24/1.3.5+24/3.5.7+24/5.7.9+...+24/25.27.29
24/1.3.5+24/3.5.7+24/5.7.9+...+24/25.27.29
ta có
\(A=6\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+..+\frac{4}{25.27.29}\right)=6\left(\frac{5-1}{1.3.5}+\frac{7-3}{3.5.7}+..+\frac{29-25}{25.27.29}\right)\)
\(=6\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+..+\frac{1}{25.27}-\frac{1}{27.29}\right)=6\left(\frac{1}{3}-\frac{1}{27.29}\right)\)
\(=2-\frac{2}{9.29}=\frac{520}{261}\)
A=24/1.3.5+24/3.5.7+24/5.7.9+...+24/25.27.29
A=24/1.3.5+24/3.5.7+24/5.7.9+...+24/25.27.29
\(A=\frac{24}{1.3.5}+\frac{24}{3.5.7}+\frac{24}{5.7.9}+...+\frac{24}{25.27.29}\)
\(A=6\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+\frac{4}{5.7.9}+...+\frac{4}{25.27.29}\right)\)
\(A=6\left(\frac{5-1}{1.3.5}+\frac{7-3}{3.5.7}+\frac{9-5}{5.7.9}+...+\frac{29-25}{25.27.29}\right)\)
\(A=6\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{25.27}-\frac{1}{27.29}\right)\)
\(A=6\left(\frac{1}{1.3}-\frac{1}{27.29}\right)=\frac{520}{261}\)
Tính A= 1.3.5+3.5.7+5.7.9+...+25.27.29
c/m : B=36/1.3.5+36/3.5.7+36/5.7.9+...+36/25.27.29 < 3
B = 9 . [ 4/1.3.5+4/3.5.7+4/5.7.9+...+4/25.27.29]
B = 9 . [ 1/3-1/783]
= 9 . [ 1/3-1/783]
= 9 . 260/783=260/87<261/87<3
Chứng minh rằng: B=36/1.3.5+36/3.5.7+36/5.7.9+...+36/25.27.29<3.
Chứng minh B=36/1.3.5+36/3.5.7+36/5.7.9+...+36/25.27.29<3
Chứng minh rằng: B=36/1.3.5+36/3.5.7+36/5.7.9+...+36/25.27.29<3.
ai đó kết bạn với mình nha mình hết lời rùi
CMR \(B=\dfrac{36}{1.3.5}+\dfrac{36}{3.5.7}+\dfrac{36}{5.7.9}+...+\dfrac{36}{25.27.29}< 3\)