So sánh 2 biểu thức:
A= \(\frac{10^{2016}+3}{10^{2016}-2}\)và B=\(\frac{10^{2016}}{10^{2016}-5}\)
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a/ Ta có
\(200-\left(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)\)
\(=1+2\left(1-\frac{1}{3}\right)+2\left(1-\frac{1}{4}\right)+...+2\left(1-\frac{1}{100}\right)\)
\(=1+2\left(\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\right)\)
\(=2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\)
Thế lại bài toán ta được:
\(\frac{200-\left(3+\frac{2}{3}+\frac{2}{4}+...+\frac{2}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}\)
\(=\frac{2\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)}{\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}}=2\)
b/ Ta có:
A - B\(=\frac{-21}{10^{2016}}+\frac{12}{10^{2016}}+\frac{21}{10^{2017}}-\frac{12}{10^{2017}}\)
\(=\frac{9}{10^{2017}}-\frac{9}{10^{2016}}< 0\)
Vậy A < B
\(A-1=\frac{10^{2016}+2}{10^{2016}-1}=\frac{3}{10^{2016}-1}\)
\(B-1=\frac{10^{2016}}{10^{2016}-3}-1=\frac{3}{10^{2016}-3}\)
Vì \(1< 3\Rightarrow10^{2016}-1>10^{2016}-3\Rightarrow\frac{3}{10^{2016}-1}< \frac{3}{10^{2016}-3}\Rightarrow A-1< B-1\Rightarrow A< B\Rightarrow\)
\(\frac{10^{2016}+2}{10^{2016}-1}=\frac{10^{2016}-1+3}{10^{2016}-1}=1+\frac{3}{10^{2016}-1}\)
\(\frac{10^{2016}}{10^{2016}-3}=\frac{10^{2016}-3+3}{10^{2016}-3}=1+\frac{3}{10^{2016}-3}\)
vì\(1< 3\Rightarrow10^{2016}-1>10^{2016}-3\Rightarrow\frac{3}{10^{2016-1}}< \frac{3}{10^{2016}-3}\Rightarrow A< B\)
Áp dung công thức \(a>b\Leftrightarrow\frac{a}{b}>\frac{a+m}{b+m}\)
\(B=\frac{10^{2017}+1}{10^{2016}+1}>\frac{10^{2017}+1+9}{10^{2016}+1+9}=\frac{10^{2017}+10}{10^{2016}+10}=\frac{10\left(10^{2016}+1\right)}{10\left(10^{2015}+1\right)}=\frac{10^{2016}+1}{10^{2015}+1}=A\)
\(\Leftrightarrow B>A\)
Xét \(A=\frac{10^{2014}+2016}{10^{2015}+2016}\Rightarrow10A=\frac{10^{2015}+20160}{10^{2015}+2016}=\frac{10^{2015}+2016+18144}{10^{2015}+2016}=1+\frac{18144}{10^{2015}+2016}\)
Xét \(B=\frac{ 10^{2015}+2016}{10^{2016}+2016}\Rightarrow10B=\frac{10^{2016}+20160}{10^{2016}+2016}=\frac{10^{2016}+2016+18144}{10^{2016}+2016}=1+\frac{18144}{10^{2016}+2016}\)
Có \(\frac{18144}{10^{2015}+2016}>\frac{18144}{10^{2016}+2016}\)
\(\Rightarrow10A>10B\Leftrightarrow A>B\)
\(\frac{10^{2016}+2^3}{9}=\frac{10^{2016}-1}{9}+\frac{2^3+1}{9}=\left(1+10+10^2+...+10^{2015}\right)+1\in N.\)