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\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{97}{48^2.49^2}+\frac{99}{49^2.50^2}\)
\(\Leftrightarrow\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{97}{2304.2401}+\frac{99}{2401.2500}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{2304}-\frac{1}{2401}+\frac{1}{2401}-\frac{1}{2500}\)
\(\Leftrightarrow\frac{1}{1}-\frac{1}{2500}=\frac{2499}{2500}< 1\left(đpcm\right)\)
ta có 1/51>1/100
1/52>1/100
..................
1/100=1/100
\(\Rightarrow\)S=1/51+1/52+...+1/100>(1/100+1/100+...+1/100)=1/100.50=1/2
\(\Rightarrow\)S>\(\frac{1}{2}\)
cái chỗ 1/100+1/100+...+1/100 có 50 số bạn nhá
chúc bạn học tốt~
\(\frac{E}{F}=\frac{5}{2}\) Chỉ nhớ kết quả thôi Hoàng Minh Đ.... à !
Giải:
\(S=\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{98}+\dfrac{1}{99}\)
\(S=\left(\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{74}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{98}+\dfrac{1}{99}\right)\)
\(\Rightarrow S>\left(\dfrac{1}{50}+\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{75}+...+\dfrac{1}{75}+\dfrac{1}{75}\right)\)
\(\Rightarrow S>\dfrac{1}{2}+\dfrac{1}{3}>\dfrac{1}{2}\)
\(\Rightarrow S>\dfrac{1}{2}\left(đpcm\right)\)
\(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{98}+\frac{1}{99}>\frac{1}{2}\)
\(\Rightarrow\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{98}+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\)(50 phân số 1/100)
\(\Rightarrow\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{98}+\frac{1}{99}>\frac{50}{100}=\frac{1}{2}\left(đpcm\right)\)
\(\frac{1}{50}+\frac{1}{51}+...+\frac{1}{98}+\frac{1}{99}>\frac{1}{2}\)
\(\Rightarrow\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{98}+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\)(50 cái như z)
\(\Rightarrow\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{98}+\frac{1}{99}>\frac{50}{100}=\frac{1}{2}\left(đpcm\right)\)
#It's the moment when you're in good mood, you accidentally click back =.=
1) Calculate
\(P=1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}....1\frac{1}{63}.1\frac{1}{80}\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{64}{63}.\frac{81}{80}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{8.8}{7.9}.\frac{9.9}{8.10}\)
\(=\frac{2.9}{10}=\frac{9}{5}\)
ta có: 10010 + 1 > 10010 - 1
⇒ A = \(\frac{100^{10}+1}{100^{10}-1}< \frac{100^{10}+1-2}{100^{10}-1-2}=\frac{100^{10}-1}{100^{10}-3}=B\)
vậy A < B
