Решите уравнение:1)9x^2=25;2) (2y-1)^2=8

1) x1=5/3;

x2=-5/3

2) x1=1/2+2^(1/2)

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

Объяснение:

1) 9x^2=25

x^2=25/9

x=+-(25/9) ^(1/2)

x1=(25^(1/2))/9^(1/2)=5/3

x2=-5/3

2) 4у^2-4у+1=8

4у^2-4у+1-8=0

4у^2-4у-7=0

D=b^2-4ac=(-4)^2-4×4×(-7)=16+112=128

x=(-b+-D^(1/2))/(2×a)=(4+-128^(1/2))/8

x1=(4+128^(1/2))/8=1/2+2^(1/2)

x2=(4-128^(1/2))/8=1/2-2^(1/2)

a² - b² = (a - b)(a + b)

9x² = 25

9x² - 5² = 0

(3x - 5)(3x + 5) = 0

3x - 5 = 0

x = 5/3

3x + 5 = 0

x = -5/3

x = {-5/3, 5/3}

(2y - 1)² = 8

|2y - 1| = √8

2y - 1 = ± 2√2

2y = 1 ± 2√2

y = 1/2 ± √2

y= { 1/2 + √2,  1/2 - √2}