(4ab-8ac)+(b²-2bc)=4a(b-2c)+b(b-2c)=(b-2c)(4a+b)
Решение
2x² - y² = 34
xy = 20
y = 20/x
2x² - (20/x)² = 34
2x² - 400/x² - 34 = 0
x⁴ - 17x² - 200 = 0
x² = t, t ≥ 0
t² - 17t - 200 = 0
D = 289 + 4*1*200 = 1089
t₁ = (17 - 33)/2
t₁ = - 8 не удовлетворяет условию t ≥ 0
t₂ = (17 + 33)/2
t₂ = 25
x² = 25
x₁ = - 5
x₂ = 5
x₁= - 5
y₁ = 20/(-5)
y₁ = - 4
x₂ = 5
y₂ = 20/5
y₂ = 4
Ответ: (- 5; - 4) (5; 4)
(a+b)-(a-b)-(b-a)= a+b-a+b-b+a= a+b
3m-(2m-3)+(2-m)= 3m-2m+3+2+m= 2m+5
(3y-1)-(2y-2)+(y-3)= 3y-1-2y+2+y-3= 2y-2= 2(y-1)