\(\frac{1}{3}.x-0.5.x=0.75\)
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25 tháng 2 2020
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: \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}+\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
\(=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}+\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}\)
\(=\frac{-3\left(\frac{-1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}{5\left(\frac{-1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}+\frac{3\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}\)
\(=\frac{-3}{5}+\frac{3}{5}=0\)
6 tháng 1 2018
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
SJ
4
NT
1
24 tháng 10 2016
\(A=\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)
\(A=\left(\frac{\frac{3}{2}+1-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}+\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{1}{2}-\frac{5}{11}-\frac{5}{12}}\right):\frac{378}{401}+115\)
\(A=\left(\frac{3.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}+\frac{-3.\left(\frac{-1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}{5.\left(\frac{-1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}\right).\frac{401}{378}+115\)
\(A=\left(\frac{3}{5}+\frac{-3}{5}\right).\frac{401}{378}+115\)
\(A=0.\frac{401}{378}+115=115\)
24 tháng 10 2016
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
1/3 . x - 0,5 . x = 0,75
=> ( 1/3 - 0,5)x = 0,75
=> ( 1/3 -1/2)x =0,75
=> (-1/6)x = 0,75
=> x = 0,75 : (-1/6)
=> x = -9/2
Vậy x = -9/2
x= -9/2