#LC
a, ĐKXĐ: `x^2 - 4x+4 ≥0`
⇔`(x-2)^2≥0 ∀ x∈ R`
⇔ `x∈ R`
b, ĐKXĐ: $\left \{ {{1+x≥0} \atop {16+16x≥0}} \right.$ ⇔ `x≥-1`
c, ĐKXĐ: `x+1≥0` ⇔ `x≥-1`
`3x - \sqrt(x+1) + 1=0` (1)
Đặt `\sqrt(x+1) = t` ; `t≥0`
⇒`3t^2 = 3x+3`
PT (1) ⇔ `3x+3 - \sqrt(x+1) - 2=0`
⇔`3t^2 - t -2=0`
⇔`(t-1)(3t+2)=0`
⇔\(\left[ \begin{array}{l}t=1(t/m)\\x=\frac{-2}{3}(loại)\end{array} \right.\)
⇔ `\sqrt(x+1) = 1`
⇔ `x+1= 1`
⇔ `x=0` (t/m)
