\(\frac{1978\times1999-1}{1978\times1998+1997}=\frac{1978\times1998+\left(1998-1\right)}{1978\times1998+1997}=\frac{1978\times1998+1997}{1978\times1998+1997}=1\)

3992006 phan 7980012

1000 nha

cảm ơn bạn nha

\(=\left(1999\times1998+1998\times1997\right)\times\left(1+\dfrac{1}{2}:1\dfrac{1}{2}-1\dfrac{1}{3}\right)\)

\(=\left(1999\times1998+1998\times1997\right)\times\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)\)

\(=\left(1999\times1998+1998\times1997\right)\times\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\)

\(=\left(1999\times1998+1998\times1997\right)\times\left(\dfrac{4}{3}-\dfrac{4}{3}\right)\)

\(=\left(1999\times1998+1998\times1997\right)\times0\)

\(=0\)

(1999×1998+1998+1997)×(1/1+1/2:3/2-4/3)

(1999×1998+1998+1997)×(1/1+1/2×2/3-4/3)

1999×1998+1998+1997)×(1/1+1/3-4/3)

(1999×1998+1998+1997)×(4/3-4/3)

(1999×1998+1998+1997)×0

0

bằng 0 nha bn

Ta có: 

Vậy giá trị biểu thức:  ( 1999   x   1998   +   1998   +   1997 )   x 1 + 1 2 : 1 1 2 - 1 1 3 = 0

\(\left(1999.1998+1998+1997\right).\left(1+\frac{1}{2}:1\frac{1}{2}-1\frac{1}{3}\right)\)

\(=\left(1999.1998+1998+1997\right).\left(1+\frac{1}{2}:\frac{3}{2}-\frac{4}{3}\right)\)

\(=\left(1999.1998+1998+1997\right).\left(1+\frac{1}{3}-\frac{4}{3}\right)\)

\(=\left(1999.1998+1998+1997\right).\left(\frac{4}{3}-\frac{4}{3}\right)\)

\(=\left(1999.1998+1998+1997\right).0=0.\)