a) প্রমাণ কর যে, (A - B) - C = A - (B∩C), যেখানে A,B ও C তিনটি সেট।

b) সমাধান করঃ   $\frac{1}{\left \lvert 3 x + 1 \right \rvert} \ge 5$  

: L.S $=(\mathrm{A}-\mathrm{B})-\mathrm{C}$

$=(\mathrm{A}-\mathrm{B}) \cap \mathrm{C}^{\prime} \quad$ (law of difference)

$=\mathrm{A} \cap\left(\mathrm{B}^{\prime} \cap \mathrm{C}^{\prime}\right)$ (law of difference)

$=\mathrm{A} \cap(\mathrm{B} \cup \mathrm{C})^{\prime}$ (De Morgan's low)

$=\mathrm{A}-(\mathrm{B} \cup \mathrm{C})($ law of difference )

$\frac{1}{|3 x+1|} \geq 5$

বা, $|3 x+1| \leq \frac{1}{5}$ বा,

$-\frac{1}{5} \leq 3 x+1 \leq \frac{1}{5}$ বा, $-\frac{6}{5} \leq 3 x \leq-\frac{4}{5}$ বা, $-\frac{2}{5} \leq x \leq-\frac{4}{15}$

আবার, $3 \mathrm{x}+1 \neq 0 \quad \therefore \mathrm{x} \neq-\frac{1}{3} \quad \therefore$ निর্ণেয় সমাধান,

$-\frac{2}{5} \leq \mathrm{x} \leq-\frac{4}{15}$ এবং $\mathrm{x} \neq-\frac{1}{3}$ [Ans.]