Đáp án:
`(\sqrt{x^2+4}-2)(\sqrt{x^2+4}+2)(x+\sqrt{x}+1)\sqrt{x-2\sqrt{x}+1}=x^2.|x\sqrt{x}-1|\ \text{với}\ x\ge 0`
Giải thích các bước giải:
`(\sqrt{x^2+4}-2)(\sqrt{x^2+4}+2)(x+\sqrt{x}+1)\sqrt{x-2\sqrt{x}+1}\quad (x\ge 0)`
`=[(\sqrt{x^2+4})^2-2^2].(x+\sqrt{x}+1).\sqrt{(\sqrt{x}-1)^2}`
`=(x^2+4-4)(x+\sqrt{x}+1).|\sqrt{x}-1|`
`=x^2.(x+\sqrt{x}+1).|\sqrt{x}-1|`
Vì `x\ge 0=>x+\sqrt{x}+1\ge 1>0`
`=>x^2.(x+\sqrt{x}+1).|\sqrt{x}-1|`
`=x^2.|(x+\sqrt{x}+1)(\sqrt{x}-1)|`
`=x^2.|(\sqrt{x})^3-1^3|=x^2.|x\sqrt{x}-1|`
Vậy: `(\sqrt{x^2+4}-2)(\sqrt{x^2+4}+2)(x+\sqrt{x}+1)\sqrt{x-2\sqrt{x}+1}=x^2.|x\sqrt{x}-1| \ \text{với}\ \x\ge 0`
