\(\frac{1989\times1990+3978}{1992\times1991-3984}=\frac{1989\times1990+1989\times2}{1992\times1991-1992\times2}=\frac{1989\times\left(1990+2\right)}{1992\times\left(1991-2\right)}=\frac{1989\times1992}{1992\times1989}=1\)

Đáp số: 1.

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AI NHANH MÌNH K CHO

=1.004520344

Thấy đúng k cho tui

1 nha bạn

này có phải 3979 là 3980 không

1998*1990+3978=1998*1990+1989*2=1998*(1990+2)=1998*1992

1992*1991-3984=1992*1991-1992*2=1992*(1991-2)=1992*1989

Sau đó rút gọn lại đáp án là 1

\(=\dfrac{1989\left(1990+2\right)}{1992\left(1991-2\right)}=\dfrac{1989}{1989}\cdot\dfrac{1992}{1992}=1\)

$\left(1989.1990+3978\right):\left(1992.1991-3984\right)$

$=\left[1989.\left(1990+2\right)\right]:\left[1992\left(1991-2\right)\right]=\left(1989.1992\right):\left(1992.1989\right)=1$

\(\left(1989.1990+3978\right):\left(1992.1991-3984\right)\)

\(=\left[1989.\left(1990+2\right)\right]:\left[1992\left(1991-2\right)\right]=\left(1989.1992\right):\left(1992.1989\right)=1\)

1989 . 1990 + 3978 1992 . 1991 − 3984  ⇒  1989 . 1990 + 1989 . 2 1992 . 1991 − 1992 . 2  ⇒  1989 . ( 1990 + 2 ) 1992 . ( 1991 + 2 )    ⇒  1989 . 1992 1992 . 1989  ⇒  1

dấu chấm là dấu nhân

\(\dfrac{1989\times1990+3978}{1992\times1991-3984}\)

\(=\dfrac{1989\times1990+1989\times2}{1992\times1991-1992\times2}\)

\(=\dfrac{1989\times\left(1990+2\right)}{1992\times\left(1991+2\right)}\)

\(=\dfrac{1989\times1992}{1992\times1989}\)

\(=1\)

\(\left(1989\times1990+3978\right):\left(1992\times1991-3984\right)\\ =\left(1989\times1990+1989\times2\right):\left(1992\times1991-1992\times2\right)\\ =\left(1989\times1992\right):\left(1992\times1989\right)\\ =1\)

\(\frac{-315}{540}\)

\(=\frac{-7\cdot45}{12\cdot45}\)

\(=\frac{-7}{12}\)

\(\frac{25\cdot13}{26\cdot35}=\frac{5\cdot5\cdot13}{13\cdot2\cdot7\cdot5}=\frac{5}{2\cdot7}=\frac{5}{14}\)

= 1 nhé bn

a) ( 1989 x 1990 + 3978 ) : ( 1992 x 1991 - 3984 )

= 3 962 088 : 3 962 088

= 1