Question

va rog mult ex 10. Matematica

Answer 1

$\div\: a,b,c < = > b = \frac{a + c}{2}$

$e)a = x + 4$

$b = 2x + 5$

$c = x + 8$

$b = \frac{a + c}{2}$

$2x + 5 = \frac{x + 4 + x + 8}{2}$

$2x + 5 = \frac{2x + 12}{2}$

$2x + 5 = \frac{2(x + 6)}{2}$

$2x + 5 = x + 6$

$2x - x = 6 - 5$

$x = 1$

$f)a = x + 1$

$b = 3x$

$c = 2x + 5$

$b = \frac{a + c}{2}$

$3x = \frac{ x + 1 + 2x + 5 }{2}$

$3x = \frac{3x + 6}{2}$

$3x \times 2= 3x + 6$

$6x = 3x + 6$

$6x - 3x = 6$

$3x = 6 \: | \div 3$

$x = 2$

$g)a = {x}^{2} - 1$

$b = 3x - 4$

$c = x - 1$

$b = \frac{a + c}{2}$

$3x - 4 = \frac{ {x}^{2} - 1 + x - 1 }{2}$

$3x - 4 = \frac{ {x}^{2} + x - 2 }{2}$

$2(3x - 4) = {x}^{2} + x - 2$

$6x - 8 = {x}^{2} + x - 2$

$- {x}^{2} + 6x - x - 8 + 2 = 0$

$- {x}^{2} + 5x - 6 = 0 \: | \times ( - 1)$

${x}^{2} - 5x + 6 = 0$

${x}^{2} - 2x - 3x + 6 = 0$

$x(x - 2) - 3(x - 2) = 0$

$(x - 3)(x - 2) = 0$

$x - 3 = 0 = > x = 3$

$x - 2 = 0 = > x = 2$

$h)a = x - 2$

$b = {x}^{2}$

$c = x + 6$

$b = \frac{a + c}{2}$

${x}^{2} = \frac{x - 2 + x + 6}{2}$

${x}^{2} = \frac{2x + 4}{2}$

${x}^{2} = \frac{2(x + 2)}{2}$

${x}^{2} = x + 2$

${x}^{2} - x - 2 = 0$

${x}^{2} - 2x + x - 2 = 0$

$x(x - 2) +1( x - 2 )= 0$

$(x - 2)(x + 1) = 0$

$x - 2 = 0 = > x = 2$

$x + 1 = 0 = > x = - 1$

Answer 2

Nr date sunt in progresie aritmetica daca cel din mijloc este media aritmetica a vecinilor.

e) 2x+5 = (x+4+x+8)/2

2(2x+5) = 2x + 12

4x + 10 = 2x+12

2x = 2

x = 1

=> 5, 7, 9

f) 3x = (x+1+2x+5)/2

6x = 3x + 6

3x = 6

x = 2

=> 3, 6, 9

g) 3x-4 = (x²-1+x-1)/2

6x-8 = x²+x-2

-x²+5x - 6 = 0

x² - 5x + 6 = 0

Delta = 25-4*1*6 = 1

x1 = (5+1)/2 = 3

x2 = (5-1)/2 = 2

=> 8, 5, 2 (pt x = 3)

Sau 3, 2, 1 (pt x = 2)

h) x² = (x-2+x+6)/2

2x² = 2x + 4

2x² - 2x -4 = 0

x² - x - 2 = 0

Delta = 1-4*1*(-2) = 9

x1 = (1+3)/2 = 2

x2 = (1-3)/2 = -1

=> 0, 4, 8 (pt x = 2)

sau -3, 1, 5 (pt x = -1)