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Lời giải:
$\frac{37}{67}=1-\frac{30}{67}$
$\frac{377}{677}=1-\frac{300}{677}$
Mà $\frac{300}{677}< \frac{300}{670}=\frac{30}{67}$
$\Rightarrow 1-\frac{300}{677}>1-\frac{30}{67}$
Hay $\frac{377}{677}> \frac{37}{67}$
Ta có : \(\frac{37}{67}=\frac{370}{670}\)
Mà \(670>370\)
=> \(\frac{370}{670}< 1\)
=> \(\frac{370}{670}< \frac{7}{7}\)
Mà \(\frac{370}{670}=\frac{370}{670}\)
=> \(\frac{370}{670}< \frac{377}{677}\)
=> \(\frac{37}{67}< \frac{377}{677}\)
Vậy \(\frac{37}{67}< \frac{377}{677}\) .
Ta có : 37<67
nên \(\frac{37}{67}=\frac{37.10}{67.10}< \frac{370+7}{670+7}=\frac{377}{677}\)(do a<b thì \(\frac{a}{b}< \frac{a+m}{b+m}\))
=>\(\frac{37}{67}< \frac{377}{677}\)(dpcm)
Ta có: \(A=\dfrac{-9}{10^{2010}}+\dfrac{-19}{10^{2011}}=\dfrac{-9}{10^{2010}}+\dfrac{-9}{10^{2011}}+\dfrac{-10}{10^{2011}}\)
\(B=\dfrac{-9}{10^{2011}}+\dfrac{-19}{10^{2010}}=\dfrac{-9}{10^{2011}}+\dfrac{-9}{10^{2010}}+\dfrac{-10}{10^{2010}}\)
So sánh A với B ta thấy: \(\dfrac{-9}{10^{2010}}=\dfrac{-9}{10^{2010}};\dfrac{-9}{10^{2011}}=\dfrac{-9}{10^{2011}}\)
Mà \(\dfrac{-10}{10^{2011}}>\dfrac{-10}{10^{2010}}\)
\(\Rightarrow\) \(\dfrac{-9}{10^{2010}}+\dfrac{-9}{10^{2011}}+\dfrac{-10}{10^{2011}}>\dfrac{-9}{10^{2010}}+\dfrac{-9}{10^{2011}}+\dfrac{-10}{10^{2010}}\)
\(\Rightarrow\) \(A>B\)
Vậy A > B.
\(A=\left(\frac{-9}{2016}+\frac{-9}{2017}\right)+\frac{-10}{2017}\\ B=\left(\frac{-9}{2017}+\frac{-9}{2016}\right)+\frac{-10}{2016}\\ Do\frac{-10}{2017}>\frac{10}{2016}\)
nên A>B
a)
\(10A=\frac{10^{2002}+10}{10^{2002}+1}=1+\frac{9}{10^{2002}+1}\)
\(10B=\frac{10^{2003}+10}{10^{2003}+1}=1+\frac{9}{10^{2003}+1}\)
=> 10A > 10B => A > B
làm được bài nào hay bài ấy!
Bài 2:
Ta có: \(A=\frac{-9}{10^{2010}}+\frac{-19}{10^{2011}}=\frac{-90}{10^{2011}}+\frac{-19}{10^{2011}}=\frac{-109}{10^{2011}}\)
\(B=\frac{-9}{10^{2011}}+\frac{-19}{10^{2010}}=\frac{-9}{10^{2011}}+\frac{-190}{10^{2011}}=\frac{-199}{10^{2011}}\)
Vì \(\frac{-109}{10^{2011}}>\frac{-199}{10^{2011}}\) nên A > B
Vậy A > B
Câu 1:
a) A tự tính
b) gợi ý: \(\frac{131313}{565656}=\frac{10101.13}{10101.56}=\frac{13}{56}\)
c) Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}=\frac{2a+3b+4c+5d}{2a+3b+4c+5d}=1\)
Ta có: \(C=\frac{2a}{3b}+\frac{3b}{4c}+\frac{4c}{5d}+\frac{5d}{2a}=1+1+1+1=4\)
Vậy C = 4
Câu 2:
a) \(\frac{x+1}{2}=\frac{8}{x+1}\)
\(\Rightarrow\left(x+1\right)^2=16\)
\(\Rightarrow\left[\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Vậy \(x\in\left\{3;-5\right\}\)
b) \(x:\left(9\frac{1}{2}-\frac{3}{2}\right)=\frac{0,4+\frac{2}{9}-\frac{2}{11}}{1,6+\frac{8}{9}-\frac{8}{11}}\)
\(\Rightarrow x:\left(\frac{19}{2}-\frac{3}{2}\right)=\frac{2\left(0,2+\frac{1}{9}-\frac{1}{11}\right)}{8\left(0,2+\frac{1}{9}-\frac{1}{11}\right)}\)
\(\Rightarrow\frac{x}{8}=\frac{2}{8}\)
\(\Rightarrow x=2\)
Vậy x = 2
Câu 3:
a) tìm x, y bằng cách \(\left[\begin{matrix}\overline{34x5y}⋮9\\\overline{34x5y}⋮4\end{matrix}\right.\)
Lưu ý: số chia hết cho 4 có 2 chữ số cuối của nó chia hết cho 4
b) Ta có: \(A=\frac{-9}{10^{2010}}+\frac{-19}{10^{2011}}=\frac{-90}{10^{2011}}+\frac{-19}{10^{2011}}=\frac{-109}{10^{2011}}\)
\(B=\frac{-9}{10^{2011}}+\frac{-19}{10^{2010}}=\frac{-9}{10^{2011}}+\frac{-190}{10^{2011}}=\frac{-199}{10^{2011}}\)
Vì \(\frac{-109}{10^{2011}}>\frac{-199}{10^{2011}}\) nên A > B
Vậy A > B