b) Ta có : \(\sqrt{50}+\sqrt{26}+1>\sqrt{49}+\sqrt{25}+1=7+5+1=13\)
\(\sqrt{168}< \sqrt{169}=13\)
Vì \(\sqrt{50}+\sqrt{49}+1>13>\sqrt{168}\)
nên \(\sqrt{50}+\sqrt{49}+1>\sqrt{168}\)
\(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,265+0,5-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\\ =\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\left(\dfrac{5}{8}-\dfrac{5}{10}+\dfrac{5}{11}+\dfrac{5}{12}\right)}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\\ =\dfrac{3\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{-5\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}+\dfrac{3\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{5\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}\\ =\dfrac{-3}{5}+\dfrac{3}{5}=0\)