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Lời giải:
$\frac{7}{10^{2015}}+\frac{15}{10^{2016}}-(\frac{7}{10^{2016}}+\frac{15}{10^{2015}})$
$=\frac{-8}{10^{2015}}+\frac{8}{10^{2016}}=8(\frac{1}{10^{2016}}-\frac{1}{10^{2015}})<0$
$\Rightarrow \frac{7}{10^{2015}}+\frac{15}{10^{2016}}< \frac{7}{10^{2016}}+\frac{15}{10^{2015}}$
Ta có :
\(A=\frac{2016^{2016}+2}{2016^{2016}-1}=\frac{2016^{2016}-1+3}{2016^{2016}-1}=1+\frac{3}{2016^{2016}-1}\)
\(B=\frac{2016^{2016}}{2016^{2016}-3}=\frac{2016^{2016}-3+3}{2016^{2016}-3}=1+\frac{3}{2016^{2016}-3}\)
Do \(\frac{3}{2016^{2016}-1}< \frac{3}{2016^{2016}-3}\)
\(\Rightarrow1+\frac{3}{2016^{2016}-1}< 1+\frac{3}{2016^{2016}-3}\)
\(\Rightarrow A< B\)
Vậy \(A< B\)
Chúc bạn học tốt !!!
\(A=\frac{2016^{2016}+2}{2016^{2016}-1}=\frac{2016^{2016}-1+3}{2016^{2016}-1}=\frac{2016^{2016}-1}{2016^{2016}-1}+\frac{3}{2016^{2016}-1}=1+\frac{3}{2016^{2016}-1}\)
\(B=\frac{2016^{2016}}{2016^{2016}-3}=\frac{2016^{2016}-3+3}{2016^{2016}-3}=\frac{2016^{2016}-3}{2016^{2016}-3}+\frac{3}{2016^{2016}-3}=1+\frac{3}{2016^{2016}-3}\)
Vì \(1=1;\frac{3}{2016^{2016}-1}<\frac{3}{2016^{2016}-3}\)nên \(1+\frac{3}{2016^{2016}-1}<1+\frac{3}{2016^{2016}-3}\)
\(=>\)\(A\)\(<\)\(B\)
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 \(A>B\)
A= (2016^2016+2)/(2016^2016-1)=(2016^2016-1+3)/(2016^2016-1)=(2016^2016-1)/2016^2016-1)+(3/2016^2016-1)=1+(3/2016^2016-1) B=( 2016^2016)/(2016^2016-3)=(2016^2016-3+3)/(2016^2016-3)=(2016^2016-3)/(2016^2016-3) +(3/2016^2016-3)=1+(3/2016^2016-3) Vì 3/(2016^2016-1)<3/(2016^2016-3) Nên A<B
A=2016^2016+2/2016^2016-1>1
=>(2016^2016)+2/(2016^2016)-1<(2016^2016)+2-2/(2016^2016)-1-2=2016^2016/(2016^2016)-3=B
\(C=\dfrac{15^{2016}-3+5}{15^{2016}-3}=1+\dfrac{5}{15^{2016}-3}\)
\(D=\dfrac{15^{2016}-2+5}{15^{2016}-2}=1+\dfrac{5}{15^{2016}-2}\)
mà \(15^{2016}-3< 15^{2016}-2\)
nên C>D