Đáp án:
Giải thích các bước giải:
a) `\sqrt{x^2+4x+4}=1-2x`
`ĐK: x \le 1/2`
`⇔ x^2+4x+4=1-4x+4x^2`
`⇔ 3x^2-8x-3=0`
`⇔ (x-3)(3x+1)=0`
`⇔` \(\left[ \begin{array}{l}x-3=0\\3x+1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=3\ (loại)\\x=-\dfrac{1}{3}\ (TM)\end{array} \right.\)
Vậy `S={-\frac{1}{3}}`
b) `x-2-2\sqrt{x-2}=3`
`⇔ 2\sqrt{x-2}=x-5`
`ĐK: x \ge 5`
`⇔ 4(x-2)=x^2-10x+25`
`⇔ 4x-8=x^2-10x+25`
`⇔ x^2-14x+33=0`
`⇔ (x-11)(x-3)=0`
`⇔ \(\left[ \begin{array}{l}x-3=0\\x-11=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=3\ (loại)\\x=11\ (TM)\end{array} \right.\)
Vậy `S={11}`
