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vi ve A va ve B deu co (-12)/10^2017 nen ta chi viec so sanh (-21)/10^2017 voi (-12)/10^2017.Ma (-21)/10^2017<(-12)/10^2016 nen A < B
Lời giải:
$\frac{7}{10^{2015}}+\frac{15}{10^{2016}}-(\frac{7}{10^{2016}}+\frac{15}{10^{2015}})$
$=\frac{-8}{10^{2015}}+\frac{8}{10^{2016}}=8(\frac{1}{10^{2016}}-\frac{1}{10^{2015}})<0$
$\Rightarrow \frac{7}{10^{2015}}+\frac{15}{10^{2016}}< \frac{7}{10^{2016}}+\frac{15}{10^{2015}}$
\(A=\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}\)
\(=\frac{-7}{10^{2005}}+\frac{-7}{10^{2006}}+\frac{-8}{10^{2006}}\)
\(B=\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}\)
\(=\frac{-7}{10^{2005}}+\frac{-8}{10^{2005}}+\frac{-7}{10^{2006}}\)
Ta thấy: A và B đều có chung 2 hạng tử: \(\frac{-7}{10^{2006}};\frac{-7}{10^{2005}}\)
=> Muốn so sánh A và B thì ta so sánh: \(\frac{-8}{10^{2006}}\)và \(\frac{-8}{10^{2005}}\)
Mà \(\frac{-8}{10^{2006}}>\frac{-8}{10^{2005}}\)
=> A > B
\(A=\frac{10^{2016}+1}{10^{2017}+1}\)
\(A=\frac{10^{2016}+1}{10^{2017}+1}+\frac{10^{2017}+1}{10^{2017}+1}\)
\(A=\frac{10^{2016}+1+10^{2017}+1}{10^{2017}+1}\)
\(A=\frac{10^{2016}+10^{2017}+1+1}{10^{2016}.10+1}\)
\(A=\frac{10^{2016}.\left(1+10\right)+2}{10^{2016}.10+1}\)
\(A=\frac{10^{2016}.11+2}{10^{2016}.10+1}\)
\(A=\frac{11+2}{10+1}\)
\(A=\frac{13}{11}\)(1)
Làm tương tự phần B
Từ 1 và 2
\(\Leftrightarrow\)\(\frac{13}{11}=\frac{13}{11}\)
\(\Leftrightarrow\)A = B
Ta có:
\(A-B=-\dfrac{7}{10^{2016}}-\dfrac{15}{10^{2017}}+\dfrac{15}{10^{2016}}+\dfrac{7}{10^{2017}}\)
\(=\left(\dfrac{15}{10^{2016}}-\dfrac{7}{10^{2016}}\right)+\left(\dfrac{7}{10^{2017}}-\dfrac{15}{10^{2017}}\right)\)
\(=\dfrac{9}{10^{2016}}-\dfrac{9}{10^{2017}}=9\left(\dfrac{1}{10^{2016}}-\dfrac{1}{10^{2017}}\right)>0\)
Vậy A > B
\(A=\dfrac{-7}{10^{2016}}+\dfrac{-15}{10^{2017}}=\dfrac{-7}{10^{2016}}+\dfrac{-8}{10^{2017}}+\dfrac{-7}{10^{2017}}\\ B=\dfrac{-15}{10^{2016}}+\dfrac{-7}{10^{2017}}=\dfrac{-7}{10^{2016}}+\dfrac{-8}{10^{2016}}+\dfrac{-7}{10^{2017}}\)
Vì \(\dfrac{-8}{10^{2017}}>\dfrac{-8}{10^{2016}}\) nên \(A>B\)