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Ta có :
\(\frac{11}{15}=\frac{11\times14}{15\times14}=\frac{154}{210}\);
\(\frac{13}{14}=\frac{13\times15}{14\times15}=\frac{195}{210}\)
Vì : \(\frac{154}{210}< \frac{195}{210}\)nên \(\frac{11}{15}< \frac{13}{14}\)
Đặt \(A=\frac{2^{15}+1}{2^{16}+1}\)
\(\Rightarrow2A=\frac{2^{16}+2}{2^{16}+1}=\frac{2^{16}+1+1}{2^{16}+1}=1+\frac{1}{2^{16}+1}\)
Đặt \(B=\frac{2^{14}+1}{2^{15}+1}\)
\(\Rightarrow2B=\frac{2^{15}+2}{2^{15}+1}=\frac{2^{15}+1+1}{2^{15}+1}=1+\frac{1}{2^{15}+1}\)
Vì 216+1>215+1
\(\Rightarrow\frac{1}{2^{16}+1}< \frac{1}{2^{15}+1}\)
\(\Rightarrow1+\frac{1}{2^{16}+1}< 1+\frac{1}{2^{15}+1}\)
\(\Rightarrow2A< 2B\Rightarrow A< B\)
Vậy...
a) 3 2 + − 4 3 = 1 6
1 10 + − 4 5 = − 7 10
Mà 1 6 > − 7 10 nên 3 2 + − 4 3 > 1 10 + − 4 5
b) 1 2 + 1 3 + 1 4 + 1 5 + 1 6 = 29 20 < 2
Nên 1 2 + 1 3 + 1 4 + 1 5 + 1 6 < 2
Bài 1:
Ta có:
\(\left(\frac{1}{10}\right)^{15}=\left(\frac{1}{5}\right)^{3.5}=\left(\frac{1}{125}\right)^5\)
\(\left(\frac{3}{10}\right)^{20}=\left(\frac{3}{10}\right)^{4.5}=\left(\frac{81}{10000}\right)^5\)
Lại có:
\(\frac{1}{125}=\frac{80}{10000}< \frac{81}{10000}\Rightarrow\left(\frac{1}{125}\right)^5< \left(\frac{81}{10000}\right)^5\)
\(\Rightarrow\left(\frac{1}{10}\right)^{15}< \left(\frac{3}{10}\right)^{20}\)
Bài 2:
Ta có:
\(A=\frac{13^{15}+1}{13^{16}+1}\Rightarrow13A=\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)
\(B=\frac{13^{16}+1}{13^{17}+1}\Rightarrow13B=\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)
Mà \(\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\)
\(\Rightarrow1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)
\(\Rightarrow13A>13B\Rightarrow A>B\)
\(S=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+....+\frac{1}{20}\)
\(=\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}\right)\)
\(>\frac{1}{15}\cdot5+\frac{1}{20}\cdot5\)
\(=\frac{1}{3}+\frac{1}{4}\)
\(=\frac{7}{12}>\frac{6}{12}=\frac{1}{2}\)
\(\Rightarrow S>\frac{1}{2}\)
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:
\(A=\frac{10^{15}+1}{10^{16}+1}\)
\(10A=\frac{10^{16}+10}{10^{16}+1}\)
\(B=\frac{10^{16}+1}{10^{17}+1}\)
\(10B=\frac{10^{17}+10}{10^{17}+1}\)
Ta so sánh \(10A\) và \(10B\)
Có:
\(10A:\) Mẫu - tử = 9
\(10B:\) Mẫu - tử = 9
Lại có:
\(\frac{10^{16}+10}{10^{16}+1}\) \(-1\)\(=\frac{9}{10^{16}+1}\)
\(\frac{10^{17}+10}{10^{17}+1}-1=\frac{9}{10^{17}+1}\)
Vì \(\frac{9}{10^{16}+1}\)\(>\frac{9}{10^{17}+1}\)nên \(10A>10B\)
\(\Rightarrow\)\(A>B\)
Vậy \(A>B\)
Theo bải ra ta có:
A=\(\frac{10^{15}+1}{10^{16}+1}\)=> 10A =.\(\frac{10.\left(10^{15}+1\right)}{10^{16}+1}\)= \(\frac{10.10^{15}+1.10}{10^{16}+1}\)
= \(\frac{10.10^{15}+10}{10^{16}+1}\)=\(\frac{10^{16}+1+9}{10^{16}+1}\)= \(1+\frac{9}{10^{16}+1}\)
B= \(\frac{10^{16}+1}{10^{17}+1}\)=> 10B = \(\frac{10.\left(10^{16}+1\right)}{10^{17}+1}\)=\(\frac{10.10^{16}+1.10}{10^{17}+1}\)
= \(\frac{10.10^{16}+10}{10^{17}+1}\)= \(\frac{10^{17}+1+9}{10^{17}+1}\)= \(1+\frac{9}{10^{17}+1}\)
Vì 1=1 mà \(\frac{9}{10^{16}+1}\)> \(\frac{9}{10^{17}+1}\)nên => 10A > 10B => A>B
Vậy A>B.
Ta có
\(\frac{13}{27}:\frac{13}{27}=1\)
\(\frac{7}{15}:\frac{13}{27}=\frac{63}{65}\)
Mặt khác \(\frac{63}{65}< 1\)
\(\Rightarrow\frac{13}{27}:\frac{13}{27}>\frac{7}{15}:\frac{23}{27}\)
\(\Rightarrow\frac{13}{27}:\frac{13}{27}\times\frac{13}{27}>\frac{7}{15}:\frac{23}{27}\times\frac{13}{27}\)
\(\Rightarrow\frac{13}{27}>\frac{7}{15}\)
\(\frac{-13}{14}>\frac{-15}{16}\)