Question

Answer 1

`a)x(x-2)-x^2+3x=4`

`<=>x^2-2x-x^2+3x-4=0`

`<=>x=4`

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`b)3x^2-3x=(x-1)^2`

`<=>3x(x-1)-(x-1)^2=0`

`<=>(x-1)(3x-x+1)=0`

`<=>(x-1)(2x+1)=0`

`<=>[(x=1),(x=-1/2):}`

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`c)(x+2)(x^2-2x+4)-x(x-2)^2=-12`

`<=>x^3+8-x^3+4x^2-4x+12=0`

`<=>4x^2-4x+1=-19`

`<=>(2x-1)^2=-19` (Vô lí)

  `=>` Ptr vô nghiệm

Answer 2

a: =>x^2-2x-x^2+3x=4

=>x=4

b: =>3x(x-1)-(x-1)^2=0

=>(x-1)(3x-x+1)=0

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

=>x=1 hoặc x=-1/2

c: =>x^3+8-x(x^2-4x+4)=-12

=>x^3+8-x^3+4x^2-4x=-12

=>4x^2-4x+20=0

=>\(x\in\varnothing\)

Answer 3

a) x(x - 2) - x² + 3x = 4

x² - 2x - x² + 3x = 4

x = 4

b) 3x² - 3x = (x - 1)²

3x(x - 1) = (x - 1)²

3x(x - 1) - (x - 1)² = 0

(x - 1)(3x - x + 1) = 0

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

x - 1 = 0 hoặc 2x + 1 = 0

*) x - 1 = 0

x = 1

*) 2x + 1 = 0

2x = -1

x = -1/2

Vậy x = -1/2; x = 1

c) (x + 2)(x² - 2x + 4) - x(x - 2)² = -12

x³ + 8 - x(x² - 4x + 4) = -12

x³ + 8 - x³ + 4x² - 4x + 12 = 0

4x² - 4x + 20 = 0

Ta có: 4x² - 4x + 20

= 4x² - 4x + 1 + 19

= (2x - 1)² + 19

Do (2x - 1)² >= 0

Suy ra (2x - 1)² + 19 > 0 với mọi x

Vậy không tìm được x thỏa mãn đề bài