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\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

    = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)

      = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

      = \(1-\frac{1}{7}\)

      =   \(\frac{7}{7}-\frac{1}{7}\)

        = \(\frac{6}{7}\)

2) \(\frac{7}{4}-x.\frac{4}{3}=\frac{5}{19}\)

              \(x.\frac{4}{3}=\frac{7}{4}-\frac{5}{19}\)

         \(x.\frac{4}{3}=\frac{133}{76}-\frac{20}{76}\)

\(x.\frac{4}{3}=\frac{113}{76}\)

      \(x=\frac{113}{76}:\frac{4}{3}\)

      \(x=\frac{399}{304}\)

VẬY \(x=\frac{399}{304}\)

b) \(\left(x+\frac{3}{4}\right).\frac{5}{7}=\frac{10}{9}\)

      \(\left(x+\frac{3}{4}\right)=\frac{10}{9}:\frac{5}{7}\)

     \(x+\frac{3}{4}=\frac{14}{9}\)

              \(x=\frac{14}{9}-\frac{3}{4}\)

               \(x=\frac{29}{36}\)

Vậy \(x=\frac{29}{36}\)

c) \(x.\frac{1}{2}+\frac{3}{2}.x=\frac{4}{5}\)

\(x.\left(\frac{1}{2}+\frac{3}{2}\right)=\frac{4}{5}\)

\(x.2=\frac{4}{5}\)

     \(x=\frac{4}{5}:2\)

      \(x=\frac{2}{5}\)

Vậy \(x=\frac{2}{5}\)

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\(x\cdot\frac{22}{9}=\frac{22}{5}-\frac{11}{7}\)

\(x\cdot\frac{22}{9}=\frac{99}{35}\)

\(x=\frac{99}{35}:\frac{22}{9}\)

\(x=\frac{81}{70}\)

Vậy \(x=\frac{81}{70}\).

\(x:\frac{56}{5}=\frac{26}{5}+\frac{52}{5}\)

\(x:\frac{56}{5}=\frac{78}{5}\)

\(x=\frac{78}{5}\cdot\frac{56}{5}\)

\(x=\frac{4368}{25}\)

Vậy \(x=\frac{4368}{25}\).

\(x:\frac{19}{3}=4\)

\(x=4\cdot\frac{19}{3}\)

\(x=\frac{76}{3}\)

Vậy \(x=\frac{76}{3}\).

x + y = x.y

=> xy - x - y = 0

=> (xy - x) - y + 1 = 1

=> x(y - 1) - (y - 1) = 1

=> (x - 1)(y - 1) = 1

=> x - 1 = y - 1 = 1 hoặc x - 1 = y - 1 = -1

=> x = y = 2 hoặc x = y = 0

a) \(\dfrac{x}{15}\)=\(\dfrac{x-2}{30}\)

=>30x=15 (x-2)

=>30x=15x-30

=>30x-15x=-30

=>15x =-30

=>x =-2

b)\(\dfrac{7}{8}\)<\(\dfrac{x}{80}\)<\(\dfrac{9}{10}\)

=>\(\dfrac{70}{80}\)<\(\dfrac{x}{80}\)<\(\dfrac{72}{80}\)

=> 70<x<72

=>x=71

c)\(\dfrac{1}{6}\)<\(\dfrac{5}{x}\)<\(\dfrac{1}{4}\)

=>\(\dfrac{5}{30}\)<\(\dfrac{5}{x}\)<\(\dfrac{5}{20}\)

=>20<x<30

=>x={21;22;23;24;25;26;27;28;29}

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ừ đúng dễ mà

\(a,\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}+\dfrac{5}{7}=9\dfrac{5}{7}\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=\dfrac{68}{7}-\dfrac{5}{7}\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=9\)

\(\Leftrightarrow x-\dfrac{1}{3}=3\)

\(\Leftrightarrow x=3+\dfrac{1}{3}\)

\(\Leftrightarrow x=\dfrac{9}{3}+\dfrac{1}{3}\)

\(\Leftrightarrow x=\dfrac{10}{3}\)

\(b,x+30\%x=-1,31\)

\(\Leftrightarrow x+\dfrac{3}{10}.x=-\dfrac{131}{100}\)

\(\Leftrightarrow x.\left(1+\dfrac{3}{10}\right)=-\dfrac{131}{100}\)

\(\Leftrightarrow x.\dfrac{13}{10}=-\dfrac{131}{100}\)

\(\Leftrightarrow x=-\dfrac{131}{100}.\dfrac{10}{13}\)

\(\Leftrightarrow x=-\dfrac{131}{130}\)

\(c,-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{2}{10}\)

\(\Leftrightarrow-\dfrac{2}{3}x=-\dfrac{1}{10}\)

\(\Leftrightarrow x=-\dfrac{1}{10}.\left(-\dfrac{3}{2}\right)\)

\(\Leftrightarrow x=\dfrac{3}{20}\)