So sánh
A=\(\frac{30^{31}+1}{30^{32}+1}\) và B=\(\frac{30^{32}+1}{30^{33}+1}\)
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So sánh
A=\(\frac{30^{31}+1}{30^{32}+1}\) và B=\(\frac{30^{32}+1}{30^{33}+1}\)
Help me!Mk đg cần rất gấp ai nhanh mk tick cho
22 tháng 7 2023 lúc 13:17
Cho S = 1/30 + 1/31 + 1/32 + 1/33 + ... + 1/49 . So sánh S với 2/3
1 cho 4 số nguyên a;b;c;d thỏa mãn: a+bc+d và cd+1abtính a^{10}-b^{10}2 Cho Afrac{30}{45}+frac{30}{105}+frac{30}{189}+...+frac{30}{1197} Bleft(frac{31}{2}.frac{32}{2}.frac{33}{2}......frac{60}{2}right):left(1.3.5...59right)a) tính Ab) so sánh Avà B
Đọc tiếp
1 cho 4 số nguyên a;b;c;d thỏa mãn:
a+b=c+d và cd+1=ab
tính \(a^{10}-b^{10}\)
2 Cho
\(A=\frac{30}{45}+\frac{30}{105}+\frac{30}{189}+...+\frac{30}{1197}\)
\(B=\left(\frac{31}{2}.\frac{32}{2}.\frac{33}{2}......\frac{60}{2}\right):\left(1.3.5...59\right)\)
a) tính \(A\)
b) so sánh \(A\)và \(B\)
A) \(\frac{1}{30+31}\)+ \(\frac{1}{31+32}\)+ \(\frac{1}{32+33}\)+ ... \(\frac{1}{42+43}\)
B) \(\frac{1}{3.5}\)+ \(\frac{1}{5.7}\)+ \(\frac{1}{7.9}\)+ \(\frac{1}{9.11}\)
A)
\(\frac{1}{30}\)-\(\frac{1}{31}\)+\(\frac{1}{31}\)-\(\frac{1}{32}\)+\(\frac{1}{32}\)-\(\frac{1}{33}\)+...+\(\frac{1}{42}\)-\(\frac{1}{43}\)
=\(\frac{1}{30}\)-\(\frac{1}{43}\)
=\(\frac{13}{1290}\)
B)
=\(\frac{2}{2}\)X(\(\frac{1}{3.5}\)+\(\frac{1}{5.7}\)+\(\frac{1}{7.9}\)+\(\frac{1}{9.11}\))
=\(\frac{1}{2}\)X(\(\frac{2}{3.5}\)+\(\frac{2}{5.7}\)+\(\frac{2}{7.9}\)+\(\frac{2}{9.11}\))
=\(\frac{1}{2}\)X(\(\frac{1}{3}\)-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)-\(\frac{1}{11}\))
=\(\frac{1}{2}\)X(\(\frac{1}{3}\)-\(\frac{1}{11}\))
=\(\frac{1}{2}\)X\(\frac{8}{33}\)
=\(\frac{8}{66}\)=\(\frac{4}{33}\)
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Bài 1: So sánh
\(M=\frac{20^{30}+2}{20^{31}+2}\) và \(N=\frac{20^{31}+2}{20^{32}+2}\)
Vì N<1
=> N= 20^31+2/20^32+2
<20^31+2+38/ 20^32+2+38
=20^31+40/ 20^32+40
=20.(20^30+2) / 20.(20^31+2)
=20^30+2 / 20^32+2 = M
Vậy N<M
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\(N=\frac{20^{31}+2}{20^{32}+2}=\frac{20^{31}+2+18}{20^{32}+2+18}=\frac{20^{31}+20}{20^{32}+20}=\frac{10.\left(20^{30}+2\right)}{10.\left(20^{31}+2\right)}\)\(=M\)
\(\Rightarrow M=N\)
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\(\left(\frac{1}{3}\right)^{^{30}}.x+\left(\frac{1}{3}\right)^{31}=\left(\frac{1}{3}\right)^{32}\)
\(\left(\frac{1}{3}\right)^{30}.x+\left(\frac{1}{3}\right)^{31}=\left(\frac{1}{3}\right)^{32}\)
\(\left(\frac{1}{3}\right)^{30}.\left(x+\frac{1}{3}\right)=\left(\frac{1}{3}\right)^{32}\)
\(x+\frac{1}{3}=\left(\frac{1}{3}\right)^{32}:\left(\frac{1}{3}\right)^{30}\)
\(x+\frac{1}{3}=\left(\frac{1}{3}\right)^2\)
\(x+\frac{1}{3}=\frac{1}{9}\)
\(x=\frac{1}{9}-\frac{1}{3}=\frac{1}{9}-\frac{3}{9}\)
\(x=-\frac{2}{9}\)
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1) Tính hợp lýa)frac{31}{23}-left(frac{7}{32}+frac{8}{32}right)b)left(frac{1}{3}+frac{12}{67}+frac{13}{41}right)-left(frac{79}{67}-frac{28}{41}right)c)frac{38}{45}-left(frac{8}{45}-frac{17}{51}-frac{3}{11}right)d)left(frac{17}{28}+frac{18}{29}-frac{19}{30}-frac{30}{31}right).left(-frac{5}{12}+frac{1}{4}+frac{1}{6}right)
Đọc tiếp
1) Tính hợp lý
a)\(\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{32}\right)\)
b)\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
c)\(\frac{38}{45}-\left(\frac{8}{45}-\frac{17}{51}-\frac{3}{11}\right)\)
d)\(\left(\frac{17}{28}+\frac{18}{29}-\frac{19}{30}-\frac{30}{31}\right).\left(-\frac{5}{12}+\frac{1}{4}+\frac{1}{6}\right)\)
Trả lời
b)(1/3+12/67+13/41)-(79/67-28/41)
=1/3+12/67+13/41-79/67+28/41
=1/3+(12/67-79/67)+(13/41+28/41)
=1/3+(-67/67)+41/41
=1/3+(-1)+1
=1/3+0
=1/3.
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c)38/45-(8/45-17/51-3/11)
=38/45-8/45+17/51+3/11
=30/45+1/3+3/11
=2/3+1/3+3/11
=3/3+3/11
=1+3/11
=1 3/11.
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d)(17/28+18/29-19/30-30/31).(-5/12+1/4+1/6)
=(17/28+18/29+19/30+30/31).(-5+3+2/12)
=(17/28+18/29+19/30+30/31).0
=0.
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So sánh 1/30+1/32+1/33+...+1/60 và 1/2
so sánh:
M=10^30+1/10^31+1 và N=10^31+1/10^32+1
\(M=\frac{10^{30}+1}{10^{31}+1}\)
\(\Rightarrow10M=\frac{10\cdot(10^{30}+1)}{10^{31}+1}\)
\(\Rightarrow10M=\frac{10^{31}+10}{10^{31}+1}\)
\(\Rightarrow10M=\frac{10^{31}+1+9}{10^{31}+1}\)
\(\Rightarrow10M=1+\frac{9}{10^{31}+1}\)
\(N=\frac{10^{31}+1}{10^{32}+1}\)
\(\Rightarrow10N=\frac{10\cdot(10^{31}+1)}{10^{32}+1}\)
\(\Rightarrow10N=\frac{10^{32}+10}{10^{32}+1}\)
\(\Rightarrow10N=\frac{10^{32}+1+9}{10^{32}+1}\)
\(\Rightarrow10N=1+\frac{9}{10^{32}+1}\)
Mà\(1+\frac{9}{10^{31}+1}>1+\frac{9}{10^{32}+1}\)
Nên \(10M>10N\)
Hay \(M>N\)
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