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Ta có:
\(A=\frac{67^{2016}}{67^{2016}-11}=1+\frac{11}{67^{2016}-11}\)
\(B=\frac{67^{2016}+13}{67^{2016}+2}=1+\frac{11}{67^{2016}+2}\)
Vì \(67^{2016}-11< 67^{2016}+2\) nên \(\frac{11}{67^{2016}-11}>\frac{11}{67^{2016}+2}\Rightarrow1+\frac{11}{67^{2016}-11}>1+\frac{11}{67^{2016}+2}\)
Vậy A > B
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\frac{79}{67}+\frac{28}{41}\) \(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\) \(=\frac{1}{3}+\left(-1\right)+1=\frac{1}{3}\)
Chúc bạn hc giỏi ,Trần Phúc Đông❣ღ!
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{7.9}{67}-\frac{27}{41}\right)\)
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{63}{67}-\frac{27}{41}\right)\)
\(=\frac{6836}{8241}-\frac{774}{2747}\)
\(=\frac{4514}{8241}\)
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
=\(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
=\(\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
=\(\frac{1}{3}+\left(\frac{-67}{67}\right)+\left(\frac{41}{41}\right)\)
=\(\frac{1}{3}-1+1\)
=\(\frac{1}{3}\)
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
=\(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
=\(\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
=\(\frac{1}{3}+\left(\frac{-67}{67}\right)+\left(\frac{41}{41}\right)\)
=\(\frac{1}{3}-1+1\)
=\(\frac{1}{3}\)
\(A=\frac{67^{2016}}{67^{2016}}-11=1-11=-10< 0\)
\(\Rightarrow A< 0\)
\(B=\frac{67^{2016}+3}{67^{2016}+2}>0\)
\(\Rightarrow B>0\)
Vậy \(B>A\)
\(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}-\frac{28}{41}\)
\(=\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)+\frac{1}{3}\)
\(=\left(-1\right)+1+\frac{1}{3}\)
\(=0+\frac{1}{3}\)
\(=\frac{1}{3}\)
=1/3+12/67+13/41-79/67+28/41
=1/3+(12/67-79/67)+(13/41+28/41)
=1/3+(-1)+1
=1/3+0
=1/3
67x6
67 + 67 + 67 + 67 + 67 + 67
= 67 × 6
= 402
Đáp số: 402.