Question

dau cornoana cine ma ajutaa urgentttt​

Answer 1

Răspuns:

$1)a)7y + 3y - 4y = 10 - 4y = 6y \\ b) - 12x + 7 + 15x + 2 = 3x + 9 \\ c)x \sqrt{2} + 2x \sqrt{3} - 4x \sqrt{12} + 5x \sqrt{18} = \\ = x \sqrt{2} + 2x \sqrt{3} - 4x \times 2 \sqrt{3} + 5x \times 3 \sqrt{2} = \\ = x \sqrt{2} + 2x \sqrt{3} - 8x \sqrt{3} + 15x \sqrt{2} = 16x \sqrt{2} - 6x \sqrt{3}$

$d)x \sqrt{3} - y \sqrt{8} + x \sqrt{27} + 2y \sqrt{32} = \\ = x \sqrt{3} - y \times 2 \sqrt{2} + x \times 3 \sqrt{3} + 2y \times 4 \sqrt{2} = \\ = x \sqrt{3} - 2y \sqrt{2} + 3x \sqrt{3} + 8y \sqrt{2} = \\ = 4x \sqrt{3} + 6y \sqrt{2}$

$e)7x + 5y - 2x - 6y = 5x - y \\ f)7a - (2b - 8a) + (2a - b) = \\ = 7a - 2b + 8a + 2a - b = 17a - 3b \\ g)7b - (2a - 8b) + (2b - a) = 7b - 2a + 8b + 2b - a = 17b - 3a$

$h)5x(2x + 3x) = 5x \times 5x = 25 {x}^{2} \\ i)2(3x - 1) - 3(2x - 1) = 2 \times 3x + 2 \times( - 1) - 3 \times 2x - 3 \times ( - 1) = 6x - 2 - 6x + 3 = 1$

$j)3x(x - 4) - x(x + 1) - 2 {x}^{2} = \\ = 3 {x}^{2} - 12x - {x}^{2} - x - 2 {x}^{2} = - 13x \\ k)(2x + 8x - 10x) \div 3 = (10x - 10x) \div 3 = 0 \div 3 = 0 \\ l)(24 {x}^{2} - 18 {x}^{3} + 12x) \div 6x = \\ = 24 {x}^{2} \div 6x - 18 {x}^{3} \div 6x + 12x \div 6x = 4x - 3 {x}^{2} + 2 = - 3 {x}^{2} + 4x + 2$

$2)n = \sqrt{ ({x}^{4} ) ^{5} \div {(x \times {x}^{3} \times {x}^{5}) }^{2} } = \\ = \sqrt{ {x}^{20} \div {( {x}^{9} )}^{2} } = \sqrt{ {x}^{20} \div {x}^{18} } = \sqrt{ {x}^{2} } = |x| = x \: oricare \: ar \: fi \: x \: nr \: natural \:$

Deci N este natural pt orice x întreg nenul

$3)n = \sqrt{( x \times {x}^{3} \times {x}^{5} \times {x}^{9}) \div ( { {x}^{4} })^{4} } = \sqrt{ {x}^{18} \div {x}^{16} } = \sqrt{ {x}^{2} } = |x| = x \: pt \: orice \: nr \: intreg \: nenul$

N este natural pt orice x întreg nenul