1 и 2 задание, пожалуйста!!!. Created by MariaRo10. algebra-ru

1 и 2 задание, пожалуйста!!!...

GarveL I
1. 4x² - 12xy + 9y² - 4a² + 4ab - b² = ( (2x)² - 2*2x*3y + (3y)²) - ( (2a)² - 2*2a*b + b²) = (2x - 3y)² - (2a - b)² = (2x - 3y - 2a + b)(2x - 3y + 2a - b)
2 / (x-y)(x+y) + 2(2x - y) + 3 = x² - y² + 4x - 2y + 4 -1 = (x+2)² - (y+1)² = (x+y+3)(x-y+1)
II
1/ x³ - 3x² - 4x + 12 =0
x²(x-3) - 4(x-3)=0
(x²-4)(x-3)=0
(x-2)(x+2)(x-3)=0
x=2
x=-2
x=3
2/ x^4 + 2x^3 + 8x + 16=0
x³(x+2) + 8(x+2) = 0
(x+2)(x³+8)=0
(x+2)(x+2)(x²-2x+4) = 0
x=-2
x²-2x+4=0 D=4-16<0 решений нет

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mmb1

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2)
1)x^3-3x^2-4x+12=0
x^2×(x-3)-4(x-3)0
(x^2-4)×(x-3)=0
x^2-4=0
x-3=0
Ответ:
x2=2
x1=-2
x3=3
2)x^4+2x^3+8x+16=0
x^3×(x+2)+8(x+2)=0
(x^3+8)×(x+2)=0
x^+8=0
x+2=0
Ответ:x=-2

2 votes Thanks 3

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