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mk lam luon nhe!
Bot vao moi ve 3 don vi, ta co
\(\left(\frac{x-7}{50}-1\right)+\left(\frac{x-6}{51}-1\right)+\left(\frac{x-5}{52}-1\right)=\left(\frac{x-52}{5}-1\right)+\left(\frac{x-51}{6}-1\right)+\left(\frac{x-50}{7}-1\right)\)
Quy dong len ,ta co
\(\frac{x-57}{50}+\frac{x-57}{51}+\frac{x-57}{52}=\frac{x-57}{5}+\frac{x-57}{6}+\frac{x-57}{7}\)
\(\frac{x-57}{50}+\frac{x-57}{51}+\frac{x-57}{52}-\frac{x-57}{5}-\frac{x-57}{6}-\frac{x-57}{7}=0\)
(x-57).\(\left(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}-\frac{1}{5}-\frac{1}{6}-\frac{1}{7}\right)=0\)
Ma \(\left(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}-\frac{1}{5}-\frac{1}{6}-\frac{1}{7}\right)\) khac 0 nen => x-57=0
x=0+57 =57
Vay x =57.
Mk chac chan 100% bai nay dung
S = \(\frac{1}{50}+\frac{1}{51}+...+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{1}{100}.50=\frac{1}{2}\)
Kết luận vậy S > 1/2
\(y=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)=\left(1+\frac{1}{2}+\frac{1}{3}...+\frac{1}{100}\right)-2\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
=>y=B
=>y-B=0
y=\(y=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}......+\frac{1}{99}-\frac{1}{100}=\frac{1}{1}-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)
vay ket =\(\frac{99}{100}\)
Sửa đề: \(7^{52}+7^{51}-7^{50}\)
\(=7^{50}\left(7^2+7-1\right)=7^{50}\cdot55⋮55\)
Biến đổi vp của đẳng thức :
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{50}\)
\(=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}-2\left[\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right]\)
\(=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}-\frac{1}{2}-\frac{1}{4}-...-\frac{1}{200}\)
\(A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}+\frac{1}{100}\)
\(A=\frac{1}{51}+\frac{1}{52}+\frac{1}{51}+\frac{1}{99}+\frac{1}{100}\)
\(A=\left(\frac{1}{51}+\frac{1}{51}\right)+\frac{1}{52}+\frac{1}{99}+\frac{1}{100}\)
\(A=\frac{2}{51}+\frac{1}{52}+\frac{1}{99}+\frac{1}{100}\)
\(A=\frac{155}{1652}+\frac{1}{99}+\frac{1}{100}\)
\(A=\frac{5999}{87516}+\frac{1}{100}\)
\(A=0.078547465\)
\(A=\frac{34}{7.13}+\frac{51}{13.22}+\frac{85}{22.37}+\frac{68}{37.49}\)
\(A=17.\left(\frac{2}{7.13}+\frac{3}{13.22}+\frac{5}{22.37}+\frac{4}{37.49}\right)\)
\(A=\frac{17}{3}.\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)
\(A=\frac{17}{3}.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+\frac{1}{22}-\frac{1}{37}+\frac{1}{37}-\frac{1}{49}\right)\)
\(A=\frac{17}{3}.\left(\frac{1}{7}-\frac{1}{49}\right)\)
\(A=\frac{17}{3}.\frac{6}{49}\)
\(B=\frac{39}{7.16}+\frac{65}{16.31}+\frac{52}{31.43}+\frac{26}{43.49}\)
\(B=13.\left(\frac{3}{7.16}+\frac{5}{16.31}+\frac{4}{31.43}+\frac{2}{43.49}\right)\)
\(B=\frac{13}{3}.\left(\frac{9}{7.16}+\frac{15}{16.31}+\frac{12}{31.43}+\frac{4}{43.49}\right)\)
\(B=\frac{13}{3}.\left(\frac{1}{7}-\frac{1}{16}+\frac{1}{16}-\frac{1}{31}+\frac{1}{31}-\frac{1}{43}+\frac{1}{43}-\frac{1}{49}\right)\)
\(B=\frac{13}{3}.\left(\frac{1}{7}-\frac{1}{49}\right)=\frac{13}{3}.\frac{6}{49}\)
\(\frac{A}{B}=\frac{\frac{17}{3}.\frac{6}{49}}{\frac{13}{3}.\frac{6}{49}}=\frac{17}{13}\)
E = (252 + 251 + 250 + ... + 52 + 51 + 50) / (252 - 251 + 250 - 249 + ... + 52 - 51 + 50)
Số số hạng của dãy số là:
252-50+1=202+1=203(số)
252+251+...+51+50
\(=\left(252+50\right)\cdot\frac{203}{2}=302\cdot\frac{203}{2}=151\cdot203\)
Ta có: 252-251+250-249+...+52-51+50
=(252-251)+(250-249)+...+(52-51)+50
=1+1+...+1+50
\(=\frac{202}{2}+50=101+50=151\)
Ta có: E=\(\frac{252+251+...+51+50}{252-251+\cdots+52-51+50}\)
\(=\frac{151\cdot203}{151}=203\)