So sánh:
E=\(\frac{1995^3+1}{1995^2-1994}\)và F=\(\frac{1996^3-1}{^{1996^2}+1997}\)
Giúp mk nha
# Thầy Soobin
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(F=\frac{1996^3-1}{1996^2+1997}=\frac{\left(1996-1\right)\left(1996^2+1996+1\right)}{1996^2+1997}=\frac{1995.\left(1996^2+1997\right)}{1996^2+1997}=1995\)
E = \(\frac{1995^3}{1995^2-1994}=\frac{1995^3+1-1}{1995^2-1994}=\frac{\left(1995+1\right)\left(1995^2-1995+1\right)-1}{1995^2-1994}\)
=\(\frac{1996\left(1995^2-1994\right)-1}{1995^2-1994}=1996-\frac{1}{1995^2-1994}\)
Vì \(1995^2-1994>0\) => \(\frac{1}{1995^2-1994}-1\) => \(1996-\frac{1}{1995^2-1994}>1996-1\)
HAy E > F
a)37/53x23/48x535353/373737x242424/232323
=37/53x23/48x53/37x24/23
=851/2544x1272/851
=1/2
\(\frac{2}{7}\)+ \(\frac{5}{14}\)+\(\frac{1}{7}\)+ \(\frac{3}{14}\)=\(\frac{4}{14}\)+\(\frac{5}{14}\)+\(\frac{2}{14}\)+\(\frac{3}{14}\)=\(\frac{14}{14}\)=1
469x281+489x719=469x281+(469+20)x719=469x281+469x719+20x719=469x(281+719)+1438=469x1000+1438=469000+1438=470438
a\(\frac{2}{5}\)+\(\frac{5}{14}\)+\(\frac{1}{7}\)+\(\frac{3}{14}\)=\(\frac{53}{70}\)+\(\frac{1}{7}\)=\(\frac{9}{10}\)+\(\frac{3}{14}\)=\(\frac{39}{35}\)
b\(\frac{1995.1997-1}{1996.1995+1994}\)=3984008001
c 469x281+489x719
=(489-469)x(281+719)
=20x1000
=20000
\(\dfrac{1995x1997-1}{1996x1995+1994}=\dfrac{1995x\left(1996+1\right)-1}{1996x1995+1994}\)
\(=\dfrac{1995x1996+1995-1}{1996x1995+1994}=\dfrac{1995x1996+1994}{1996x1995+1994}=1\)
\(\frac{1995.1997-1}{1996.1995+1994}\)
\(=\frac{1995.1996+1995-1}{1996.1995+1994}\)
\(=\frac{1995.1996+1994}{1996.1995+1994}=1\)
#
-\(\frac{1995\cdot1997-1}{1995\cdot1996+1994}\)
=\(\frac{1995\cdot\left(1996+1\right)-1}{1995\cdot1996+1994}\)
=\(\frac{1995\cdot1996+1995\cdot1-1}{1995\cdot1996+1994}\)
=\(\frac{1995\cdot1996+1994}{1995\cdot1996+1994}\)
=1