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Đây nha
Ta có:
(1−�2)(1−�)>0(1−a2)(1−b)>0
⇔1+�2�>�2+�>�3+�3(1)⇔1+a2b>a2+b>a3+b3(1)
(Vì 0<�,�<10<a,b<1)
Tương tự ta có:
\hept{1+�2�>�3+�3(2)�+�2�>�3+�3(3)\hept{1+b2c>b3+c3(2)a+c2a>c3+a3(3)
Cộng (1), (2), (3) vế theo vế ta được
2(�3+�3+�3)<3+�2�+�2�+�2�2(a3+b3+c3)<3+a2b+b2c+c2a
Đúng(0)
\(B=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{19}{9^2.10^2}\)
\(B=\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{19}{81.100}\)
\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{81}-\frac{1}{100}\)
\(B=1-\frac{1}{100}< 1\left(đpcm\right)\)
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
=> đpcm
Ủng hộ mk nha ^_-
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{19}{9^2.10^2}=\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{19}{81.100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{81}-\frac{1}{100}=1-\frac{1}{100}<1\)
Vậy \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{19}{9^2.10^2}<1\)
a)ta có 3B=1+1/3+1/3^2+........+1/3^2003+1/3^2004
B= 1/3+1/3^2+........+1/3^2003+1/3^2004+1/3^2005
suy ra 2B=1-1/3^2005
suy ra B=\(\frac{1-\frac{1}{3}^{2005}}{2}\)
suy ra B=1/2-1/3^2005/2 bé hơn 1/2
từ đấy suy ra B bé hơn 1/2
2
a) (2x - 1)4 = 81
<=> \(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=4\\2x=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
\(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
= \(\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
= \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{81}-\frac{1}{100}\)
= \(\frac{1}{1}-\frac{1}{100}=\frac{99}{100}< 1\)
=) \(A< 1\) (ĐPCM)
\(B=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}\)
\(B=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+....+\frac{19}{81.100}\)
\(B=\frac{4-1}{1.4}+\frac{9-4}{4.9}+\frac{16-9}{9.16}+....+\frac{100-81}{81.100}\)
\(B=\frac{4}{1.4}-\frac{1}{1.4}+\frac{9}{4.9}-\frac{4}{4.9}+\frac{16}{9.16}-\frac{9}{9.16}+...+\frac{100}{81.100}-\frac{81}{81.100}\)
\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+....+\frac{1}{81}-\frac{1}{100}\)
\(B=1-\frac{1}{100}< 1\)
=> B < 1 (Đpcm)
B = 3/12.22 + 5/22.32 + 7/32.42 + ... + 19/92.102
B = 3/1.4 + 5.4.9 + 7/9.16 + ... + 19/81.100
B = 1 - 1/4 + 1/4 - 1/9 + 1/9 - 1/16 + ... + 1/81 - 1/100
B = 1 - 1/100 < 1 ( đpcm)