CMR1/10^2+1/15^2+...+1/500^2<1/15
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CMR:1/10^2+1/15^2+...+1/500^2<1/15
CMR1×2-1/2!+2×3-1/2!+3×4-1/4!+...+2023×2024/2024!<2
TH1
42:x=6
x= 42 :6
X= 7
TH 2
36:x = 6
X = 36: 6
X= 6
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chung minh rang 1/10^2+1/15^2+1/20^2+...+1/500^2<1/15
500-{5[409-(2^3.3-21)^2]+10^3}:15
{[(3^2+1).10-(8:2+6)]:2}+55-10(10:5)^3
cho (x+z)(y+z)=1
Cmr1/(x-y)2+1/(x+z)2+1/(y+z)2 ≥4
1-3/2×10 - 3/4×15 - 3/8×15 -...-3/198×500
1/1 x 10 + 1/2 x 15 + 1/3 x 20+...........................................+1/98 x 495 + 1/99 x 500
cho tong S =4/3+1/4+1/5+...+1/8+1/9
CMR1 <S<2 chu y / la phan so
tính giá trị của tổng sau :
1/1 x 10 + 1/2 x 15 + 1/3 x 20 + ... + 1/89 x 495 + 1/ 99 x 500
\(\frac{1}{1\times10}+\frac{1}{2\times15}+\frac{1}{3\times20}+...+\frac{1}{98\times495}+\frac{1}{99\times500}\)
\(=\frac{1}{1\times2\times5}+\frac{1}{2\times3\times5}+\frac{1}{3\times4\times5}+...+\frac{1}{98\times99\times5}+\frac{1}{99\times100\times5}\)
\(=\frac{1}{5}\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\right)\)
\(=\frac{1}{5}\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{5}\times\left(1-\frac{1}{100}\right)=\frac{1}{5}\times\frac{99}{100}=\frac{99}{500}\)
\(\frac{1}{1\times10}+\frac{1}{2\times15}+\frac{1}{3\times20}+...+\frac{1}{98\times495}+\frac{1}{99\times500}\)
\(=\frac{1}{1\times2\times5}+\frac{1}{2\times3\times5}+\frac{1}{3\times4\times5}+...+\frac{1}{98\times90\times5}+\frac{1}{90\times100\times5}\)
\(=\frac{1}{5}\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\right)\)
\(=\frac{1}{5}\times\left(\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+...+\frac{99-98}{98\times99}+\frac{100-99}{99\times100}\right)\)
\(=\frac{1}{5}\times\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(=\frac{1}{5}\times\left(1-\frac{1}{100}\right)=\frac{99}{500}\)
cảm ơn bạn Xyz và bạn Đoàn Đức Hà nhìu (^-^) thank you !!!!!!!!!!!!!!!!!