \(\left(\sqrt{10}-\sqrt{2}\right)\) . \(\sqrt{12+4\sqrt{5}}\)
= \(\left(\sqrt{10}-\sqrt{2}\right)\) . \(\left(\sqrt{10}+\sqrt{2}\right)\)
= \(\left(\sqrt{10}\right)^2-\left(\sqrt{2}\right)^2\)
= 10 - 2
= 8
6. \(\left(4\sqrt{2}+\sqrt{30}\right)\). \(\left(\sqrt{5}-\sqrt{3}\right)\). \(\sqrt{4-\sqrt{15}}\)
= \(\sqrt{2}\)\(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\) . \(\sqrt{4-\sqrt{15}}\)
= \(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\) . \(\sqrt{8-2\sqrt{15}}\)
= \(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)\)
= \(\left(4+\sqrt{15}\right)\) . \(\left(\sqrt{5}-\sqrt{3}\right)^2\)
= \(\left(4+\sqrt{15}\right)\). \(\left(8-2\sqrt{15}\right)\)
= 32 - \(8\sqrt{15}\) + \(8\sqrt{15}\) - 30
= 2
7. \(\left(7-\sqrt{14}\right)\) . \(\sqrt{9-2\sqrt{14}}\)
= \(\sqrt{7}\) \(\left(\sqrt{7}-\sqrt{2}\right)\). \(\left(\sqrt{7}-\sqrt{2}\right)\)
= \(\sqrt{7}\). \(\left(\sqrt{7}-\sqrt{2}\right)^2\)
= \(\sqrt{7}\) . \(\left(9-2\sqrt{14}\right)\)
= 9\(\sqrt{7}\) - 14\(\sqrt{2}\)
TICK MÌNH NHA!
a, \(\left(\sqrt{2006}-\sqrt{2005}\right).\left(\sqrt{2006}+\sqrt{2005}\right)=\left(2006-2005\right)=1\)
b.
=\(\frac{7+4\sqrt{3}+14-8\sqrt{3}}{49-48}\left(21+4\sqrt{3}\right)\)
=\(\left(21-4\sqrt{3}\right)\left(21+4\sqrt{3}\right)\)
=441-48
393
vậy.......
hc tốt
Ý anh là so sánh đúng ko ạ?
15) Bình phương hai vế,ta cần so sánh: \(\left(\frac{5}{4}\sqrt{2}\right)^2\text{ và }\left(\frac{2}{3}\sqrt{7}\right)^2\Leftrightarrow\frac{25}{8}\text{ và }\frac{28}{9}\)
Dễ thấy \(\frac{25}{8}>\frac{28}{9}\Rightarrow\frac{5}{4}\sqrt{2}>\frac{2}{3}\sqrt{7}\)
16) \(\sqrt{15}-\sqrt{14}=\frac{1}{\sqrt{15}+\sqrt{14}}< \frac{1}{\sqrt{14}+\sqrt{13}}=\sqrt{14}-\sqrt{13}\)
Xíu em làm tiếp,tắm đã
17/ Tương tự câu 16,18
18) \(\sqrt{9}-\sqrt{7}=\frac{2}{\sqrt{9}+\sqrt{7}};\sqrt{7}-\sqrt{5}=\frac{2}{\sqrt{7}+\sqrt{5}}\)
Dễ thấy \(\sqrt{9}+\sqrt{7}>\sqrt{7}+\sqrt{5}\Rightarrow\sqrt{9}-\sqrt{7}< \sqrt{7}-\sqrt{5}\)
13)Ta có: \(2\sqrt{6}=\sqrt{4.6}=\sqrt{24}>\sqrt{23}\Rightarrow-2\sqrt{6}< -\sqrt{23}\)
14)\(\sqrt{111}-7< \sqrt{121}-7=11-7=4\)
:v Thứ tự ngộ nhỉ?