\(n=\frac{1995\times1994-1}{1993\times1995+1994}\)

\(n=\frac{1995\times\left(1993+1\right)-1}{1995\times1993+1994}\)

\(n=\frac{1995\times1993+1995\times1-1}{1995\times1993+1994}\)

\(n=\frac{1995\times1993+1994}{1995\times1993+1994}\)

\(n=1\)(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

Ông Thắng nhanh ghê

An à ông đừng khen bạn bè cứ phải khách sao =))

= \(\frac{1996\times\left(1994+1\right)-996}{1000+1996\times1994}=\frac{1996\times1994+1996-996}{1000+1996\times1994}\)

= \(\frac{1996\times1994+1000}{1000+1996\times1994}\)

= 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

bằng 1/2 mới đúng

phải =1 chứ nhỉ

\(A=\dfrac{1995\times1994-1}{1993\times1995+1994}\)

\(A=\dfrac{1995\times\left(1993+1\right)-1}{1993\times1995+1994}\)

\(A=\dfrac{1995\times1993+1995-1}{1995\times1993+1994}\)

\(A=\dfrac{1995\times1993+1994}{1995\times1993+1994}\)

\(A=1\)

mik cần gấp

(1995 × 1994 - 1)/(1993 × 1995 + 1994)

= 3978029/3978029

= 1

\(=\dfrac{1994\left(1994+1\right)-1}{1994^2-1+1994}=\dfrac{1994^2+1993}{1994^2+1993}=1\)

=3980022,999

=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 \(C=\frac{1995}{1997}.\frac{1990}{1993}.\frac{1997}{1994}.\frac{1993}{1995}.\frac{997}{995}\)

      \(C=\frac{1995.1990.1997.1993.997}{1997.1993.1994.1995.995}\)

      \(C=\frac{1990.997}{1994.995}\)

      \(C=\frac{995.2+997}{997.2+995}=1\)

\(B=\frac{3+33+333+3333+ 33333}{4+44+444+4444+44444}\)

\(\Rightarrow B=\frac{3\left(1+11+111+1111+11111\right)}{4\left(1+11+111+1111+11111\right)}=\frac{3}{4}\)

B=\(\frac{3+33+333+3333+33333}{4+44+444+4444+44444}\)

B= \(\frac{3.\left(1+11+111+1111+11111\right)}{4.\left(1+11+111+1111+11111\right)}\)

B=\(\frac{3}{4}\)

Sau mình làm tiếp vội quá! k mình nha