Tính
(1989x1990+3978):(1992x1991-3984)
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a/\(\frac{2^3\cdot3^4}{2^2\cdot3^2\cdot5}=\frac{18}{5}\)\(\frac{2^4\cdot5^2\cdot11^2\cdot7}{2^3\cdot5^3\cdot7^2\cdot11}=\frac{2\cdot11}{5\cdot7}=\frac{22}{35}\)
b/\(\frac{121\cdot75\cdot130\cdot169}{39\cdot60\cdot11\cdot198}=\frac{11^2\cdot5^3\cdot13^3\cdot2\cdot3}{2^3\cdot3^4\cdot5\cdot11^2\cdot13}=\frac{5^2\cdot13^2}{2^2\cdot3^3}=\frac{4225}{108}\)
c/\(\frac{1998\cdot1990+3978}{1992\cdot1991-3984}=\frac{2^2\cdot3^3\cdot37\cdot5\cdot199+2\cdot3^2\cdot13\cdot17}{2^3\cdot3\cdot83\cdot11\cdot181-2^4\cdot3\cdot83}=\frac{2\cdot3^2\cdot11\cdot20101}{2^3\cdot3^3\cdot13\cdot17\cdot83}=\frac{11\cdot20101}{2^2\cdot3\cdot13\cdot17\cdot83}\)
\(\left(1989.1990+3978\right):\left(1992.1991-3984\right)\)
\(=\left[1989.\left(1990+2\right)\right]:\left[1992\left(1991-2\right)\right]=\left(1989.1992\right):\left(1992.1989\right)=1\)
Mik sửa lại đề
\(\frac{1989.1990+3978}{1992\cdot1991-3984}\)
Trả lời
\(\frac{1989.1990+3978}{1992.1991-3984}\)
\(=\frac{1989.1990+1989\cdot2}{1992.1991-1992.2}\)
\(=\frac{1989.\left(1992-2\right)}{1992\left(1991-2\right)}\)
\(=\frac{1989.1990}{1992.1989}\)
\(=\frac{1990}{1992}\)
\(=\frac{995}{996}\)
\(=\dfrac{1989\left(1990+2\right)}{1992\left(1991-2\right)}=\dfrac{1989}{1989}\cdot\dfrac{1992}{1992}=1\)
= ( 1989.1990 + 1989.2) : ( 1992.1991 - 1992.2 )
= 1989.(1990 + 2 ) : 1992.( 1991 - 2 )
= 1989.1992 : 1992.1989
\(\Rightarrow\)= 1
(1989 x 1990 + 3978): (1992 x 1991 - 3984)
= (1989 x 1990 + 1990 x 2 - 2) : (1992x1991-3984)
= [ 1990 x ( 1989 +2) -2] : [ 1992 x1991 - 1991 x2 - 2]
= (1990 x 1991 -2): [ 1991x (1992-2) -2]
= (1990 x 1991 - 2):( 1991 x 1990 -2)
= 1