\(\frac{1995\times1994-1}{1993\times1995+1994}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n=\frac{1995\times1994-1}{1993\times1995+1994}\)
\(n=\frac{1995\times\left(1993+1\right)-1}{1995\times1993+1994}\)
\(n=\frac{1995\times1993+1995\times1-1}{1995\times1993+1994}\)
\(n=\frac{1995\times1993+1994}{1995\times1993+1994}\)
\(n=1\)(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
\(\frac{1995.1994-1}{1993.1995+1994}=\frac{1995\left(1993+1\right)-1}{1995.1993+1994}\)
\(=\frac{1995.1993+1995.1-1}{1995.1993+1994}=\frac{1995.1993+1994}{1995.1993+1994}\)
=1
= \(\frac{1996\times\left(1994+1\right)-996}{1000+1996\times1994}=\frac{1996\times1994+1996-996}{1000+1996\times1994}\)
= \(\frac{1996\times1994+1000}{1000+1996\times1994}\)
= 1
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
\(A=\dfrac{1995\times1994-1}{1993\times1995+1994}\)
\(A=\dfrac{1995\times\left(1993+1\right)-1}{1993\times1995+1994}\)
\(A=\dfrac{1995\times1993+1995-1}{1995\times1993+1994}\)
\(A=\dfrac{1995\times1993+1994}{1995\times1993+1994}\)
\(A=1\)
\(=\dfrac{1994\left(1994+1\right)-1}{1994^2-1+1994}=\dfrac{1994^2+1993}{1994^2+1993}=1\)
=1995*1994-1/(1994-1)*1995+1994
=1995*1994-1/1994*1995-1995+1994
=1995*1994-1/1994*1995-1
=1
\(\dfrac{1995.1994-1}{1993.1995+1994}=\dfrac{1995.1994-1}{1993.1995+1994}=\dfrac{1993.1995+1995-1}{1993.1995+1994}=1\)
\(B=\)\(\frac{3+33+333+3333+33333}{4+44+444+4444+44444}\)
\(B=\frac{3.1+3.11+3.111+3.1111+3.11111}{4.1+4.11+4.111+4.1111+4.11111}\)
\(B=\frac{3.\left(1+11+111+1111+11111\right)}{4.\left(1+11+111+1111+11111\right)}\)
\(B=\frac{3}{4}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\)
\(A.2=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right).2\)
\(A.2=\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\)
=>\(A.2-A=\left(\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}\right)-\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}\right)\)
\(A=\frac{2}{3}-\frac{1}{192}\)
\(A=\frac{127}{192}\)
\(\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)
Đặt \(C=\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)
\(C=\frac{1995.1990.1997.1993.997}{1997.1993.1994.1995.995}\)
\(C=\frac{1990.997}{1994.995}\)
\(C=\frac{995.2+997}{997.2+995}=1\)
\(B=\frac{3+33+333+3333+ 33333}{4+44+444+4444+44444}\)
\(\Rightarrow B=\frac{3\left(1+11+111+1111+11111\right)}{4\left(1+11+111+1111+11111\right)}=\frac{3}{4}\)
\(\frac{1995\cdot1994-1}{1993\cdot1995+1994}=\frac{1995\cdot\left(1993+1\right)-1}{1993\cdot1995+1994}=\frac{1995\cdot1993+1995-1}{1993\cdot1995+1994}=\frac{1995\cdot1993+1994}{1995\cdot1993+1994}=1\)