Question

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Answer 1

Răspuns:

1. a2=4, a6=16

a2=a1+r

a6=a1+5r

4=a1+r

6=a1+5r

-------------- (-)

6-4=5r-r

2=4r

r=1/2

4=a1+1/2=>8=2a1+1 ==> 2a1=7 => a1=7/2

S10= [2a1+(n-1)r]n/2 = [2 × 7/2 +9 ×1/2]9 / 2=( 7+9/2)×9/2=

$\frac{7 \times 9 + \frac{9 \times 9}{2} }{2} = \frac{2 \times 7 \times 9 + 9 \times 9}{2} \times \frac{1}{2} = \frac{207}{4}$

2.

$x + 2 = \sqrt{(x - 4) \times (x + 4)} = \sqrt{ {x}^{2} - 16}$

${x}^{2} + 4x + 4 = {x}^{2} - 16$

$4x = - 20 = > x = - 5$

${(x + 2)}^{2} = {x}^{2} - 16$

3. f(x) =x-2, g(x)=7-2x

a)

$f(x) \geqslant g(x) \\ x - 2 \geqslant 7 - 2x \\ x + 2x \geqslant 7 + 2 \\ 3x \geqslant 9 \\ x \geqslant 3$

x aparține intervalului [3, infinit)

b) g(f(x)) =7-2(x-2)=7-2x+2=9-2x

(g°f) (3)=9-2×3=9-6=3 care este nr natural

c) M apartine Gf => f(0)=m

f(0)=-2 =>m=-2

N aparține Gf => f(n) =0

n-2=0 => n=3

d)

$a = g( \frac{1}{ \sqrt{2} } ) = 7 - 2 \times \frac{1}{ \sqrt{2} } = \\ 7 - \frac{2 \sqrt{2} }{2} = 7 - \sqrt{2} \\ b = g( \frac{1}{ \sqrt{3} } ) = 7 - 2 \times \frac{1}{ \sqrt{3} } = \\ 7 - \frac{2 \sqrt{3} }{3} = \frac{21 - 2 \sqrt{3} }{3}$

$\sqrt{2 } = 1.41 \\ \sqrt{3 } = 1.73$

$a = 7 - 1.41 = 5.59$

$b = 5.84$

a<b

e)m=-2, n=2

O(0,0), M(0,-2), N(2,0)

Aria OMN = 1/2 | 0 0 1 | =

| 0 - 2 1|

| 2 0 1|

= 1/2(0+0+0-(-4)-0-0)=1/2×4=2