Question

Calculați: a) (3x+2)²-12x b) (2x-5)(2x+5)-4x(x-6) c) (x-3)²+3(2x-3) d) 4(x+1)²-(2x+3)(2x-3) e) (3x+2)(3x-2)-9(x+1)(x-1) f) (radical din 2+radical din 3)²-(radical din 6+1)² VA ROOOOOG AM NEVOIE URGENT​

Answer 1

Răspuns:

$a) {(3x + 2)}^{2} - 12x = (3 {x})^{2} + 2 \times 3x \times 2 + {2}^{2} = \\ = 9 {x}^{2} + 12x + 4$

$b)(2x - 5)(2x + 5) - 4x(x - 6) = \\ = (2 {x})^{2} - {5}^{2} - 4x \times x - 4x \times ( - 6) = \\ = 4 {x}^{2} - 25 - 4 {x}^{2} + 24x = \\ = 24x - 25$

$c)( {x - 3)}^{2} + 3(2x - 3) = \\ = {x}^{2} - 2 \times x \times 3 + {3}^{2} + 3 \times 2x + 3 \times ( - 3) = \\ = {x}^{2} - 6x + 9 + 6x - 9 = {x}^{2}$

$d)4( {x + 1)}^{2} - (2x + 3)(2x - 3) = \\ = 4( {x}^{2} + 2 \times x \times 1 + {1}^{2} ) - (( {2x)}^{2} - {3}^{2} ) = \\ = 4( { {x}^{2} + 2x + 1)} - (4 {x}^{2} - 9) = \\ = 4 {x}^{2} + 4 \times 2x + 4 \times 1 - 4 {x}^{2} + 9 = \\ = 4 {x}^{2} + 8 {x} + 4 - 4 {x}^{2} + 9 = \\ = 8x + 13$

$e)(3x + 2)(3x - 2) - 9(x + 1)(x - 1) = \\ = (3 {x})^{2} - {2}^{2} - 9( {x}^{2} - {1}^{2} ) = \\ = 9 {x}^{2} - 4 - 9 ({x}^{2} - 1) = \\ = 9 {x}^{2} - 4 - 9 {x}^{2} + 9 = 5$

$f)( { \sqrt{2} + \sqrt{3} })^{2} - ( { \sqrt{6} + 1 })^{2} = \\ = ( { \sqrt{2}) }^{2} + 2 \times \sqrt{2} \times \sqrt{3} + ( { \sqrt{3} })^{2} - (( { \sqrt{6} })^{2} + 2 \times \sqrt{6} \times 1 + {1}^{2} ) = \\ = 2 + 2 \sqrt{6} + 3 - (6 + 2 \sqrt{6} + 1) = \\ = 5 + 2 \sqrt{6} - 6 - 2 \sqrt{6} - 1 = \\ = 5 - 6 - 1 = - 1 - 1 = - 2$