Question

Answer 1

`a)` Thay `x=25` (t/m đk) vào `B` có: `B=[\sqrt{25}-2]/[\sqrt{25}+3]=3/8`

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`b)` Với `x >= 0,x \ne 9,x \ne 16` có:

`A=[2\sqrt{x}+1-(\sqrt{x}+3)(\sqrt{x}-3)+(2\sqrt{x}+1)(\sqrt{x}-4)]/[(\sqrt{x}-3)(\sqrt{x}-4)]`

`A=[2\sqrt{x}+1-x+9+2x-8\sqrt{x}+\sqrt{x}-4]/[(\sqrt{x}-3)(\sqrt{x}-4)]`

`A=[x-5\sqrt{x}+6]/[(\sqrt{x}-3)(\sqrt{x}-4)]`

`A=[(\sqrt{x}-2)(\sqrt{x}-3)]/[(\sqrt{x}-3)(\sqrt{x}-4)]=[\sqrt{x}-2]/[\sqrt{x}-4]`

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`c)` Với `x >= 0,x \ne 9;16` có:

`M=A:B=[\sqrt{x}-2]/[\sqrt{x}-4]:[\sqrt{x}-2]/[\sqrt{x}+3]=[\sqrt{x}+3]/[\sqrt{x}-4]`

  `=>M > 8<=>[\sqrt{x}+3]/[\sqrt{x}-4] > 8`

       `<=>[\sqrt{x}+3-8\sqrt{x}+32]/[\sqrt{x}-4] > 0`

       `<=>[35-7\sqrt{x}]/[\sqrt{x}-4] > 0`

     `<=>[\sqrt{x}-5]/[\sqrt{x}-4] < 0<=>4 < \sqrt{x} < 5<=>16 < x < 25` (t/m)