Question

Giải các phương trình sau:

1) \(\sqrt{x^2-4}-x+2=0\)

2) \(\sqrt{1-2x^2}=x-1\)

Answer 1

`1)\sqrt{x^2-4}-x+2=0`               `ĐK: x >= 2,x <= -2`

`<=>\sqrt{(x-2)(x+2)}-(x-2)=0`

`<=>\sqrt{x-2}(\sqrt{x+2}-\sqrt{x-2})=0`

`<=>[(\sqrt{x-2}=0),(\sqrt{x+2}=\sqrt{x-2}):}`

`<=>[(x-2=0<=>x=2),(x+2=x-2<=>0x=-4\text{ (Vô lí)}):}`

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`2)\sqrt{1-2x^2}=x-1`          `ĐK: x >= 1,x <= -\sqrt{2}/2`

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

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

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

`<=>[(x=0),(x=2/3):}`  (ko t/m)

  `=>` Ptr vô nghiệm

Answer 2

x - 2 - x + 2=0

0=0(vô nghiệm)

1 - 2x=x - 1

-x - 2x=-1+1

-x =0

x=0