Cho B=\(\sqrt{20+\sqrt{20+\sqrt{20+...+\sqrt{20}}}}\) , C=\(\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+...+\sqrt[3]{24}}}}\)
Chứng minh rằng 7<B+C<8
Cho \(A=\sqrt{20+\sqrt{20+\sqrt{20+...+\sqrt{20}}}}\)
\(B=\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+...+\sqrt[3]{24}}}}\)
Chứng minh rằng 7<A+B<8. tìm [A+B]
Cho \(A=\sqrt{20+\sqrt{20+\sqrt{20}+....+\sqrt{20}}}\)
\(B=\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+...+\sqrt[3]{24}}}}\)
Chứng minh rằng 7<A+B<8 tìm [A+B]
Cho \(A=\sqrt{20+\sqrt{20+\sqrt{20+.....+\sqrt{20}}}}\)
\(B=\sqrt[3]{24+\sqrt[3]{24+....+\sqrt[3]{24}}}\)
Chứng minh rằng 7 < A + B < 8
Ta có:
\(A< \sqrt{20+\sqrt{20+...+\sqrt{20+\sqrt{25}}}}\)
\(\Leftrightarrow A< \sqrt{25}=5\)(1)
\(B< \sqrt[3]{24+\sqrt[3]{24+...+\sqrt[3]{24+\sqrt[3]{27}}}}\)
\(\Leftrightarrow B< \sqrt[3]{27}=3\)(2)
Từ (1) và (2) suy ra A+B<5+3=8
Ta có:
\(A>\sqrt{19,36}=4,4\)(3)
\(B>\sqrt[3]{17,576}=2,6\)(4)
Từ (3) và (4) suy ra A+B>4,4+2,6=7
Vậy 7<A+B<8
Cho \(A=\sqrt{20+\sqrt{20+\sqrt{20+...+\sqrt{20}}}}\); \(B=\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+...\sqrt[3]{24}}}}\)
Mỗi số đều có 2005 dấu căn. Tìm [A+B]?
\(\Rightarrow A+B< 3+5=8\)
mặt khác ta có A+B>\(\sqrt{20}+\sqrt[3]{24}=7.3566....>7\)\(\Rightarrow\left[A+b\right]=7\)
Cho \(T=\sqrt{20+\sqrt{20+...+\sqrt{20}}}+\sqrt[3]{24+\sqrt[3]{24+...+\sqrt[3]{24}}}\)
(2006 dấu căn) (2006 dấu căn)
CM: 7<T<8
\(\sqrt{20+\sqrt{20+...+\sqrt{20}}}< \sqrt{20+\sqrt{20+...+\sqrt{20+\sqrt{25}}}}=5\)
Và:\(\sqrt[3]{24+\sqrt[3]{24+...+\sqrt[3]{24}}}< \sqrt[3]{24+\sqrt[3]{24+...+\sqrt[3]{24+\sqrt[3]{27}}}}=3\)
Nên \(T=\sqrt{20+\sqrt{20+...+\sqrt{20}}}+\sqrt[3]{24+\sqrt[3]{24+...+\sqrt[3]{24}}}< 8\)(2).Từ (1) và (2), ta có: \(7< T< 8\)đpcmChứng minh rằng a,\(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}< 24\)
b,\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>10\)
b, \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{100}}>10\)
Ta có: \(1< 100\Rightarrow\sqrt{1}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{1}}< \frac{1}{\sqrt{100}}\)
\(2< 100\Rightarrow\sqrt{2}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{2}}< \frac{1}{\sqrt{100}}\)
\(3< 100\Rightarrow\sqrt{3}< \sqrt{100}\Rightarrow\frac{1}{\sqrt{3}}< \frac{1}{\sqrt{100}}\)
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\(100=100\Rightarrow\sqrt{100}=\sqrt{100}\frac{1}{\sqrt{100}}=\frac{1}{\sqrt{100}}\left(1\right)\)
Từ (1) suy ra:
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}+\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}\left(100sh\frac{1}{\sqrt{100}}\right)\)
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>\frac{1}{\sqrt{100}}.100\)
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>\frac{10}{\sqrt{100}}\)
\(\Rightarrow\frac{1}{\sqrt{10}}+\frac{1}{\sqrt{20}}+\frac{1}{\sqrt{30}}+...+\frac{1}{\sqrt{100}}>10\left(ĐPCM\right)\)
B=\(\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20}+2\sqrt{43}+24\sqrt{3}\)
\(B=\left(13-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)-8\sqrt{20+2\sqrt{43+24\sqrt{3}}}\)
\(=\left(2\sqrt{3}-1\right)^2\left(2+\sqrt{3}\right)^2-8\sqrt{20+2\sqrt{\left(4+3\sqrt{3}\right)^2}}\)
\(=\left(3\sqrt{3}+4\right)^2-8\sqrt{20+2\left(4+3\sqrt{3}\right)}\)
\(=\left(3\sqrt{3}+4\right)^2-8\sqrt{28+6\sqrt{3}}\)
\(=\left(3\sqrt{3}+4\right)^2-8\sqrt{\left(3\sqrt{3}+1\right)^2}\)
\(=43+24\sqrt{3}-8\left(3\sqrt{3}+1\right)=35\)
Chứng minh rằng : \(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}< 24\)
Cái này thì....mình mù tịt
Vì chưa học!!!!
Ai đồng ý thì cho mình xin 1 k!!!
\(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}\)\(\sqrt{42}\)= 23,75790715.
Vì vậy : \(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}\) sẽ lớn hơn 24.
Dùng máy tính là được chứ gì.
tính x=\(\sqrt{97-56\sqrt{3}}+\sqrt{52+16\sqrt{3}}\)
y=\(\sqrt{33+20\sqrt{2}}+\sqrt{24-16\sqrt{2}}\)
Ta có: \(x=\sqrt{97-56\sqrt{3}}+\sqrt{52+16\sqrt{3}}\)
\(=\sqrt{49-2\cdot7\cdot4\sqrt{3}+48}+\sqrt{48+2\cdot4\sqrt{3}\cdot2+4}\)
\(=\sqrt{\left(7-4\sqrt{3}\right)^2}+\sqrt{\left(4\sqrt{3}+2\right)^2}\)
\(=\left|7-4\sqrt{3}\right|+\left|4\sqrt{3}+2\right|\)
\(=7-4\sqrt{3}+4\sqrt{3}+2\)
\(=9\)
Làm luôn phần y :D
y = \(\sqrt{33+20\sqrt{2}}+\sqrt{24-16\sqrt{2}}\)
y = \(\sqrt{33+2.10\sqrt{2}}+\sqrt{24-2.8\sqrt{2}}\)
y = \(\sqrt{33+2.5.2\sqrt{2}}+\sqrt{24-2.4.2\sqrt{2}}\)
y = \(\sqrt{25+2.5.\sqrt{8}+8}+\sqrt{16-2.4.\sqrt{8}+8}\)
y = \(\sqrt{\left(5+\sqrt{8}\right)^2}+\sqrt{\left(4-\sqrt{8}\right)^2}\)
y = |5 + \(\sqrt{8}\)| + |4 - \(\sqrt{8}\)|
y = 5 + \(\sqrt{8}\) + 4 - \(\sqrt{8}\) (Vì 4 > \(\sqrt{8}\) nên 4 - \(\sqrt{8}\) > 0)
y = 9
Vậy y = 9
Chúc bn học tốt!