CUET Preparation Today
If x = 7 and y = 2 then, what is the value of $\frac{x^3-y^3}{x^2-y^2}-\frac{3xy}{x+y}$ ? |
28/9 21/10 25/9 26/9 |
25/9 |
x = 7 and y = 2 $\frac{x^3-y^3}{x^2-y^2}-\frac{3xy}{x+y}$ = $\frac{7^3-2^3}{7^2-2^2}-\frac{3 × 7 × 2}{7 + 2}$ = \(\frac{343}{45}\) - \(\frac{42}{9}\) = \(\frac{125}{45}\) = \(\frac{25}{9}\) |
