Решите систему уравнений a) x+y=6 x²-y²=12 б) x²-y²=16 x²+y²=34

A) x=-y+  (y+6)^2 -y²=y²-y²+6y+36=12 6y=12-36 y=24/6 y=4 x+4=6 x=6-4 x=2

1)21x² =5x-121x²-5x+1=0 d=b²-4ac d = 25 - 4·21·1 = 25-84 = - 59  если d < 0, то действительных корней нет.2)x² -3x=0d=b²-4ac d = 9 - 4·1·0 = 9 √d =  √9 = 3 x₁ = (3+3)/2 = 6/2 = 3 x₂ = (3-3)/2 = 0/2 = 0 ответ: {0;   3} 3)2x² -72=0d=b²-4ac d = 0 - 4·2·(-72) = 576 √d =  √576 = 24 x₁ = (0+24)/4 = 24/4 = 6 x₂ = (0-24)/4 = -24/4 = -6 ответ: {-6;   6} 4)8x² -4=5x8x²-5x-4=0 d=b²-4ac d = 25 - 4·8·(-4) = 25+128= 153 √d =  √153  x₁ = (5+√153)/16  x₂ = (5-√153)/16 ответ: {(5-√153)/16 ;   (5+√153)/16 } 5)x² =18-3xx²+3x-18=0 d=b²-4ac d = 9 - 4·1·(-18) = 9+72=81 √d =  √81 = 9 x₁ = (-3+9)/2 = 6/2 = 3 x₂ = (-3-9)/2 = -12/2 = -6 ответ: {-6;   3} 6)9y² +12y+4=0d=b²-4ac d = 12² - 4·9·4 = 144-144=0  - один корень √d =  √0 = 0 y  = (-12)/18 = - ²/₃ ответ: - ²/₃

а) 48,

b1 = 48,   q = 1/4

b6 = b1*q^5 = 48/4^5 =  3/64

bn = b1*q^(n-1) =  48/4^(n-1) = 3*(4^(3-n))

б)  64/9, -32/

b1 = 64/9,   q = - 3/2

b6 = b1*q^5 = - 64*243/(9*32) =  -54

bn = b1*q^(n-1) =  (64/9)*(-3/2)^(n-1)

в)  -0,001; -0,

b1 = -0,001;   q = 10

b6 = b1*q^5 = -0,001*10^5 =  -100

bn = b1*q^(n-1)= -0,001* 10^(n-1) =  -10^(n-4) 

г) -100,  

b1= -100;   q = -0,1

b6 = b1*q^5 = 100 *(-10)^(-5) =  -0,001

bn = b1*q^(n-1) =  100*(-0,1)^(n-1)