Question

Answer 1

`a// \sqrt{2x+1}=\sqrt{x}`        `ĐK: x >= 0`

`<=>2x+1=x`

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

Vậy ptr vô nghiệm

`b// \sqrt{4x-3}=\sqrt{x+2}`          `ĐK: x >= 3/4`

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

`<=>3x=5`

`<=>x=5/3` (t/m)

Vậy `S={5/3}`

Answer 2

a). đk x > = 0

=> \(\sqrt{2x+1}-\sqrt{x}=0\)

=> \(\sqrt{x}-\sqrt{x}+\sqrt{x+1}=0\)

=> \(\sqrt{x+1}=0\)

=> x = -1 ( ktm)