Question

Ajutati-ma va rog mult e nevoie urgent.Mersi

Answer 1

1)
9 + ... + 25z⁴ = 3² + ... + (5z²)² = 3² + 2 × 3 × 5z² + (5z²)² = (3 + 5z²)²

2)
(y - 3)³ + (y + 1)³ =
= y³ - 3*3y² + 3*9y -27 + y³ +3y² + 3y + 1 = 
= y³ - 9y² + 27y -27 + y³ +3y² + 3y + 1 = 
= 2y³ - 6y² + 30y -26  = 
= 2(y³ - 3y² + 15y - 13)=
= 2(y³ - y³ - 2y² + 2y + 13y - 13)=
=2[y²(y-1) - 2y(y - 1) + 13(y - 1)]=
= 2[(y - 1)(y² - 2y + 13)] =
= 2(y - 1)(y² - 2y + 13)

3)
A = L²
A1 = 4² = 16
A2 = (x - 1)² = x² - 2x + 1
A3 = (x + 2)² = x² + 4x + 4
A1 + A3 = A4
16 + x² - 2x + 1 = x² + 4x + 4
x² - 2x + 17 = x² + 4x + 4
x² - x² - 2x - 4x = 4 - 17
- 6x = - 13   | ×(-1)
6x = 13
 x = 13 / 6
 
4)
4a)  x² + 6x + 9 = (x + 3)² ;  celalalt numitor = x+3
Numaratorul fractiei care se va rasturna *x² - x + 1) nu are solutii reale.
=>   DVA = R \ {-3}

4b)
 $E(x)= \frac{2 x^{3}+2 }{ x^{2}+6x+9} : \frac{ x^{2} -x+1}{x+3}= \\ \\ =\frac{2 (x^{3}+1) }{ (x+3)^{2}} : \frac{ x^{2} -x+1}{x+3}= \\ \\ =\frac{2 (x+1)( x^{2} -x+1) }{ (x+3)^{2}} : \frac{ x^{2} -x+1}{x+3}= \\ \\ = \frac{2 (x+1)( x^{2} -x+1) }{ (x+3)^{2}} * \frac{x+3}{ x^{2} -x+1}= \\ \\ = \frac{2(x+1)}{x+3} \\ \\ => E(x) = \frac{2(x+1)}{x+3}$
 
4c)  
$E(- \frac{1}{2})=\frac{2(- \frac{1}{2} +1)}{-\frac{1}{2}+3} = \frac{2\frac{1}{2}}{ \frac{5}{2}}=2* \frac{1}{2} * \frac{2}{5} = \frac{2}{5}$

4d)
$E(x) = \frac{2(x+1)}{x+3}=\frac{2x+2}{x+3} \\ \\ 2x+2 = x+3 \\ x = 3-2 = 1 \\ => x = 1 \;\;\text{apartine lui N} \\ \\ E(1) = \frac{2+2}{1+3}= \frac{4}{4} =1 \;\;\text{apartine lui N}$


5)
$4^{n}-2^{n+1} + 1= \\ =(2^{2})^{n}-2*2^{n} + 1= \\ = (2^{n})^{2} - 2*2^{n}*1+1= \\=(2^{n}-1)^{2} \\ cctd$