Toán Lớp 9: giải phương trình √(x+3)=4-√(x-1)
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Toán Lớp 9: giải phương trình √(x+3)=4-√(x-1)
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DK:x \ge 1\\
\sqrt {x + 3} = 4 – \sqrt {x – 1} \\
\to \sqrt {x + 3} + \sqrt {x – 1} = 4\\
\to x + 3 + 2\sqrt {\left( {x + 3} \right)\left( {x – 1} \right)} + x – 1 = 16\\
\to 2x + 2 + 2\sqrt {{x^2} + 2x – 3} = 16\\
\to 2\sqrt {{x^2} + 2x – 3} = 14 – 2x\\
\to \sqrt {{x^2} + 2x – 3} = 7 – x\\
\to {x^2} + 2x – 3 = 49 – 14x + {x^2}\left( {DK:7 \ge x} \right)\\
\to 16x = 52\\
\to x = \dfrac{{13}}{4}
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