\(\left(\frac{3}{4}\cdot x\right):\frac{1}{2}=\frac{4}{5}\)

\(\frac{3}{4}\cdot x=\frac{4}{5}\cdot\frac{1}{2}\)

\(\frac{3}{4}\cdot x=\frac{2}{5}\)

\(x=\frac{2}{5}:\frac{3}{4}\)

\(x=\frac{8}{15}\)

2 chữ x hả bạn

\(\frac{1}{2}.x-\frac{3}{4}=\frac{5}{6}\)

\(\frac{1}{2}.x=\frac{5}{6}+\frac{3}{4}\)

\(\frac{1}{2}.x=\frac{19}{12}\)

\(x=\frac{19}{12}:\frac{1}{2}\)

\(x=\frac{19}{6}\)

Vậy \(x=\frac{19}{6}\)

\(\frac{1}{2}\)x X = \(\frac{5}{6}\)+ \(\frac{3}{4}\)

\(\frac{1}{2}\)x X = \(\frac{19}{12}\)

X = \(\frac{19}{12}\): \(\frac{1}{2}\)

X = \(\frac{19}{6}\)

đề gi sai ròi bạn ơi

1/4 * 1/5 * x = 1/2

1/20 *x = 1/2

x          = 1/2 :1/20 

x          = 1/2 * 20/1

x          = 10

=> x + 1/2 = 5/6 + 3/4

=> x + 1/2 = 19/12

=> x = 19/12 - 1/2

=> x = 13/12

Vậy x = 13/12

Tk mk nha

a  X+1/2-3/4=5/6 suy ra :                                                                        Mỉnh phải đi ngủ rồi mai mình giải tiếp cho nhé

  X+1/2=5/6+3/4

  X+1/2=19/12

  X=19/12-1/2

  X=13/12                                                                                            

a) \(\lim \frac{{2{n^2} + 6n + 1}}{{8{n^2} + 5}} = \lim \frac{{{n^2}\left( {2 + \frac{6}{n} + \frac{1}{{{n^2}}}} \right)}}{{{n^2}\left( {8 + \frac{5}{{{n^2}}}} \right)}} = \lim \frac{{2 + \frac{6}{n} + \frac{1}{n}}}{{8 + \frac{5}{n}}} = \frac{2}{8} = \frac{1}{4}\)

b) \(\lim \frac{{4{n^2} - 3n + 1}}{{ - 3{n^3} + 6{n^2} - 2}} = \lim \frac{{{n^3}\left( {\frac{4}{n} - \frac{3}{{{n^2}}} + \frac{1}{{{n^3}}}} \right)}}{{{n^3}\left( { - 3 + \frac{6}{n} - \frac{2}{{{n^3}}}} \right)}} = \lim \frac{{\frac{4}{n} - \frac{3}{{{n^2}}} + \frac{1}{{{n^3}}}}}{{ - 3 + \frac{6}{n} - \frac{2}{{{n^3}}}}} = \frac{{0 - 0 + 0}}{{ - 3 + 0 - 0}} = 0\).

c) \(\lim \frac{{\sqrt {4{n^2} - n + 3} }}{{8n - 5}} = \lim \frac{{n\sqrt {4 - \frac{1}{n} + \frac{3}{{{n^2}}}} }}{{n\left( {8 - \frac{5}{n}} \right)}} = \frac{{\sqrt {4 - 0 + 0} }}{{8 - 0}} = \frac{2}{8} = \frac{1}{4}\).

d) \(\lim \left( {4 - \frac{{{2^{{\rm{n}} + 1}}}}{{{3^{\rm{n}}}}}} \right) = \lim \left( {4 - 2 \cdot {{\left( {\frac{2}{3}} \right)}^{\rm{n}}}} \right) = 4 - 2.0 = 4\).

e) \(\lim \frac{{{{4.5}^{\rm{n}}} + {2^{{\rm{n}} + 2}}}}{{{{6.5}^{\rm{n}}}}} = \lim \frac{{{{4.5}^{\rm{n}}} + {2^2}{{.2}^{\rm{n}}}}}{{{{6.5}^{\rm{n}}}}} = \lim \frac{{{5^n}.\left[ {4 + 4.{{\left( {\frac{2}{5}} \right)}^{\rm{n}}}} \right]}}{{{{6.5}^n}}} = \lim \frac{{4 + 4.{{\left( {\frac{2}{5}} \right)}^{\rm{n}}}}}{6} = \frac{{4 + 4.0}}{6} = \frac{2}{3}\).

g) \(\lim \frac{{2 + \frac{4}{{{n^3}}}}}{{{6^{\rm{n}}}}} = \lim \left( {2 + \frac{4}{{{{\rm{n}}^3}}}} \right).\lim {\left( {\frac{1}{6}} \right)^{\rm{n}}} = \left( {2 + 0} \right).0 = 0\).

a, 3/4 + 1/4.x=2

    1/4.x        = 2-3/4

     1/4.x      =5/4

     x           = 5/4:1/4

     x           = 5

b, x-2/3.9/4=2,5-1/2

    x-2/3.9/4=2

    x-2/3     =2:9/4

    x-2/3   =8/9

     x        = 8/9+2/3

     x         = 14/9

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

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

=>\(\frac{3}{4}x=\frac{2}{5}\)

=>\(x=\frac{2}{5}:\frac{3}{4}\)

=>\(x=\frac{8}{15}\)

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

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

\(\Rightarrow x=\frac{2}{5}:\frac{3}{4}=\frac{8}{15}\)

k nha