Sửa đề: \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}+\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)

Ta có: \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}+\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)

\(=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}+\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}\)

\(=\frac{-3\left(\frac{-1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}{5\left(\frac{-1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}+\frac{3\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}\)

\(=\frac{-3}{5}+\frac{3}{5}=0\)

x là 1,6

0.7 / x = 0.75 - 0.5 / x

0.75 = 0.5/x + 0.7/x

0.75 = (0.5+0.7)/x

0.75 = 1.2/x

x = 1.2/0.75

x = 1,6

0,7 / x = 0,75 - 0,5 / x

0,75 = 0,7 / x + 0,5 / x

0,75 = ( 0,7 + 0,5 ) / x

0,75 = 1,2 / x

     x = 1,2 : 0,75

     x = 1,6

Vậy x = 1,6

Tìm x à?

x+x=0,125+x:(0,5+0,25+0,75)=1/2

x+x=0,125+          x:1,5        =1/2

x+x=                     x:1,5         =1/2-0,125

x+x=                     x:1,5           =0,375

x+x=                     x                 =0,375x1,5

x+x=                     x            =     0,5625

x+x=0,5625x2

x+x=1,125

=x*1+x:0,125+x:0,5+x:0,25-x:0,75=1/2

=x*[1+0,125+0,5+0,25-0,75]=1/2

=x*0,125=1/2

=x         =1/2*0,125

 x=0.0625

0 nha bạn k mình nha

0,7:x=0,75 - 0,5

0,7:x = 0,25

      x=0,7:0,25

      x=2,8

Vậy x=2,8

Hình như sai đề ~ pải ko ạ

\(A=\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)

\(A=\left(\frac{\frac{3}{2}+1-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}+\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{1}{2}-\frac{5}{11}-\frac{5}{12}}\right):\frac{378}{401}+115\)

\(A=\left(\frac{3.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}+\frac{-3.\left(\frac{-1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}{5.\left(\frac{-1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}\right).\frac{401}{378}+115\)

\(A=\left(\frac{3}{5}+\frac{-3}{5}\right).\frac{401}{378}+115\)

\(A=0.\frac{401}{378}+115=115\)

A = \(\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)

= \(\left(\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}+\frac{\frac{3.125}{100}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{-\frac{5.125}{100}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)

= \(\left(\frac{3\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}+\frac{3\left(\frac{125}{100}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{-5\left(\frac{125}{100}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}\right):\frac{1890}{2005}+115\)

= \(\left(\frac{3}{5}+-\frac{3}{5}\right):\frac{1890}{2005}+115\)

= 115

\(A=\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)

\(=\left(\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}+\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\right):\frac{378}{401}+115\)

\(=\left(\frac{3\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}+\frac{3\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{-5\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}\right)}\right):\frac{378}{401}+115\)

\(=\left(\frac{3}{5}+\frac{-3}{5}\right).\frac{401}{378}+115\)

\(=0.\frac{401}{378}+115\)

\(=115\)

Vậy A = 115