không dùng máy tính, tính:
M=\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
K
Khách
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19 tháng 9 2021
\(a,\left(\sqrt{\sqrt{3}}\right)^4=3< 4=\left(\sqrt{2}\right)^4\Rightarrow\sqrt{\sqrt{3}}< \sqrt{2}\\ b,\left(\sqrt{2\sqrt{3}}\right)^4=12< 18=\left(\sqrt{3\sqrt{2}}\right)^4\Rightarrow\sqrt{2\sqrt{3}}=\sqrt{3\sqrt{2}}\\ c,\left(2+\sqrt{6}\right)^2=8+4\sqrt{6};5^2=25=8+17;\left(4\sqrt{6}\right)^2=96< 289=17^2\\ \Rightarrow4\sqrt{6}< 17\Rightarrow2+\sqrt{6}< 5\\ d,\left(7-2\sqrt{2}\right)^2=57-28\sqrt{2};4^2=16=57-41;\left(28\sqrt{2}\right)^2=1568< 41^2=1681\\ \Rightarrow28\sqrt{2}< 41\Rightarrow7-2\sqrt{2}>4\\ e,\left(\sqrt{15}+\sqrt{8}\right)^2=23+4\sqrt{30};7^2=49=23+26;\left(4\sqrt{30}\right)^2=240< 676=26^2\\ \Rightarrow4\sqrt{30}< 26\Rightarrow\sqrt{15}+\sqrt{8}< 7\)
\(f,\left(\sqrt{37}-\sqrt{14}\right)^2=51-2\sqrt{518};\left(6-\sqrt{15}\right)^2=51-12\sqrt{15};\left(2\sqrt{518}\right)^2=2072;\left(12\sqrt{15}\right)^2=2160\\ \Rightarrow2\sqrt{518}< 12\sqrt{15}\Rightarrow\sqrt{37}-\sqrt{14}>6-\sqrt{15}\)
CH
Cô Hoàng Huyền
Admin
VIP
23 tháng 6 2017
1) \(A=\left(\sqrt{7-\sqrt{21}+4\sqrt{5}}\right)^2=7-\sqrt{21}+4\sqrt{5}\)
\(B=\left(\sqrt{5}-1\right)^2=6-2\sqrt{5}\)
\(\Rightarrow A-B=1-\sqrt{21}+6\sqrt{5}=\left(1+\sqrt{180}\right)-\sqrt{21}>0\)
\(\Rightarrow A>B\Rightarrow\sqrt{7-\sqrt{21}+4\sqrt{5}}>\sqrt{5}-1\)
2) \(C=\left(\sqrt{5}+\sqrt{10}+1\right)^2=5+10+1+10\sqrt{2}+2\sqrt{5}+2\sqrt{10}\)
\(=26+10\sqrt{2}+2\sqrt{5}+2\sqrt{10}>26+10>35=\left(\sqrt{35}\right)^2\)
Vậy \(\sqrt{5}+\sqrt{10}+1>\sqrt{35}\)
3) \(\left(\frac{15-2\sqrt{10}}{3}\right)^2=\frac{225-60\sqrt{10}+40}{9}=\frac{265-60\sqrt{10}}{9}=\frac{265}{9}-\frac{20\sqrt{10}}{3}< 15\)
Vậy nên \(\frac{15-2\sqrt{10}}{3}< \sqrt{15}\)
5 tháng 10 2021
b: \(\sqrt{2017}-\sqrt{2016}=\dfrac{1}{\sqrt{2016}+\sqrt{2017}}\)
\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)
mà \(\sqrt{2016}+\sqrt{2017}< \sqrt{2016}+\sqrt{2015}\)
nên \(\sqrt{2017}-\sqrt{2016}>\sqrt{2016}-\sqrt{2015}\)
19 tháng 9 2021
a) \(1=\sqrt{1}< \sqrt{2}\)
b) \(2=\sqrt{4}>\sqrt{3}\)
c) \(6=\sqrt{36}< \sqrt{41}\)
d) \(7=\sqrt{49}>\sqrt{47}\)
e) \(2=1+1=\sqrt{1}+1< \sqrt{2}+1\)
f) \(1=2-1=\sqrt{4}-1>\sqrt{3}-1\)
g) \(2\sqrt{31}=\sqrt{4.31}=\sqrt{124}>\sqrt{100}=10\)
h) \(\sqrt{3}>0>-\sqrt{12}\)
i) \(5=\sqrt{25}< \sqrt{29}\)
\(\Rightarrow-5>-\sqrt{29}\)
17 tháng 9 2019
\(T=\frac{\sqrt{2}.\left(4+\sqrt{7}\right)}{\sqrt{2}.\left(2\sqrt{2}+\sqrt{4+\sqrt{7}}\right)}+\frac{\sqrt{2}.\left(4-\sqrt{7}\right)}{\sqrt{2}.\left(2\sqrt{2}-\sqrt{4-\sqrt{7}}\right)}\)
\(T=\frac{4\sqrt{2}+\sqrt{14}}{4+\sqrt{8+2\sqrt{7}}}+\frac{4\sqrt{2}-\sqrt{14}}{4-\sqrt{8-2\sqrt{7}}}\)
\(T=\frac{4\sqrt{2}+\sqrt{14}}{4+\sqrt{7+2\sqrt{7}+1}}+\frac{4\sqrt{2}-\sqrt{14}}{4-\sqrt{7-2\sqrt{7}+1}}\)
\(T=\frac{4\sqrt{2}+\sqrt{14}}{4+\left(\sqrt{7}+1\right)^2}+\frac{4\sqrt{2}-\sqrt{14}}{4-\left(\sqrt{7}-1\right)^2}\)\(T=\frac{4\sqrt{2}+\sqrt{14}}{4+|\sqrt{7}+1|}+\frac{4\sqrt{2}-\sqrt{14}}{4-|\sqrt{7}-1|}\)
\(T=\frac{4\sqrt{2}+\sqrt{14}}{4+\sqrt{7}+1}+\frac{4\sqrt{2}-\sqrt{14}}{4-\sqrt{7}+1}\)
\(T=\frac{4\sqrt{2}+\sqrt{14}}{5+\sqrt{7}}+\frac{4\sqrt{2}-\sqrt{14}}{5-\sqrt{7}}\)
\(T=\frac{\left(4\sqrt{2}+\sqrt{14}\right).\left(5-\sqrt{7}\right)}{\left(5+\sqrt{7}\right).\left(5-\sqrt{7}\right)}+\frac{\left(4\sqrt{2}-\sqrt{14}\right).\left(5+\sqrt{7}\right)}{\left(5+\sqrt{7}\right).\left(5-\sqrt{7}\right)}\)
\(T=\frac{20\sqrt{2}-\sqrt{98}}{9}\)
\(T=\frac{13\sqrt{2}}{9}\)
TD
14 tháng 10 2019
B = \(\frac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\frac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
=> \(\frac{2}{\sqrt{2}}B=\frac{8+2\sqrt{7}}{6+\sqrt{8+2\sqrt{7}}}+\frac{8-2\sqrt{7}}{6-\sqrt{8-2\sqrt{7}}}\)
=> \(\frac{2}{\sqrt{2}}B=\frac{\left(\sqrt{7}+1\right)^2}{6+\sqrt{7}+1}+\frac{\left(\sqrt{7}-1\right)^2}{6-\sqrt{7}+1}\)
=> \(\frac{2}{\sqrt{2}}B=\frac{\left(\sqrt{7}+1\right)^2}{\sqrt{7}\left(\sqrt{7}+1\right)}+\frac{\left(\sqrt{7}-1\right)^2}{\sqrt{7}\left(\sqrt{7}-1\right)}\)
=> \(\frac{2}{\sqrt{2}}B=\frac{\sqrt{7}+1}{\sqrt{7}}+\frac{\sqrt{7}-1}{\sqrt{7}}=\frac{2\sqrt{7}}{\sqrt{7}}=2\)
=> B = \(\sqrt{2}\)
CH
Cô Hoàng Huyền
Admin
VIP
17 tháng 6 2016
a. Có nhiều cách nhé. Với lớp 9 cô dùng cách này. Cô hướng dẫn nhé :)
Giả thiệt cho như hình vẽ. Gỉa sử AB = 1cm, khi đó do góc ADB = 30độ nên \(\frac{AB}{BD}=\frac{1}{2};\frac{AB}{AD}=\frac{\sqrt{3}}{3}\)
Vậy \(AC=AD+DC=AD+DB=2+\sqrt{3}\)
Vậy \(tan15=\frac{AB}{AC}=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}\)
b. Dựa vào công thức : \(tan^215+1=\frac{1}{cos^215}\)
không dùng máy tính sao tính dc hỏi ngu
Ta có :
\(\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)^2=4+\sqrt{7}-2\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}+4-\sqrt{7}\)
\(8-2\sqrt{16-7}=8-2\sqrt{9}=8-2.3=8-6=2\)
\(\Rightarrow\)\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\sqrt{2}\) ( vì \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}>0\) )
\(\Rightarrow\)\(M=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}=\sqrt{2}-\sqrt{2}=0\)
Vậy \(M=0\)
Chúc bạn học tốt ~