So sánh: \(\frac{2\sqrt{3}+3}{2\sqrt{3}-3}\) với \(3+\sqrt{3}\)
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17 tháng 8 2017
Ta có:
\(\frac{1-\sqrt{n}+\sqrt{n+1}}{1+\sqrt{n}+\sqrt{n+1}}=\frac{\left(1-\sqrt{n}+\sqrt{n+1}\right)^2}{\left(1+\sqrt{n}+\sqrt{n+1}\right)\left(1-\sqrt{n}+\sqrt{n+1}\right)}=\frac{2n+2-2\sqrt{n}+2\sqrt{n+1}-2\sqrt{n\left(n+1\right)}}{2\left(1+\sqrt{n+1}\right)}\)
\(=\frac{\left[2n+2-2\sqrt{n}+2\sqrt{n+1}-2\sqrt{n\left(n+1\right)}\right]\left(1-\sqrt{n+1}\right)}{2\left(1+\sqrt{n+1}\right)\left(1-\sqrt{n+1}\right)}=\frac{-2n\sqrt{n+1}+2n\sqrt{n}}{-2n}=\sqrt{n+1}-\sqrt{n}\)
Suy ra:
\(Q=\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2017}-\sqrt{2016}=\sqrt{2017}-\sqrt{2}< \sqrt{2017}-1=R\)
Vậy Q < R.
VH
1
24 tháng 7 2016
\(\frac{2\sqrt{3+3}}{2\sqrt{3-3}}=3,0621......\)
\(\sqrt{3}=1,7320.....\)
=> \(\frac{2\sqrt{3+3}}{2\sqrt{3-3}}\) > 3 > \(\sqrt{3}\)
18 tháng 7 2016
\(tacó:...\frac{1}{3.\left(\sqrt{1}+\sqrt{2}\right)}>\frac{1}{3.2}=\frac{1}{\left(1+2.1\right).2.1}\)
\(\frac{1}{5.\left(\sqrt{2}+\sqrt{3}\right)}>\frac{1}{5.4}=\frac{1}{\left(1+2.2\right).2.2}\)
\(\frac{1}{7.\left(\sqrt{3}+\sqrt{4}\right)}>\frac{1}{7.6}=\frac{1}{\left(1+2..3\right).2.3}\)
....
\(\frac{1}{49.\left(\sqrt{48}+\sqrt{49}\right)}>\frac{1}{49.48}=\frac{1}{\left(1+2.48\right).2.48}\)
cộng vế theo vế ta đươc S =\(\frac{1}{\left(1+2.1\right).2}+\frac{1}{\left(1+2.2\right).2.2}+...+\frac{1}{\left(1+2.48\right).48.2}\)
\(=\frac{1}{2}.\left(\frac{1}{3}+\frac{1}{10}+\frac{1}{21}+\frac{1}{36}+...+\frac{1}{4656}\right)\) < \(\frac{1}{2}.\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{4656}\right)\)
mà lại có : \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+..+\frac{1}{4656}\)
=> \(\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{9312}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{96.97}\)
= \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...-\frac{1}{97}=\frac{1}{2}-\frac{1}{97}=\frac{95}{194}\)
vậy S < \(\frac{95}{194}\)
mà \(\frac{95}{194}< \frac{3}{7}\)
=> S < \(\frac{3}{7}\)
KẾT LUẬN : S <\(\frac{3}{7}\)
24 tháng 6 2015
B= 0,5 <=> \(\frac{2-5\sqrt{x}}{\sqrt{x}+3}=0,5\)
<=> \(2.\left(2-5\sqrt{x}\right)=\sqrt{x}+3\) <=> 4 - 10\(\sqrt{x}\) = \(\sqrt{x}\) + 3
<=> 11\(\sqrt{x}\) = 1 <=> x = \(\frac{1}{11^2}=\frac{1}{121}\)(thỏa mãn)
c) Xét hiệu: B - \(\frac{2}{3}\) = \(\frac{2-5\sqrt{x}}{\sqrt{x}+3}-\frac{2}{3}=\frac{6-15\sqrt{x}-2\left(\sqrt{x}+3\right)}{3\left(\sqrt{x}+3\right)}=\frac{-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}\le0\) Với mọi x > = 0
=> \(B\le\frac{2}{3}\)
13 tháng 8 2020
Xét phân số tổng quát là:
\(A=\frac{1}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{1\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(2n+1\right)\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n+1}}< \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n}}\)
=> \(A< \frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n}.\sqrt{n+1}}=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)
Thay từng số 1; 2; ....; 48 vào phân số tổng quát A
=> \(S< \frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{48}}-\frac{1}{\sqrt{49}}\right)\)
=> \(S< \frac{1}{2}\left(1-\frac{1}{7}\right)=\frac{1}{2}.\left(\frac{6}{7}\right)=\frac{3}{7}\)
VẬY \(S< \frac{3}{7}\)
\(\frac{2\sqrt{3}+3}{2\sqrt{3}-3}=\frac{\left(2\sqrt{3}+3\right)^2}{\left(2\sqrt{3}-3\right)\left(2\sqrt{3}+3\right)}=\frac{12+12\sqrt{3}+9}{12-3^2}=\frac{21+12\sqrt{3}}{3}=7+4\sqrt{3}>3+\sqrt{3}\)
thank you!!!!!!!