Trong \(\Delta ABC\), nếu \(\dfrac{{\tan A}}{{\tan C}} = \dfrac{{{{\sin }^2}A}}{{{{\sin }^2}C}}\) thì \(\Delta ABC\) là tam giác gì?
A.Tam giác vuông.
B.Tam giác cân.
C.Tam giác đều.
D.Tam giác vuông hoặc cân.

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Quy đồng các phân thức \(\dfrac{{3{x^2} + 2x - 4}}{{{x^3} - 7x + 6}};\,\)\(\dfrac{{2x - 1}}{{{x^2} + x - 6}};\)\(\dfrac{3}{{x + 3}}\)
A.\(\dfrac{3x^2 + 2x - 4}{\left ( x + 3 \right )\left ( x - 2 \right )\left ( x - 1 \right )}\,;\,\,\dfrac{\left ( 2x - 1 \right )\left ( x - 1 \right )}{\left ( x + 3 \right )\left ( x - 2 \right )\left ( x - 1 \right )}\,;\,\,\dfrac{3\left ( x - 2 \right )\left ( x - 1 \right )}{\left ( x + 3 \right )\left ( x - 2 \right )\left ( x - 1 \right )}\)
B.\(\dfrac{3x^2 + 2x - 4}{\left ( x - 3 \right )\left ( x + 2 \right )\left ( x - 1 \right )}\,;\,\,\dfrac{\left ( 2x - 1 \right )\left ( x - 1 \right )}{\left ( x - 3 \right )\left ( x + 2 \right )\left ( x - 1 \right )}\,;\,\,\dfrac{3\left ( x + 2 \right )\left ( x - 1 \right )}{\left ( x - 3 \right )\left ( x + 2 \right )\left ( x - 1 \right )}\)
C.\(\dfrac{3x^2 + 2x - 4}{\left ( x - 3 \right )\left ( x + 2 \right )\left ( x + 1 \right )}\,;\,\,\dfrac{\left ( 2x - 1 \right )\left ( x + 1 \right )}{\left ( x - 3 \right )\left ( x + 2 \right )\left ( x + 1 \right )}\,;\,\,\dfrac{3\left ( x + 2 \right )\left ( x + 1 \right )}{\left ( x - 3 \right )\left ( x + 2 \right )\left ( x + 1 \right )}\)
D.\(\dfrac{3x^2 + 2x - 4}{\left ( x + 3 \right )\left ( x - 2 \right )\left ( x + 1 \right )}\,;\,\,\dfrac{\left ( 2x - 1 \right )\left ( x + 1 \right )}{\left ( x + 3 \right )\left ( x - 2 \right )\left ( x + 1 \right )}\,;\,\,\dfrac{3\left ( x - 2 \right )\left ( x + 1 \right )}{\left ( x + 3 \right )\left ( x - 2 \right )\left ( x + 1 \right )}\)