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\(\left(\frac{1}{4}+35\%+0,65+75\%\right)x\)\(\left(1\frac{6}{3}+4\frac{8}{24}\right)\)
\(=\left(\frac{1}{4}+\frac{7}{20}+\frac{13}{20}+\frac{3}{4}\right)\)\(x\left(3+\frac{13}{3}\right)\)
\(=\left(\frac{5}{20}+\frac{7}{20}+\frac{13}{20}+\frac{15}{20}\right)\)\(x7\)
\(=2x7\)
\(=14\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left[\left(35\frac{5}{7}+2\frac{3}{4}\right)-5\frac{5}{7}+\frac{1}{4}\right]:\left(11+x\right)=3\)
\(\left[\left(35\frac{5}{7}-5\frac{5}{7}\right)+2\frac{3}{4}+\frac{1}{4}\right]:\left(11+x\right)=3\)
\(\left[30+3\right]:\left(11+x\right)=3\)
\(33:\left(11+x\right)=3\)
\(11+x=11\)
\(x=0\)
\(\left[\left(35\frac{5}{7}+2\frac{3}{4}\right)-5\frac{5}{7}+\frac{1}{4}\right]\div\left(11+x\right)=3\)
\(\left[\left(\frac{35\times7+5}{7}+\frac{2\times4+3}{4}\right)-\frac{5\times7+5}{7}+\frac{1}{4}\right]\div\left(11+x\right)=3\)
\(\left[\left(\frac{250}{7}+\frac{11}{4}\right)-\frac{40}{7}+\frac{1}{4}\right]\div\left(11+x\right)=3\)
\(\left[\frac{1077}{28}-\frac{167}{28}\right]\div\left(11+x\right)=3\)
\(32,5\div\left(11+x\right)=3\)
\(11+x=32,5\div3\)
\(11+x=\frac{65}{6}\)
\(x=\frac{65}{6}-11=-\frac{1}{6}\)
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A = 2017 - ( 1 -\(\frac{3}{4}\)) - ( 1 - \(\frac{3}{5}\)) -...- ( 1 - \(\frac{3}{2020}\)) : Còn lại
A = 2017 - 1 + \(\frac{3}{4}\) - 1 + \(\frac{3}{5}\) -... - 1 +\(\frac{3}{2020}\) : Còn lại
A = 0 + \(\frac{3}{4}\)+\(\frac{3}{5}\)+\(\frac{3}{2020}\) : CÒn lại
A = 3 x ( \(\frac{1}{4}\)+\(\frac{1}{5}\)+...+ \(\frac{1}{2020}\)) : ( \(\frac{1}{4x5}\)+\(\frac{1}{5x5}\) + ... )
A = 3 : \(\frac{1}{5}\)
A = 15
= 15 nha
đúng 1000000000000000000000000000000% đó cái này ở violympic cấp tỉnh lớp 5 đó....
60,625 la dap an
\(\left(35\%+\frac{1}{4}+75\%+0,65\right)\cdot105+\left(\frac{1}{2}+50\%\right)\)
\(=\left(\frac{7}{20}+\frac{1}{4}+\frac{3}{4}+\frac{13}{20}\right)\cdot105+\left(\frac{1}{2}+\frac{1}{2}\right)\)
\(=2\cdot105+1\)
\(=211\)
\(=210x+1\)