So Sánh \(\frac{12}{13}và\frac{11}{14}\)
So sánh giá trị của các biểu thức sau:
A=\(\frac{11}{\text{(12+13)}}\) +\(\frac{12}{\text{(13+14) }}\)+\(\frac{13}{\text{(14+12)}}\) ; B= \(\frac{\text{(11+12+13)}}{(12+13+14)}\)
Ai nhanh 3 tick nha
Trả lời
\(A=\frac{11}{12+13}+\frac{12}{13+14}+\frac{1}{14+15}\)
Hay
\(A=\frac{11}{12+13}+\frac{12}{13+14}+\frac{13}{14+15}\)
mong xem lại hộ cái
So sánh:
\(\frac{11^{13}+1}{11^{14}+1}\) và \(\frac{11^{14}+1}{11^{15}+1}\)
Vì 1113 . 1115 = 1114 . 1114 = 1128 nên \(\frac{11^{13}+1}{11^{14}+1}=\frac{11^{14}+1}{11^{15}+1}\)
Cho \(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{99}+\frac{1}{100}\)SO SÁNH A VỚI 1
a, Cho A=\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{99}+\frac{1}{100}\) . So Sánh A với 1
b, B=\(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\). So sánh B với \(\frac{1}{2}\)
c, cho M=\(\frac{2013}{2014}+\frac{2014}{2015}\)và N=\(\frac{2013+2014}{2014+2015}\). So sánh M và N
Câu a, p/s cuối cùng là \(\frac{1}{100}\)nha mí bn
a) Ta có :
\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}\)
\(>\frac{1}{10}+\frac{1}{100}.90=\frac{1}{10}+\frac{90}{100}=1\)
vậy A > 1
b) \(B=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\)
\(>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{1}{20}.10=\frac{1}{2}\)
Vậy B > \(\frac{1}{2}\)
\(\frac{1+2+3+4+5}{6+7+8+9+10}\)và \(\frac{11+12+13+14+15}{5+6+7+8+9}\)
So sánh nhé các bạn
Gợi ý: Rút gọn 2 ps, quy đồng rồi so sánh.
Bài 1: so sánh:
a, \(\frac{11^{13}+1}{11^{14}+1}\)và \(\frac{11^{14}+1}{11^{15}+1}\)
giúp mk nha!!!
1113+1/ 1114+1 = 1114+1/1115+1
\(S=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}\)
Hãy so sánh S và \(\frac{1}{2}\)
\(S=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+....+\frac{1}{20}\)
\(=\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}\right)\)
\(>\frac{1}{15}\cdot5+\frac{1}{20}\cdot5\)
\(=\frac{1}{3}+\frac{1}{4}\)
\(=\frac{7}{12}>\frac{6}{12}=\frac{1}{2}\)
\(\Rightarrow S>\frac{1}{2}\)
Bài làm
Ta có:
\(\frac{1}{11}>\frac{1}{20}\), \(\frac{1}{12}>\frac{1}{20}\), \(\frac{1}{13}>\frac{1}{20}\), \(\frac{1}{14}>\frac{1}{20}\), \(\frac{1}{15}>\frac{1}{20}\), \(\frac{1}{16}>\frac{1}{20}\), \(\frac{1}{17}>\frac{1}{20}\), \(\frac{1}{18}>\frac{1}{20}\),\(\frac{1}{19}>\frac{1}{20}\)
=> \(S=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}\)
hay \(\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}\)
=> \(S=\frac{1}{20}.10=\frac{10}{20}=\frac{1}{2}\)
Do đó: \(S=\frac{1}{2}\)
# Chúc bạn học tốt #
Ta có các phân số : \(\frac{1}{11};\frac{1}{12};\frac{1}{13};\frac{1}{14};\frac{1}{15};\frac{1}{16};\frac{1}{17};\frac{1}{18};\frac{1}{19}>\frac{1}{20}\)
Do đó : \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)có 10 phân số \(\frac{1}{20}\)
\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}>\frac{10}{20}\)
\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}>\frac{1}{2}\)
Vậy : \(S>\frac{1}{2}\)
Cho A =\(\frac{1}{10}\)+ \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{100}\)Hãy so sánh A với \(\frac{1}{2}\)
\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{100}\)
\(A< \frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{100.101}\)
\(A< \frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{101}\)
\(A< \frac{1}{10}-\frac{1}{101}=\frac{101}{1010}-\frac{10}{1010}=\frac{91}{1010}< \frac{505}{1010}\)
\(A< \frac{1}{2}\)
So sánh:
\(\frac{9}{10};\frac{3}{4};\frac{11}{14};\frac{13}{18}\)