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Lời giải:
Ta có:
\(N=\dfrac{-7}{10^{2005}}+\dfrac{-15}{10^{2006}}=\dfrac{-7}{10^{2005}}+\dfrac{-7}{10^{2006}}+\dfrac{-8}{10^{2006}}\)
\(M=\dfrac{-15}{10^{2005}}+\dfrac{-7}{10^{2006}}=\dfrac{-7}{10^{2005}}+\dfrac{-8}{10^{2005}}+\dfrac{-7}{10^{2006}}\)
Xét \(N\) và \(M\) có \(\dfrac{-7}{10^{2005}}+\dfrac{-7}{10^{2006}}\) chung.
Mà \(\dfrac{-8}{10^{2005}}>\dfrac{-8}{10^{2006}}\) nên \(N>M\).
\(a,\frac{20132013}{20142014}=\frac{2013.10001}{2014.10001}=\frac{2013}{2014}=1-\frac{1}{2014};\frac{131313}{141414}=\frac{13.10101}{14.10101}=\frac{13}{14}=1-\frac{1}{14}.\text{Vì: 14 bé hơn 2014 nên:}\frac{1}{14}>\frac{1}{2014}\Rightarrow\frac{20132013}{20142014}>\frac{131313}{141414}\)
\(C=2013^9+2013^9.2013=2013^9\left(2013+1\right)=2013^9.2014;D=2014^9.2014\text{ vì: 2013^9< 2014^9 nên: C bé thua D }\)
\(c,M=\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}=\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2006}};N=\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2005}}.Vì:10^{2006}>10^{2005}.Nên:\frac{-8}{10^{2006}}>\frac{-8}{10^{2005}}\Rightarrow M>N\)
a) Vì A=\(\dfrac{15^{16}+1}{15^{17}+1}\) < 1
\(\Rightarrow\dfrac{15^{16}+1}{15^{17}+1}< \dfrac{15^{16}+1+14}{15^{17}+1+14}=\dfrac{15^{16}+15}{15^{17}+15}\) \(=\dfrac{15\left(15^{15}+1\right)}{15\left(15^{16}+1\right)}\) \(=\dfrac{15^{15}+1}{15^{16}+1}\)
Vậy A<B
b) A=\(\dfrac{2006^{2007}+1}{2006^{2006}+1}>1\)
\(\Rightarrow\dfrac{2006^{2007}+1+2005}{2006^{2006}+1+2005}\)
= \(\dfrac{2006^{2007}+2006}{2006^{2006}+2006}\)
= \(\dfrac{2006\left(2006^{2006}+1\right)}{2006\left(2006^{2005}+1\right)}\)
= \(\dfrac{2006^{2006+1}}{2006^{2005}+1}\)
Vậy A>B
a, \(A-B=\frac{3}{8^3}+\frac{7}{8^4}-\frac{7}{8^3}-\frac{3}{8^4}==\left(\frac{7}{8^4}-\frac{3}{8^4}\right)-\left(\frac{7}{8^3}-\frac{3}{8^3}\right)=\frac{4}{8^4}-\frac{4}{8^3}< 0\)
Vậy A < B
b, \(A=\frac{10^7+5}{10^7-8}=\frac{10^7-8+13}{10^7-8}=1+\frac{13}{10^7-8}\)
\(B=\frac{10^8+6}{10^8-7}=\frac{10^8-7+13}{10^8-7}=1+\frac{13}{10^8-7}\)
Vì \(10^7-8< 10^8-7\Rightarrow\frac{1}{10^7-8}>\frac{1}{10^8-7}\Rightarrow\frac{13}{10^7-8}>\frac{13}{10^8-7}\Rightarrow A>B\)
c,Áp dụng nếu \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{a+n}\) có:
\(B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)
Vậy A < B
mình ghi nhầm nên các bạn cứ hết hai phân số là một câu nhé ví dụ như \(\dfrac{-5}{8}\):\(\dfrac{15}{4}\)
\(\dfrac{2006\times2005-1}{2004\times2006+2005}=\dfrac{2006\times\left(2004+1\right)-1}{2004\times2006+2005}\)
\(=\dfrac{2004\times2006+2006-1}{2004\times2006+2005}=\dfrac{2004\times2006+2005}{2004\times2006+2005}\)
\(=1\)
\(18\times\left(\dfrac{19191919+88888}{21212121+99999}\right)=18\times\left(\dfrac{19}{21}+\dfrac{8}{9}\right)\)
\(=18\times\dfrac{113}{63}=\dfrac{226}{7}=32\dfrac{2}{7}\)
d, Vì B=10^1993+1/10^1992+1 > 1 =>10^1993+1/10^1992+1>10^1993+1+9/10^1992+1+9 = 10^1993+10/10^1992+10= 10. (10^1992+1)/10. (10^1991+1) = 10^1992+1/10^1991+1=A Vậy A=B
cau d B>1 ta co tinh chat (\(\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ) B> \(\dfrac{10^{1993}+1+9}{10^{1992}+1+9}\)\(=\dfrac{10^{1993}+10}{10^{1992}+10}\)=\(\dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\dfrac{10^{1992}+1}{10^{1991}+1}\)=A
Suy ra B>A(chuc ban hoc goi nhe)
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
Chị sử dụng cách làm lớp 7 ở câu 3 nha em
em cũng tự quy đồng và suy ra cách làm của cô giáo dạy em nha
chữ cj xấu thì mong em thông cảm
1, \(\dfrac{1717}{8585}=\dfrac{17.101}{85.101}\&\dfrac{1313}{5151}=\dfrac{13.101}{51.101}\)
\(\Leftrightarrow\dfrac{1}{5}\&\dfrac{13}{51}\)
Ta thấy \(\dfrac{1}{5}< \dfrac{13}{51}\Rightarrow\dfrac{1717}{8585}< \dfrac{1313}{5151}\)