\(\frac{3a-b}{2a+7}+\frac{3b-a}{2b-7}=\frac{3a-b}{2a+a-b}+\frac{3b-a}{2b-\left(a-b\right)}=\frac{3a-b}{3a-b}+\frac{3b-a}{3b-a}=2\)

a,\(3:7+\left(-5:2\right)+\left(-3:5\right)\)

\(=\dfrac{3}{7}-\dfrac{5}{2}-\dfrac{3}{5}\)

\(=-\dfrac{187}{70}\)

b,\(-8:18-15:17\)

\(=-\dfrac{8}{18}-\dfrac{15}{17}\)

\(=-\dfrac{203}{153}\)

c,\(4:5-\left(-2:7\right)-7:10\)

\(=\dfrac{4}{5}-\left(-\dfrac{2}{7}\right)-\dfrac{7}{10}\)

\(=\dfrac{27}{10}\)

d,\(3,5-\left(-2:7\right)\)

\(=3,5+\dfrac{2}{7}\)

\(=\dfrac{53}{14}\)

\(P=-\left(2x+1\right)^2-7\left(y-3,5\right)^2+\dfrac{2}{3}\)

 vì \(\left\{{}\begin{matrix}-\left(2x+1\right)^2\le0,\forall x\\-7\left(y-3,5\right)^2\le0,\forall y\end{matrix}\right.\)

\(\Rightarrow P=-\left(2x+1\right)^2-7\left(y-3,5\right)^2+\dfrac{2}{3}\le\dfrac{2}{3}\)

Dấu "=" xảy ra khi

\(\left\{{}\begin{matrix}2x+1=0\\y-3,5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=3,5=\dfrac{7}{2}\end{matrix}\right.\)

Vậy \(GTLN\left(P\right)=\dfrac{2}{3}\left(tạix=-\dfrac{1}{2};y=\dfrac{7}{2}\right)\)

???????

|3,5-x|\(+\dfrac{2}{7}=\dfrac{16}{7}\)

\(\Leftrightarrow\left|3,5-x\right|=\dfrac{16}{7}-\dfrac{2}{7}\)

\(\Leftrightarrow\left\{{}\begin{matrix}3,5-x=2\\3,5-x=-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}1,5\\x=5,5\end{matrix}\right.\)

\(|3,5-x|=\dfrac{16}{7}-\dfrac{2}{7}\)

\(|3,5-x|=2\)

\(\Rightarrow|\left[{}\begin{matrix}3,5-x=2\\3,5-x=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3,5-2\\x=3,5-\left(-2\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1,5\\x=5,5\end{matrix}\right.\)

\(\Rightarrow x\in\left\{1,5;5,5\right\}\)

\(\dfrac{4}{5}-\left(-\dfrac{2}{7}\right)-\dfrac{7}{10}\\ =\dfrac{4}{5}+\dfrac{2}{7}-\dfrac{7}{10}\\ =\dfrac{56}{70}+\dfrac{20}{70}-\dfrac{49}{70}\\ =\dfrac{27}{70}\\ 3,5-\left(-\dfrac{2}{7}\right)\\ =\dfrac{7}{2}+\dfrac{2}{7}\\ =\dfrac{49}{14}+\dfrac{4}{14}\\ =\dfrac{53}{14}\\ \left(-3\right).\left(-\dfrac{7}{12}\right)\\ =\dfrac{21}{12}\\ =\dfrac{7}{4}\)