Tìm \(x\) biết:
\(a)\,\,\frac{{\sqrt {2x - 1} }}{{\sqrt {x - 1} }} = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b)\,\,\frac{{\sqrt {{x^2} - 4} }}{{\sqrt {x - 2} }} = 3\)
A.(begin{array}{l}a),,,S = left{ {frac{3}{2}} right}.\b),,S = left{- 7 right}.end{array})
B.(begin{array}{l}a),,,S = left{ {frac{3}{2}} right}.\b),,S = left{ 7 right}.end{array})
C.(begin{array}{l}a),,,S = left{ {frac{3}{2}} right}.\b),,S = left{ 2 right}.end{array})
D.(begin{array}{l}a),,,S = left{ {frac{3}{2}} right}.\b),,S = left{ 2;,7 right}.end{array})
\(a)\,\,\frac{{\sqrt {2x - 1} }}{{\sqrt {x - 1} }} = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b)\,\,\frac{{\sqrt {{x^2} - 4} }}{{\sqrt {x - 2} }} = 3\)
A.(begin{array}{l}a),,,S = left{ {frac{3}{2}} right}.\b),,S = left{- 7 right}.end{array})
B.(begin{array}{l}a),,,S = left{ {frac{3}{2}} right}.\b),,S = left{ 7 right}.end{array})
C.(begin{array}{l}a),,,S = left{ {frac{3}{2}} right}.\b),,S = left{ 2 right}.end{array})
D.(begin{array}{l}a),,,S = left{ {frac{3}{2}} right}.\b),,S = left{ 2;,7 right}.end{array})
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