Đáp án:
$\sqrt[]{x^2+1}=x+4$
ĐKXĐ : $x ≥ -4$
$⇔(\sqrt[]{x^2+1})^2 =(x+4)^2$
$⇔x^2+1 =x^2+8x+16$
$⇔x^2-x^2-8x=16-1$
$⇔-8x=15$
$⇔x=-\dfrac{15}{8}$ (TM)
Vậy pt có 1 nghiệm $x=-\dfrac{15}{8}$
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$\sqrt[]{x^2+2x+1} =\dfrac{1}{2}$
$⇔\sqrt[]{(x+1)^2} =\dfrac{1}{2}$
$⇔| x+1 | =\dfrac{1}{2}$
$⇔$\(\left[ \begin{array}{l}x+1=\dfrac{1}{2}\\x+1=-\dfrac{1}{2}\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\x=-\dfrac{3}{2}\end{array} \right.\)
Vậy pt có 2 nghiệm x ∈ {$-\dfrac{1}{2}; -\dfrac{3}{2}$}
