tính nhanh
\(\frac{1995\times1997-1}{1996\times1995+199}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1997.1996+1=1997.(1995+1)+1=1997.1995+19971+1=1997.1995+1998
\(\frac{1997x1996+1}{1995x1997+1996}\)=\(\frac{1996+1}{1995+1996}\)=\(\frac{1}{1995}\)
Đặt A = 1996 x 1995 - 996
Đặt B = 1000 + 1996 x 1994
Ta có:
A = (1996 - 996) x 1995
= 1000 x 1995
1995000
B = 1000 x (1996 + 1994)
= 1000 x 3990
= 3990000
Ta rút gọn:1995000/3990000 = 2
Mình tính ra = 2 nhưng ko biết có đúng ko
\(\frac{1996X1997+1998X3}{1997X1999-1997X1997}\)=\(\frac{3986012+5994}{3992003-3988009}\)=\(\frac{3992006}{3994}\)=\(\frac{1996003}{1997}\)
1996 x 1997 + 1998 x \(\frac{3}{1997}\)x 1999 - 1997 x 1997
= 3986012 + \(\frac{5994}{1997}\)x 1999 - 1997 x 1997
= 3986012 + 6000003005 x 1999 - 1997 x 1997
= 3986012 + 1199400601 - 3988009
= 1203386613 - 3988009
= 1199398604
= \(\frac{1996\times\left(1994+1\right)-996}{1000+1996\times1994}=\frac{1996\times1994+1996-996}{1000+1996\times1994}\)
= \(\frac{1996\times1994+1000}{1000+1996\times1994}\)
= 1
\(\frac{1995.1994-1}{1993.1995+1994}=\frac{1995\left(1993+1\right)-1}{1995.1993+1994}\)
\(=\frac{1995.1993+1995.1-1}{1995.1993+1994}=\frac{1995.1993+1994}{1995.1993+1994}\)
=1
\(\frac{1995\cdot1994-1}{1993\cdot1995+1994}=\frac{1995\cdot\left(1993+1\right)-1}{1993\cdot1995+1994}=\frac{1995\cdot1993+1995-1}{1993\cdot1995+1994}=\frac{1995\cdot1993+1994}{1995\cdot1993+1994}=1\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{1}-\frac{1}{8}\)
\(=\frac{7}{8}\)'
\(n=\frac{1995\times1994-1}{1993\times1995+1994}\)
\(n=\frac{1995\times\left(1993+1\right)-1}{1995\times1993+1994}\)
\(n=\frac{1995\times1993+1995\times1-1}{1995\times1993+1994}\)
\(n=\frac{1995\times1993+1994}{1995\times1993+1994}\)
\(n=1\)(vì TS = MS)
\(\frac{1995.1994-1}{1993.1995+1994}=\frac{1995.1994-1}{\left(1994-1\right)1995+1994}=\frac{1995.1994-1}{1994.1005-1995+1994}=\frac{1995.1994-1}{1994.1995-1}=1\)
Vậy n = 1
1995(1996+1)-1/1996.1995+1994
1995.1996+1995-1/1996.1995+1994
1995.1996+1994/1996.1995+1994=1
\(\frac{1995.1997-1}{1996.1995+199}\)= \(\frac{1995.\left(1996+1\right)-1}{1996.1995+199}\)
= \(\frac{1995.1996+1995.1-1}{1996.1995+199}\)
= \(\frac{1995.1996+1994}{1996.1995+199}\)
làm đến đây thì mình chịu