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1)
\(\frac{1994.1993-1992.1993}{1992.1993+1944.7+1986}\)
\(=\frac{\left(1994-1992\right).1993}{2.1985028+2.6804+2.993}\)
\(=\frac{2.1993}{2.\left(1985028+6804+993\right)}\)
\(=\frac{2.1993}{2.1992825}\)
\(=\frac{193}{1992825}\)
2)
a) \(\frac{399.45+55.399}{1995.1996-1991.1995}\)
\(=\frac{399.\left(45+55\right)}{1995.\left(1996-1991\right)}\)
\(=\frac{399.100}{1995.5}\)\(=4\)
Câu b bài 2 mình không biết làm nữa, xin lỗi nhé!
\(\frac{1995.1993-18}{1993.1994+1975}=\frac{\left(1994+1\right).1993-18}{1993.1994+1975}\)
\(=\frac{1994.1993+1993-18}{1993.1994+1975}=\frac{1994.1993+1975}{1993.1994+1975}=1.\)
\(\frac{1995\cdot1993-18}{1993\cdot1994+1975}\)
\(=\frac{\left(1994+1\right)\cdot1993-18}{1993\cdot1994+1975}\)
\(=\frac{1994\cdot1993+\left(1993-18\right)}{1993\cdot1994+1975}\)
\(=\frac{1994\cdot1993+1975}{1993\cdot1994+1975}\)
\(=1\)
A=15x(1/7-1/12-1/98)/18(1/7-1/12-1/98)
A=15/18
A=5/6
h nhe!!!
\(A=\frac{15\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}{18\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}\)
\(A=\frac{15}{18}=\frac{5}{6}\)
\(A=\frac{15\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}{18\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{18}\right)}=\frac{15}{18}\)
\(A=\frac{15\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}{18\left(1-\frac{1}{7}-\frac{1}{12}-\frac{1}{98}\right)}=\frac{15}{18}=\frac{5}{6}\)
18 x \(\left(\frac{1919}{2121}+\frac{888}{999}\right)\)
= 18 x \(\frac{113}{63}\)
= \(\frac{226}{7}\)
\(18\left(\frac{19191919}{21212121}+\frac{88888}{99999}\right)\)
\(=18\left(\frac{19}{21}+\frac{8}{9}\right)\)
\(=18\left(\frac{57}{63}+\frac{56}{63}\right)\)
\(=18\left(\frac{113}{63}\right)\)
\(=\frac{3^2.2.113}{3^2.7}\)
\(=\frac{226}{7}\)
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