\(B=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+...+\frac{19}{9^2.10^2}\)
\(B=\frac{3}{1.4}+\frac{5}{4.9}+...+\frac{19}{81.100}\)
\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{81}-\frac{1}{100}\)
\(B=1-\frac{1}{100}< 1\left(đpcm\right)\)
CHỨNG MINH RẰNG BIỂU THỨC NÀY CHIA HẾT CHO 3:
1+3+5+7+9-9-11-13-15-17-19:3
Chứng minh:
\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}< 1\)
Chứng minh rằng: \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+......+\frac{19}{9^2.10^2}< 1\)
Chứng minh rằng:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\)
1.Chứng minh rằng: \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^3.4^2}+...+\frac{19}{9^2.10^2}< 1\)
2.Chứng minh rằng: \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
Làm nhanh giúp mình nhé mọi người !!!
Chứng minh rằng: \(\frac{3}{1^2.2^2}\)+ \(\frac{5}{2^2.3^2}\)+ \(\frac{7}{3^2.4^2}\)+ ... + \(\frac{19}{9^2.10^2}\)<1
Chứng minh rằng:
\(\dfrac{7}{1^3.2^3}+\dfrac{19}{2^3.3^3}+\dfrac{37}{3^3.4^3}+...+\dfrac{29701}{99^3.100^3}< 1\)
TÍNH : 2/3/19+1/5-5/3/19+(-5/3/19-6/5)
tính
2/2/19+1/5-5/3/19+(-5/3/19-6/5)
