Với giá trị nào của a thì phương trình có nghiệm. Tính x theo a.
A.\(\begin{array}{l}a \ge - \dfrac{1}{2}\\x = {\left( {\sqrt {a + \dfrac{1}{2}} - \dfrac{1}{2}} \right)^2} - \dfrac{3}{4}\end{array}\)
B.\(\begin{array}{l}a \ge - \dfrac{1}{4}\\x = {\left( {\sqrt {a + \dfrac{1}{2}} + \dfrac{1}{2}} \right)^2} - \dfrac{3}{4}\end{array}\)
C.\(\begin{array}{l}a \ge - \dfrac{1}{4}\\x = {\left( {\sqrt {a + \dfrac{1}{2}} - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}\end{array}\)
D.\(\begin{array}{l}a \ge - \dfrac{1}{4}\\x = {\left( {\sqrt {a + \dfrac{1}{2}} - \dfrac{1}{2}} \right)^2} - \dfrac{3}{4}\end{array}\)
A.\(\begin{array}{l}a \ge - \dfrac{1}{2}\\x = {\left( {\sqrt {a + \dfrac{1}{2}} - \dfrac{1}{2}} \right)^2} - \dfrac{3}{4}\end{array}\)
B.\(\begin{array}{l}a \ge - \dfrac{1}{4}\\x = {\left( {\sqrt {a + \dfrac{1}{2}} + \dfrac{1}{2}} \right)^2} - \dfrac{3}{4}\end{array}\)
C.\(\begin{array}{l}a \ge - \dfrac{1}{4}\\x = {\left( {\sqrt {a + \dfrac{1}{2}} - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4}\end{array}\)
D.\(\begin{array}{l}a \ge - \dfrac{1}{4}\\x = {\left( {\sqrt {a + \dfrac{1}{2}} - \dfrac{1}{2}} \right)^2} - \dfrac{3}{4}\end{array}\)
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