So sanh A va B:
\(A=\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+...+\frac{1}{43}+\frac{1}{44}\)
\(B=\frac{5}{6}\)
Chứng minh rằng
\(\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+...+\frac{1}{43}+\frac{1}{44}>\frac{5}{6}\)
a) \(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\)
b) \(\frac{\frac{15}{12}+\frac{3}{4}-1}{3-\frac{5}{6}+\frac{2}{3}}+\frac{\frac{16}{5}+\frac{16}{7}-\frac{16}{9}}{\frac{17}{5}+\frac{17}{7}-\frac{17}{9}}\)
\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\)
\(=\frac{71}{19}+\frac{13}{17}+\frac{35}{43}+6\)
\(=\frac{1454}{323}+\frac{35}{43}+6\)
\(=5,...+6\)
\(=11,...\)
\(Bai2a\)\(A=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\)
\(=\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}\)
\(=\sqrt{3}-2\)
\(VayA=\sqrt{3}-2\)
So sanh A va B biet rang:
\(A=\frac{10^{15}+1}{10^{16}+1}\) \(B=\frac{10^{16}+1}{10^{17}+1}\)
TOÁN 6 NHA
Trước hết ta so sánh 10A và 10B
Ta có:
\(10A=\frac{10^{16}+10}{10^{16}+1}=1+\frac{9}{10^{16}+1}\) \(10B=\frac{10^{17}+10}{10^{17}+1}=1+\frac{9}{10^{17}+1}\)
Vì: \(\frac{9}{10^{16}+1}>\frac{9}{10^{17}+1}\) nên 10A > 10B, do đó A>B
Ta thấy:B<1 vì 1015+1<1016+1
Theo quy tắc :\(\frac{a}{b}\)<\(\frac{a+m}{b+m}\)nên ta có: B =\(\frac{10^{16}+1}{10^{17}+1}\)<\(\frac{10^{16}+1+9}{10^{17}+1+9}\)<\(\frac{10^{16}+10}{10^{17}+10}\)<\(\frac{10\left(10^{15}+1\right)}{10\left(10^{16}+1\right)}\)=A
Suy ra B<A
trước hết ta so sánh 10A và 10B
10A =10^16+10/10^16+1 10B=10^17+10/10^17+1
10A=1+9/10^16+1 10B=1+9/10^17 +1
mà 1=1;9/10^16+1>9/10^17+1 nên 10A>10B nên A>B
so sanh A=\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}\)va B=\(\frac{7}{12}\)
So sanh \(A=\frac{1}{42}+\frac{1}{43}+.......+\frac{1}{80}\)va \(B=\frac{7}{12}\)
Ta có
\(\frac{7}{12}=\frac{3}{12}+\frac{4}{12}=\frac{1}{4}+\frac{1}{3}=\frac{20}{80}+\frac{20}{60}=\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}\)
\(>\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}\)
hay B>A
nhớ tick mình nha
Ta chia A làm hai phần mỗi phần 20 số hạng
\(C=\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\)với \(D=\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}\)
Xét \(C=\frac{1}{41}+\frac{1}{42}+...+\frac{1}{61}\)\(>\frac{1}{61}+\frac{1}{61}+\frac{1}{61}+...+\frac{1}{61}\)\(=\frac{1}{61}.20=\frac{1}{3}\)
Xét \(D=\frac{1}{61}+\frac{1}{62}+...+\frac{1}{80}>\frac{1}{80}+\frac{1}{80}+...+\frac{1}{80}=\frac{1}{80}.20=\frac{1}{4}\)
Mà A = C + D > \(\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
=> A > B
So sanh A và B :
\(A=\frac{10^{15}+1}{10^{16}+1}\); \(B=\frac{10^{16}+1}{10^{17}+1}\)
\(10A=\frac{10^{16}+10}{10^{16}+1}=\frac{10^{16}+1+9}{10^{16}+1}=\frac{1+9}{10^{16}+1}\)
\(10B=\frac{10^{17}+10}{10^{17}+1}=\frac{10^{17}+1+9}{10^{17}+1}=\frac{1+9}{10^{17}+1}\)
Vì\(\frac{1+9}{10^{16}+1}\)>\(\frac{1+9}{10^{17}+1}\)
=>10A>10B
=>A>B
so sanh A va B bit
A=\(\frac{1}{3}+\frac{1}{4}+...+\frac{1}{17}\)
B=\(\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2010}\)
Cho A=\(\frac{1}{4}.\frac{3}{6}.\frac{5}{8}....\frac{43}{46}.\frac{45}{48}\) và B=\(\frac{2}{5}.\frac{4}{7}.\frac{6}{9}.....\frac{44}{47}.\frac{46}{49}\).So sánh A và B
so sánh:
1)\(\frac{10^{11}-1}{10^{12}-1}\)và \(\frac{10^{10}+1}{10^{11}+1}\)
2) \(\frac{54.107-53}{53.107-54}\)và \(\frac{135.269-133}{135.269+135}\)
3)\(\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+...+\frac{1}{43}+\frac{1}{44}và\frac{5}{6}\)