Tìm x không âm, biết :
a) \(\sqrt x = 2\);
b) \(\sqrt x + 1 = 5\);
c) \(\sqrt {x - 1} + 1 = 4\);
d) \(\sqrt {x - 1} = \sqrt 3 \).
Sử dụng công thức: \(\sqrt {f\left( x \right)} = A \Leftrightarrow {\left[ {\sqrt {f\left( x \right)} } \right]^2} = {A^2} \Leftrightarrow f\left( x \right) = {A^2}.\)
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\(\begin{array}{l}a)\;\;\sqrt x = 2\;\;\left( {DK\;\;x \ge 0} \right)\\ \Leftrightarrow x = 4\;\;\left( {tm} \right)\end{array}\) Vậy \(x = 4.\) \(\begin{array}{l}c)\;\sqrt {x - 1} + 1 = 4\;\;\;\left( {DK:\;\;x \ge 1} \right)\\ \Leftrightarrow \sqrt {x - 1} = 3\\ \Leftrightarrow x - 1 = 9\\ \Leftrightarrow x = 10\;\;\left( {tm} \right).\end{array}\) Vậy \(x = 10.\) |
\(\begin{array}{l}b)\;\;\sqrt x + 1 = 5\;\;\;\;\left( {DK:\;\;x \ge 0} \right)\\ \Leftrightarrow \sqrt x = 5 - 1 = 4\\ \Leftrightarrow x = 16.\end{array}\) Vậy \(x = 16.\) \(\begin{array}{l}d)\;\;\sqrt {x - 1} = \sqrt 3 \;\;\;\left( {x \ge 1} \right)\\ \Leftrightarrow x - 1 = 3\\ \Leftrightarrow x = 4.\end{array}\) Vậy \(x = 4.\) |