121.75.130.169/39.60.11.198
1998.1990+3978/1992.1991-3984
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PM
8
27 tháng 2 2017
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: \(\frac{1998.1990+3978}{1992.1991-3984}\)= \(\frac{3976020}{3962088}\)= \(\frac{994005}{990522}\)
19 tháng 2 2018
\(a.\frac{2\cdot\left(-13\right)\cdot9\cdot10}{\left(-3\right)\cdot4\cdot\left(-5\right)\cdot26}\)
\(=\frac{2\cdot\left(-13\right)\cdot3\cdot3\cdot2\cdot5}{\left(-3\right)\cdot2\cdot2\cdot\left(-5\right)\cdot13\cdot2}\)
\(=-\frac{3}{2}\)
b) \(\frac{2^3\cdot3^4}{2^2\cdot3^2\cdot5}=\frac{2\cdot3^2}{5}=\frac{2\cdot9}{5}=\frac{18}{5}\)
\(\frac{2^4\cdot5^2\cdot11^2\cdot7}{2^3\cdot5^3\cdot7^2\cdot11}=\frac{2\cdot1\cdot11\cdot1}{1\cdot5\cdot7\cdot1}=\frac{22}{35}\)
c) \(\frac{121\cdot75\cdot130\cdot169}{39\cdot60\cdot11\cdot198}=\frac{11\cdot11\cdot13\cdot10\cdot169}{13\cdot3\cdot6\cdot10\cdot11\cdot11\cdot6\cdot3}\)
\(=\frac{169}{3\cdot6\cdot6\cdot3}=\frac{169}{324}\)
d) \(\frac{1998\cdot1990+3978}{1992\cdot1991-3984}\)
NT
2
RS
1
NT
0
5 tháng 2 2018
\(a,\frac{121.75.130.169}{39.60.11.198}=\frac{11^2.5^2.3.13.2.5.13^2}{13.3.5.2^2.3.11.3^2.2.11}\)\(=\frac{11^2.5^3.3.13^3.2}{13.3^4.5.2^3.11^2}=\frac{5^2.13^2}{3^3.2^2}=\frac{4425}{108}\)
\(b,\frac{1989.1990+3978}{1992.1991-3984}=\frac{1989.1990+1989.2}{1992.1991-1992.2}\)\(=\frac{1989\left(1990+2\right)}{1992\left(1991-2\right)}=\frac{1989.1992}{1992.1989}=\frac{1}{1}=1\)
\(c.\frac{135.350+135.550}{900.100+35.900}=\frac{135\left(350+550\right)}{900\left(100+35\right)}=\)\(\frac{135.900}{900.135}=\frac{1}{1}=1\)
\(d.\frac{243.650-243.350}{600.200+600.43}=\frac{243\left(650-350\right)}{600\left(200+43\right)}\)\(=\frac{243.300}{600.243}=\frac{300}{600}=\frac{1}{2}\)
Ta có:
\(\frac{121.75.130.169}{39.60.11.198}=\frac{11^2.3.5^2.2.5.13.13^2}{3.13.2^2.3.5.11.2.3^2.11}=\frac{11^2.3.5^3.2.13^3}{3^4.13.2^3.5.11^2}=\frac{5^2.13^2}{3^3.2^2}=\frac{4225}{108}\)
\(\frac{1998.1990+3978}{1992.1991-3984}=\frac{2.3^3.37.2.5.199+2.3^2.13.17}{2^3.3.83.11.181-2^4.3.83}=\frac{2.3^2\left(3.37.2.5.199+13.17\right)}{2^3.3.83\left(11.181-2\right)}=\frac{3.221111}{2^2.83.1989}\)
b)\(\frac{1989.1990+3978}{1992.1991-3984}=\frac{1989.1990+1989.2}{1992.1991-1992.2}=\frac{1989.\left(1990+2\right)}{1992.\left(1991-2\right)}=\frac{1989.1992}{1992.1989}=1\)
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