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B2 :P Ta có : \(B=\frac{2^{10}+1}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)
\(C=\frac{2^{10}-1}{2^{10}-3}=1+\frac{2}{2^{10}-3}\)
Nên : B > C
Ta có : các phân số từ 1/11 ; 1/12 đến 1/19 đều lớn hơn phân số 1/20
Từ đó lại có : 1/11 + 1/12 + 1/13 + ... + 1/19 + 1/20 > 1/20 + 1/20 + 1/20+ ...+ 1/20 ( số số hạng gồm 10 phân số 1/20)
=> 1/11+ 1/12+ 1/13+...+ 1/20 > 10/20
=> 1/11+1/12+1/13+...+1/20 > 1/2
<=> S > 1/2 .
Ta có :
\(S=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\) ( 10 số \(\frac{1}{20}\) )
\(S>\frac{1}{20}.10=\frac{10}{20}=\frac{1}{2}\)
Vậy \(S>\frac{1}{2}\)
Ta có \(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{10^{12}-1-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}\)
\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{10^{11}+1+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}\)
Vì \(\frac{9}{10^{12}-1}< \frac{9}{10^{11}+1};1=1\Rightarrow1-\frac{9}{10^{12}-1}< 1+\frac{9}{10^{11}+1}\Rightarrow\frac{10^{11}-1}{10^{12}-1}< \frac{10^{10}+1}{10^{11}+1}\)
Suy ra\(A< B\)
\(A=\frac{10^{11}-1}{10^{12}-1}\) => \(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{10^{12}-1-9}{10^{12}-1}\)
=> \(10A=1-\frac{9}{10^{12}-1}\)=> 10A < 1
\(B=\frac{10^{10}+1}{10^{11}+1}\) => \(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{10^{11}+1+9}{10^{11}+1}\)
=> \(10B=1+\frac{9}{10^{11}+1}\)=> 10B > 1
=> 10B > 10A => B > A
ĐS: B > A
a) \(\frac{7}{5}.\frac{-31}{125}.\frac{1}{2}.\frac{10}{17}.\frac{-1}{2^3}=\frac{7.\left(-31\right).1.10.\left(-1\right)}{5.2.125.17.2^3}=\frac{31.7}{17.125.2^3}=\frac{217}{17000}\)
b) \(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).\left(\frac{-5}{12}+\frac{1}{4}+\frac{1}{6}\right)=\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{20}{31}\right).0=0\)
c) \(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)...\left(\frac{1}{99}+1\right)=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{100}{99}=\frac{3.4.5...100}{2.3.4...99}=\frac{100}{2}=50\)
d) \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}...\frac{-99}{100}=\frac{-\left(1.2.3..99\right)}{2.3.4...100}=-\frac{1}{100}\)
e) \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{29.31}{30.30}=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}=\frac{\left(1.2.3..29\right).\left(3.4.5...31\right)}{\left(2.3.4...30\right).\left(2.3.4...30\right)}\)
\(=\frac{1.31}{30.2}=\frac{31}{60}\)
(x+1/4-1/3).(13/6-1/4)=7/46
(x+1/4-1/3).23/12=7/46
(x+1/4-1/3)=7/46:23/12
(x+1/4-1/3)=7/46.12/23
(x+1/4-1/3)=42/529
x+1/4=42/529+1/3
x+1/4=655/1587
x=655/1587-1/4
x=1033=/6348
vậy x=1033/6348
Chọn P/S trung gian rồi đi so sánh
Mặc dù mk học rồi nhưng quên mất
Bạn thông cảm nhé
~~~~~ Chúc bạn học tốt ~~~~~~
B= 2761/2520
=> B>1
k nha
Ta có :
1/4+1/5<1/4+1/4=1/2
1/6+1/7<1/6+1/6=1/3
1/8+1/9<1/8+1/8=1/4
và 1/10
vậy B=1/2+1/3+1/4+1/10
vì : 1/2+1/3+1/4>1
nên B>1