Question

Ajutati-ma va rog la exercitiul de mai jos!

Answer 1

a) 4x²+12x+9-4x²+9+2x²-4x+2 = 2x²+8x+20
b) Daca in ultima paranteza este 0, rezolvarea este urmatoarea:
18+12√6+12+2(5-2√6) = 18+12√6+12+10-4√6 = 40+8√6
Daca in ultima paranteza avem √7-√5, rezolvarea este urmatoarea: 18+12√6+12+2(5-2√6)-(7-5) = 18+12√6+12+10-4√6-2 = 38+8√6

Answer 2

a)

(2x+3)²-(2x-3)(2x+3)+2(x-1)²=

(2x)²+2•2x•3+3²-(2x)²+3²+2x²-4x+2=

2²x²+2•2x•3+3²-2²x²+3²+2x²-4x+2=

4x²+2•2x•3+9-4x²+9+2x²-4x+2=

4x²+12x+9-4x²+9+2x²-4x+2=

(4x²-4x²+2x²)+(12x-4x)+(9+9+2)=

2x²+8x+20

$b) \\ \\ (3 \sqrt{2} + 2 \sqrt{3}) ^{2} + 2( \sqrt{3} - \sqrt{2}) ^{2} - ( \sqrt{7} + \sqrt{5})( \sqrt{5} - \sqrt{5}) = \\ \\ {(3 \sqrt{2} + 2 \sqrt{3}) }^{2} + 2 {( \sqrt{3} - \sqrt{2}) }^{2} - ( \sqrt{7} + \sqrt{5}) \times 0 = \\ \\ {(3 \sqrt{2} + 2 \sqrt{3}) }^{2} + 2 {( \sqrt{3} - \sqrt{2}) }^{2} - 0 = \\ \\(3 \sqrt{2} + 2 \sqrt{3}) ^{2} + 2 {( \sqrt{3} - \sqrt{2}) }^{2} = \\ \\ {(3 \sqrt{2}) }^{2} + 2 \times 3 \sqrt{2} \times 2 \sqrt{3} + {(2 \sqrt{3}) }^{2} + 6 - 4 \sqrt{6} + 4 = \\ \\ {3}^{2} { \sqrt{2} }^{2} + 2 \times 3 \sqrt{2} \times 2 \sqrt{3} + (2 \sqrt{3}) ^{2} + 6 - 4 \sqrt{6} + 4 = \\ \\ 9 { \sqrt{2} }^{2} + 2 \times 3 \sqrt{2} \times 2 \sqrt{3} + {(2 \sqrt{3}) }^{2} + 6 - 4 \sqrt{6} + 4 = \\ \\ 9 \times 2 + 2 \times 3 \sqrt{2} \times 2 \sqrt{3} + ( {2 \sqrt{3}) }^{2} + 6 - 4 \sqrt{6} + 4 = \\ \\ 9 \times 2 + 2 \times 3 \sqrt{2} \times 2 \sqrt{3} + {2}^{2} { \sqrt{3} }^{2} + 6 - 4 \sqrt{6} + 4 = \\ \\ 9 \times 2 + 2 \times 3 \sqrt{2} \times 2 \sqrt{3} + 4 { \sqrt{3} }^{2} + 6 - 4 \sqrt{6} + 4 = \\ \\ 18 + 2 \times 3 \sqrt{2} \times 2 \sqrt{3} + 4 \times 3 - 6 - 4 \sqrt{6} + 4 = \\ \\ 18 + 2 \times 3 \times 2 \sqrt{2 \times 3} + 4 \times 3 + 6 - 4 \sqrt{6} + 4 = \\ \\ 18 + 6 \times 2 \sqrt{6} + 4 \times 3 + 6 - 4 \sqrt{6} + 4 = \\ \\ 18 + 12 \sqrt{6} + 12 + 6 - 4 \sqrt{6} + 4 = \\ \\ (18 + 12 + 6 + 4) + (12 \sqrt{6} - 4 \sqrt{6}) = \\ \\ 40 + 8 \sqrt{6}$