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đề phải là \(\frac{1989.1990+3978}{1992.1991-3984}\)
ta có
1989.1990+3978=1989.1990+1989.2=1989.1992
1992.1991-3984=1992.1991-1992.2=1992.1989
gộp cả 2 cái ta có \(\frac{1989.1992}{1992.1989}\)=1
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Câu 1: \(\frac{121.75.130.169}{39.60.11.198}\)=\(\frac{11x11x15x5x130x13x13}{13x13x15x4x11x11x18}\)= \(\frac{5x130}{4x18}\)= \(\frac{650}{72}\)= \(\frac{325}{36}\)
Câu 2: \(\frac{1998.1990+3978}{1992.1991-3984}\)= \(\frac{3976020}{3962088}\)= \(\frac{994005}{990522}\)
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a,\(\frac{2^3.3^4}{2^2.3^2.5}=\frac{2.3^2}{5}=\frac{2.9}{5}=\frac{18}{5}\)
b, \(\frac{2^4.5^2.11^2.7}{2^3.5^3.7^2.11}=\frac{2.11}{5.7}=\frac{22}{35}\)
c,\(\frac{1998.1990+3978}{1992.1991-3984}=\frac{18.111.1990+18.221}{24.83.1991-166.24}\)
\(=\frac{18.\left(111.1990+221\right)}{24.\left(83.1991-166\right)}\)
\(=\frac{3.221111}{4.165087}=\frac{221111}{4.55029}=\frac{221111}{220116}\)
\(\)
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a) 121.75.130.169 39.60.11.198 = 11 2 .5 2 .3.13.5.2.13 2 3.13.2 2 .3.5.11.2.3 2 = 11.5 2 .13 2 2 2 .3 3
b) 1998.1990 + 3978 1992.1991 − 3984 = ( 1991 − 2 ) .1990 + 3978 ( 190 + 2 ) .1991 − 3984
= 1990.1991 − 3980 + 3978 1990.1991 + 3982 − 3984 = 1990.1991 − 2 1990.1991 − 2 = 1
á đù đúng kwqo ưqffjoisdifadlsdfrgdfgdsfgr