Toán Lớp 9: Giải các phương trình :
a, √ 2x – 1 = √ x + 1
b, x – √ x + 4 = 2
c, √ 4 – x^2 – x + 2 = 0
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Toán Lớp 9: Giải các phương trình :
a, √ 2x – 1 = √ x + 1
b, x – √ x + 4 = 2
c, √ 4 – x^2 – x + 2 = 0
Comments ( 1 )
a)x = 2\\
b)x = 5\\
c)x = 2
\end{array}\)
a)DK:x \ge \dfrac{1}{2}\\
\sqrt {2x – 1} = \sqrt {x + 1} \\
\to 2x – 1 = x + 1\\
\to x = 2\\
b)DK:x \ge – 4\\
x – \sqrt {x + 4} = 2\\
\to x – 2 = \sqrt {x + 4} \\
\to {x^2} – 4x + 4 = x + 4\left( {DK:x \ge 2} \right)\\
\to {x^2} – 5x = 0\\
\to \left[ \begin{array}{l}
x = 5\\
x = 0\left( l \right)
\end{array} \right.\\
\to x = 5\\
c)DK:2 \ge x \ge – 2\\
\sqrt {4 – {x^2}} – x + 2 = 0\\
\to \sqrt {\left( {2 – x} \right)\left( {x + 2} \right)} + \left( {2 – x} \right) = 0\\
\to \sqrt {2 – x} \left( {\sqrt {x + 2} + \sqrt {2 – x} } \right) = 0\\
\to 2 – x = 0\left( {do:\sqrt {x + 2} + \sqrt {2 – x} > 0\forall 2 \ge x \ge – 2} \right)\\
\to x = 2
\end{array}\)