so sánh :\(\frac{4}{5}\sqrt{3}+\frac{9}{13}\sqrt{2}\) với \(2.4\)
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3 tháng 8 2018
a)Ta có: \(2\sqrt{5}< 5\sqrt{2}\)\(2\sqrt{5}=\sqrt{2^2.5}=\sqrt{20}\)
\(5\sqrt{2}=\sqrt{5^2.2}=\sqrt{50}\)
Vì \(\sqrt{20}< \sqrt{50}\)
Nên \(2\sqrt{5}< 5\sqrt{2}\)
b)Ta có: \(3\sqrt{13}=\sqrt{3^2.13}=\sqrt{117}\)
\(4\sqrt{11}=\sqrt{4^2.11}=\sqrt{176}\)
Vì \(\sqrt{117}< \sqrt{176}\)
Nên \(3\sqrt{13}< 4\sqrt{11}\)
c) Ta có: \(\frac{3}{4}.\sqrt{7}=\sqrt{\left(\frac{3}{4}\right)^2.7}=\sqrt{\frac{63}{16}}\)
\(\frac{2}{5}.\sqrt{5}=\sqrt{\left(\frac{2}{5}\right)^2.5}=\sqrt{\frac{4}{5}}\)
Vì \(\sqrt{\frac{63}{16}}>1\)
\(\sqrt{\frac{4}{5}}< 1\)
Nên \(\sqrt{\frac{63}{16}}>\sqrt{\frac{4}{5}}\)
Vậy \(\frac{3}{4}.\sqrt{7}>\frac{2}{5}.\sqrt{5}\)
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 5 2016
\(0,5\sqrt{100}-\sqrt{\frac{4}{25}}=0,5.10-\frac{\sqrt{4}}{\sqrt{25}}=5-\frac{2}{5}=\frac{23}{5}=\frac{138}{30}\)
\(\left(\sqrt{1\frac{1}{9}-\sqrt{\frac{9}{16}}}\right):5=\left(\sqrt{\frac{10}{9}-\frac{3}{4}}\right):5=\sqrt{\frac{13}{36}}:5=\frac{\sqrt{13}}{6}:5=\frac{\sqrt{13}}{30}\)
Vì 13 < 138 nên \(\sqrt{13}< 138\Rightarrow\frac{\sqrt{13}}{30}< \frac{138}{30}\)
Vậy \(0,5\sqrt{100}-\sqrt{\frac{4}{25}}>\left(\sqrt{1\frac{1}{9}-\sqrt{\frac{9}{16}}}\right):5\).
FF
0
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.
NV
Nguyễn Việt Lâm
Giáo viên
24 tháng 9 2019
\(D=\sqrt{5}-\sqrt{13-4\sqrt{\left(\sqrt{5}-2\right)^2}}=\sqrt{5}-\sqrt{13-4\left(\sqrt{5}-2\right)}\)
\(=\sqrt{5}-\sqrt{21-4\sqrt{5}}=\sqrt{5}-\sqrt{\left(2\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}-2\sqrt{5}+1=1-\sqrt{5}\)
\(B=10\sqrt{5}+\left|1-\sqrt{5}\right|-\frac{4\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
\(=10\sqrt{5}+\sqrt{5}-1-\sqrt{5}+1=10\sqrt{5}\)
\(C=\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\frac{2+\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\frac{12\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
\(=\sqrt{3}-1+2+\sqrt{3}+2\left(3-\sqrt{3}\right)=7\)