a,\(\frac{30}{7}:\frac{3}{5}=\frac{30}{7}.\frac{5}{3}=\frac{50}{7}\)

b,\(\frac{2}{9}:\frac{31}{100}=\frac{2}{9}.\frac{100}{31}=\frac{200}{279}\)

a)\(1,5:2,16=15:216=5:72\)

b)\(4\dfrac{2}{7}:\dfrac{3}{5}=\dfrac{30}{7}:\dfrac{3}{5}=\dfrac{30}{7}.\dfrac{5}{3}=\dfrac{50}{7}=50:7\)

c)\(\dfrac{\dfrac{2}{9}}{0,31}=\dfrac{2}{9}:\dfrac{31}{100}=\dfrac{2}{9}.\dfrac{100}{31}=\dfrac{31}{450}=31:450\)

a) 1,5 : 2,16 

\(=\frac{15}{10}:\frac{216}{100}=\frac{3}{2}:\frac{54}{25}=\frac{3}{2}.\frac{25}{54}=\frac{25}{36}=25:36\)

Các câu còn lại lm tương tự

\(2,04:-3,12=\frac{204}{100}:\frac{-312}{100}=\frac{204}{100}.\frac{100}{-312}=\frac{204}{-312}=\frac{-17}{26}\)    

giải:

a. \(2,04:\left(-3,12\right)=\frac{204}{100}:\frac{-312}{100}=\frac{204}{100}.\frac{100}{-312}=\frac{204}{-312}=\frac{-17}{26}\)

b,\(\left(-1\frac{1}{2}\right):1,25=\frac{-3}{2}:\frac{125}{100}=\frac{-3}{2}.\frac{100}{125}=\frac{-6}{5}\) 

c,\(4:5\frac{3}{4}=\frac{4}{1}:\frac{23}{4}=\frac{4}{1}.\frac{4}{23}=\frac{16}{23}\)                                                                                                                          d,\(10\frac{3}{7}:5\frac{3}{14}=\frac{73}{7}:\frac{73}{14}=\frac{73}{7}.\frac{14}{73}=\frac{2}{1}\)

a) \(1,5:2,16=\frac{15}{10}:\frac{216}{100}=\frac{3}{2}:\frac{54}{25}=\frac{3}{2}.\frac{25}{54}=\frac{25}{36}=25:36\)

b) \(4\frac{2}{7}:\frac{3}{5}=\frac{30}{7}.\frac{5}{3}=\frac{50}{7}=50:7\)

c) \(\frac{2}{9}:0,31=\frac{2}{9}:\frac{31}{100}=\frac{2}{9}.\frac{100}{31}=\frac{200}{279}=200:279\)

\(5,5:\frac{1}{2}=\frac{55}{10}:\frac{1}{2}=11\)

Em ơi, đây là số tự nhiên nên không tỉ số được !!!

\(=\frac{11}{2}:\frac{1}{2}\)

\(=\frac{11}{2}\times2\)

\(=11\)

\(=11:1\)

Bài 1a:

\(\frac{4,6}{1,4}\) = \(\frac{11,5}{5x}\)

4,6 x 5\(x\) = 11,5 x 1,4

23\(x\) = 16,1

\(x\) = 16,1 : 23

\(x\) = 0,7

Vậy \(x\) = 0,7

Bài 1b:

\(\frac{x^2}{6}\) = \(\frac{24}{25}\)

25\(x^2\) = 24.6

25\(x^2\) = 144

\(x^2\) = 144/25

\(x\) = - 12/5 hoặc \(x\) = 12/5

Vậy \(x\) ∈ {-12/5; 12/5}

Bài 1c:

\(\frac{x+2}{2}\) = \(\frac{1}{1-x}\)

(\(x\) + 2)(1 - \(x\)) = 1.2

\(x\) - \(x^2\) + 2 - 2\(x\) = 2

\(x^2\) + (2\(x\) - \(x\)) + 2 - 2 = 0

\(x^2\) + \(x\) + 0 = 0

\(x\)(\(x\) + 1) = 0

\(x\) = 0

\(x\) + 1 = 0

\(x\) = - 1

Vậy \(x\) ∈ {-1; 0}

A)1,2 : 3,24                    b) \(2\frac{1}{5}\) : \(\frac{3}{4}\)                                    c)\(\frac{2}{7}\) : 0,42

     \(\frac{12}{10}\) : \(\frac{324}{100}\)                          \(\frac{11}{5}\) : \(\frac{3}{4}\)                                    \(\frac{2}{7}\) : \(\frac{42}{100}\)