Đáp án+Giải thích các bước giải:
`a) sqrt{x^2+2x+1}=7` ĐK: `x≥-1/2`
`⇔x^2+2x+1=7^2`
`⇔(x+1)^2=49`
`⇔x+1=±7`
⇔$\left[\begin{matrix} x+1=7\\ x+1=-7\end{matrix}\right.$
⇔$\left[\begin{matrix} x=6 (TM) \\ x=-8 (loại)\end{matrix}\right.$
Vậy `S∈{6}`
`b) x-3sqrtx=0` ĐK: `x≥0`
`⇔x=3sqrtx`
`⇔x^2=9x`
`⇔x^2-9x=0`
`⇔x(x-9)=0`
⇔$\left[\begin{matrix} x=0\\ x-9=0\end{matrix}\right.$
⇔$\left[\begin{matrix} x=0 (TM)\\ x=9 (TM)\end{matrix}\right.$
Vậy `S∈{0; 9}`
`c) x-sqrt{x-2}=4` ĐK: x≥2
`⇔x-4=sqrt{x-2}`
`⇔(x-4)^2=x-2` `(x≥4)`
`⇔x^2-8x+16-x+2=0`
`⇔x^2-9x+18=0`
`⇔x^2-3x-6x+18=0`
`⇔x(x-3)-6(x-3)=0`
`⇔(x-6)(x-3)=0`
⇔$\left[\begin{matrix} x-6=0\\ x-3=0\end{matrix}\right.$
⇔$\left[\begin{matrix} x=6 (TM)\\ x=3 (loại)\end{matrix}\right.$
Vậy `S∈{6}`
`a) sqrt{x^2+6x+9}=3` ĐK: `x≥-3`
`⇔ x^2+6x+9=3^2`
`⇔ (x+3)^2=9`
`⇔x+3=±3`
⇔$\left[\begin{matrix} x+3=3\\ x+3=-3\end{matrix}\right.$
⇔$\left[\begin{matrix} x=0 (TM)\\ x=-6 (loại)\end{matrix}\right.$
Vậy `S∈{0}`
`b) 3x-2sqrtx-1=0` ĐK: `x≥0`
`⇔3x-1=2sqrtx`
`⇔(3x-1)^2=4x` `(x≥1/3)`
`⇔9x^2-6x+1-4x=0`
`⇔9x^2-10x+1=0`
`⇔9x^2-9x-x+1=0`
`⇔9x(x-1)-(x-1)=0`
`⇔(9x-1)(x-1)=0`
⇔$\left[\begin{matrix} 9x-1=0\\ x-1=0\end{matrix}\right.$
⇔$\left[\begin{matrix} x=1/9 (loại)\\ x=1 (TM)\end{matrix}\right.$
Vậy `S∈{1}`
`c) x+2sqrt{x+2}=1` ĐK: `x≥-2`
`⇔2sqrt{x+2}=1-x`
`⇔4(x+2)=(1-x)^2` `(x≥1)`
`⇔4x+2=1-2x+x^2`
`⇔x^2+2x-1-4x-2=0`
`⇔x^2-2x-3=0`
`⇔x^2-3x+x-3=0`
`⇔x(x-3)+(x-3)=0`
`⇔(x-3)(x+1)=0`
⇔$\left[\begin{matrix} x+1=0\\ x-3=0\end{matrix}\right.$
⇔$\left[\begin{matrix} x=-1 (loại)\\ x=1 (TM)\end{matrix}\right.$
Vậy `S∈{1}`
