6) $$\left(3\frac{25}{48}-1\frac{35}{48}\right)+\left(4\frac{65}{72}-2\frac{68}{72}\right);$$

$$\left(3\frac{25}{48}-1\frac{35}{48}\right)+\left(4\frac{65}{72}-2\frac{68}{72}\right);$$ $$3\frac{25}{48}-1\frac{35}{48} = 2\frac{73}{48} - 1\frac{35}{48} = 1 + \frac{73 - 35}{48} = 1 + \frac{38}{48} = 1 + \frac{19}{24}$$ $$4\frac{65}{72}-2\frac{68}{72} = 3\frac{137}{72} - 2\frac{68}{72} = 1 + \frac{137-68}{72} = 1 + \frac{69}{72} = 1 + \frac{23}{24}$$ $$1\frac{19}{24}+1\frac{23}{24} = 2 + \frac{19+23}{24} = 2 + \frac{42}{24} = 2 + \frac{7}{4} = 2+1\frac{3}{4}=3\frac{3}{4}$$ Ответ: $$3\frac{3}{4}$$