Question

Giúp e vs=(

Answer 1

`C_2:`

`a)\sqrt{x^2+6x+9}=5`         

`<=>\sqrt{(x+3)^2}=5`

`<=>|x+3|=5`

`<=>[(x+3=5),(x+3=-5):}<=>[(x=2),(x=-8):}`

~~~~~~~~~~~~~~

Có vẻ đề phải là: `\sqrt{9x-9}?`

`b)\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{4x-4}+\sqrt{x-1}=8`      `ĐK: x >= 1`

`<=>4\sqrt{x-1}-3\sqrt{x-1}+2\sqrt{x-1}+\sqrt{x-1}=8`

`<=>4\sqrt{x-1}=8`

`<=>\sqrt{x-1}=2`

`<=>x-1=4<=>x=5` (t/m)

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`C_3:`

`a)A` xác định `<=>{(x > 0),(\sqrt{x}-1 \ne 0),(\sqrt{x}-2 \ne 0):}<=>{(x > 0),(x \ne 1,x \ne 2):}`

Với `x > 0,x \ne 1,x \ne 2` có:

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

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

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

`A=1/\sqrt{x} .[\sqrt{x}+2]/3`

`A=[\sqrt{x}+2]/[3\sqrt{x}]`

~~~~~~~~~~~~~~~~~~~~~~

`b)` Với `x > 0,x \ne 1,x \ne 2` có:

`A=1/4<=>[\sqrt{x}+2]/[3\sqrt{x}]=1/4`

          `<=>4\sqrt{x}+8=3\sqrt{x}`

         `<=>\sqrt{x}=-8` (Vô lí)

 `=>` Ko có gtr của `x` t/m