ĐKXĐ: x>=-3/5

\(\sqrt{5x+3}=\sqrt3-\sqrt2\)

=>\(5x+3=\left(\sqrt3-\sqrt2\right)^2=5-2\sqrt6\)

=>\(5x=5-2\sqrt6-3=-2\sqrt6+2\)

=>\(x=\frac{-2\sqrt6+2}{5}\left(nhận\right)\)

căn 2 ở trong căn 3 mà

\(a,\dfrac{2}{3}.\dfrac{5}{4}-\dfrac{3}{4}.\dfrac{2}{3}=\dfrac{2}{3}.\left(\dfrac{5}{4}-\dfrac{3}{4}\right)=\dfrac{2}{3}.\dfrac{2}{4}=\dfrac{1}{3}\)

\(b,2.\left(\dfrac{-3}{2}\right)-\dfrac{7}{2}=-6.\dfrac{1}{2}-7.\dfrac{1}{2}=\left(-6-7\right).\dfrac{1}{2}=-13.\dfrac{1}{2}=\dfrac{-13}{2}\)

\(c,-\dfrac{3}{4}.5\dfrac{3}{13}-0,75.\dfrac{36}{13}=-\dfrac{3}{4}.\left(\dfrac{68}{13}-\dfrac{36}{13}\right)=-\dfrac{3}{4}.\dfrac{32}{13}=-\dfrac{24}{13}\)

a) \(\dfrac{2}{3}.\dfrac{5}{4}-\dfrac{3}{4}.\dfrac{2}{3}\)

\(=\dfrac{2}{3}.\left(\dfrac{5}{4}-\dfrac{3}{4}\right)\)

\(=\dfrac{2}{3}.\dfrac{2}{4}\)

\(=\dfrac{2}{3}.\dfrac{1}{2}\)

\(=\dfrac{1}{3}\)

b) \(2.\left(\dfrac{-3}{2}\right)^2-\dfrac{7}{2}\)

\(=2.\dfrac{9}{4}-\dfrac{7}{2}\)

\(=\dfrac{9}{2}-\dfrac{7}{2}\)

\(=\dfrac{2}{2}=1\)

c) \(-\dfrac{3}{4}.5\dfrac{3}{13}-0,75.\dfrac{36}{13}\)

\(=-\dfrac{3}{4}.\dfrac{68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)

\(=\dfrac{3}{4}.\dfrac{-68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)

\(=\dfrac{3}{4}.\left(\dfrac{-68}{13}-\dfrac{36}{13}\right)\)

\(=\dfrac{3}{4}.\dfrac{-104}{13}\)

\(=\dfrac{3}{4}.\left(-8\right)\)

\(=-6\)

\(\sqrt{x+1}⋮\sqrt{x-3}\)

\(\Rightarrow\sqrt{x-3+4}⋮\sqrt{x-3}\)

\(\text{Vì }\sqrt{x-3}⋮\sqrt{x-3}\text{ nên }\sqrt{4}⋮\sqrt{x-3}\)

\(\Rightarrow\sqrt{x-3}\inƯ\left(2\right)\)

\(\Rightarrow\sqrt{x-3}\in\left\{\pm1;\pm2\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{4;2;5;1\right\}\)

\(\Rightarrow x=2\)

Mik ko chắc đâu nha vì mik ms lp 6 thôi nên nếu có sai sót gì thì xin mọi người đừng ném đá .