Giải pt $\sqrt{x\left(x-1\right)}+\sqrt{x\left(x+2\right)}=2\sqrt{x^2}$

Đáp án:

`x in{0; 9/8}`

Giải thích các bước giải:

 `\sqrt{x.(x-1)} + \sqrt{x.(x+2)} = 2\sqrt{x^2}` (ĐK: `x>=0)`

`<=> \sqrt{x^2-x} + \sqrt{x^2+2x} = 2.|x|`

`<=> x^2- x+2\sqrt{(x^2-x)(x^2+ 2x)} + x^2+ 2x= 4.|x|^2`

`<=> x^2-x +2\sqrt{x^4+ 2x^3- x^3- 2x^2} +x^2+2x= 4x^2`

`<=> 2x^2 +x+2\sqrt{x^4+x^3- 2x^2}= 4x^2`

`<=> 2\sqrt{x^4+ x^3- 2x^2} = 4x^2- 2x^2-x`

`<=> 2\sqrt{x^4+ x^3- 2x^2} = 2x^2- x`

`<=> 4(x^4+x^3- 2x^2) = 4x^4- 4x^3+x^2`

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

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

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

`<=> 8x^3-9x^2= 0`

`<=> x^2.(8x-9)= 0`

`<=>`\(\left[ \begin{array}{l}x^2=0\\8x-9=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=0\\x=9/8\end{array} \right.\) (tm ĐK)

$Vậy$ `x in{0; 9/8}`