a: 2/7=10/35

5/9=10/18

b: 7/8=14/16

14/5=14/5

c: 4/11=60/165

5/12=60/144

3/5=60/100

$\frac{\left(3x-2\right)\left(5-x\right)}{2-7x}$=$\frac{3x^2-17x+10}{7x-2}$≥0

TH1: $\{\begin{array}{c}3x^2-17x-10\ge 0\\ 7x-2>0\end{array}$=> $\{\begin{array}{c}\{\begin{array}{c}x\le \frac{\left(17-\sqrt{409}\right)}{6}\\ x\ge \frac{\left(17+\sqrt{409}\right)}{6}\end{array}\\ x>\frac{2}{7}\end{array}$=> x≥$\frac{\left(17+\sqrt{409}\right)}{6}$

TH2:$\{\begin{array}{c}3x^2-17x-10\le 0\\ 7x-20<0\end{array}$=>$\{\begin{array}{c}\frac{17-\sqrt{409}}{6}\le x\le \frac{17+\sqrt{409}}{6}\\ x<\frac{2}{7}\end{array}$=>$\frac{17-\sqrt{409}}{6}\le x<\frac{2}{7}$

ok

$\frac{\left(3x-2\right)\left(5-x\right)}{2-7x}$=$\frac{3x^2-17x+10}{7x-2}$≥0

TH1: $\{\begin{array}{c}3x^2-17x-10\ge 0\\ 7x-2>0\end{array}$=> $\{\begin{array}{c}\{\begin{array}{c}x\le \frac{\left(17-\sqrt{409}\right)}{6}\\ x\ge \frac{\left(17+\sqrt{409}\right)}{6}\end{array}\\ x>\frac{2}{7}\end{array}$=> x≥$\frac{\left(17+\sqrt{409}\right)}{6}$

TH2:$\{\begin{array}{c}3x^2-17x-10\le 0\\ 7x-20<0\end{array}$=>$\{\begin{array}{c}\frac{17-\sqrt{409}}{6}\le x\le \frac{17+\sqrt{409}}{6}\\ x<\frac{2}{7}\end{array}$=>$\frac{17-\sqrt{409}}{6}\le x<\frac{2}{7}$

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