Question

DAU COROANĂ!!! e urgent!!! ​

Answer 1

Explicație pas cu pas:

${(x + 1)}^{3} = {x}^{3} + 3 {x}^{2} + x + 1$

${(y - 2)}^{3} = {y}^{3} - 6 {x}^{2} + 12x - 8$

${( \sqrt{2} - 1)}^{3} = { (\sqrt{2}) }^{3} - 3 {( \sqrt{2}) }^{2} + 3 \sqrt{2} - 1 = 2 \sqrt{2} - 6 + 3 \sqrt{2} - 1 = 5 \sqrt{2} - 7$

${( \sqrt{3} + \sqrt{2} )}^{3} = { (\sqrt{3}) }^{3} + 3 {( \sqrt{3} )}^{2} \sqrt{2} + 3 \sqrt{3} {( \sqrt{2} )}^{2} + {( \sqrt{2} )}^{3} = 3 \sqrt{3} + 9 \sqrt{2} + 6 \sqrt{3} + 2 \sqrt{2} = 9 \sqrt{3} + 11 \sqrt{2}$

(3x-1)³ + (3x+1)³ = 27x³ - 27x² + 9x - 1 + 27x³ + 27x² + 9x + 1 = 54x³ + 18x

(2√2 + 1)³ - (2√2 - 1)³ = (2√2)³ + 3(2√2)² + 3×2√2 + 1 - [(2√2)³ - 3(2√2)² + 3×2√2 - 1] = 16√2 + 24 + 6√2 + 1 - (16√2 - 24 + 6√2 - 1) = 22√2 + 25 - (22√2 - 25) = 22√2 + 25 - 22√2 + 25 = 50

Answer 2

Răspuns:

6)

a.

$(x+1)^{3}$

= x³ + 3x² · 1 + 3x · 1² + 1³

= x³ + 3x² + 3x · 1 + 1

$\boxed{= x³ + 3x² + 3x + 1}$

;

$(y-2)^3$

= y³ - 3y² · 2 + 3y · 2² - 2³

= y³ - 6y² + 3y · 4 - 8

$\boxed{= y³ - 6y² + 12y - 8}$

;

$(\sqrt{2} -1)^{3}\\=\sqrt{2}^3-3 \cdot \sqrt{2}^2 \cdot 1+3\sqrt{2} \cdot 1^{2} - 1^3 \\=2\sqrt{2}-3 \cdot2 \cdot 1+3\sqrt{2} -1\\=5\sqrt{2} -6-1\\\boxed{=5\sqrt{2} -7}$

;

$(\sqrt{3} + \sqrt{2} )^3\\=\sqrt{3}^3+3 \cdot \sqrt{3}^2 \cdot \sqrt{2}+3\sqrt{3} \cdot \sqrt{2}^2 + \sqrt{2}^3\\=3\sqrt{3}+9\sqrt{2}+6\sqrt{3}+2\sqrt{2}\\\boxed{=9\sqrt{3}+11\sqrt{2}}$

b.

$(3x-1)^3+(3x+1)^3\\=27x^3-27x^2+9x-1+27x^3+27x^2+9x+1\\=54x^3+9x+9x\\=\boxed{54x^3+18x}$

;

$(2\sqrt{2}+1)^3-(2\sqrt{2}-1)^3\\=16\sqrt{2}+24+6\sqrt{2}+1-(16\sqrt{2}-24+6\sqrt{2}-1)\\=16\sqrt{2}+24+6\sqrt{2}+1-(22\sqrt{2}-25)\\=16\sqrt{2}+24+6\sqrt{2}+1-22\sqrt{2}+25\\=22\sqrt{2}+25-22\sqrt{2}+25\\=0+25\\\boxed{=25}$

De Reținut:

$\boxed{(a+b)^3=a^3+3a^2b+3ab^2+b^3}\\\boxed{(a-b)^3=a^3-3a^2b+3ab^2-b^3}}$

Sper ca te-am ajutat!