Đáp án:
B2:
a) ÁP dụng đl Talet ta có:
$\begin{array}{l}
\frac{{IH}}{{FK}} = \frac{{EH}}{{EK}}\\
\Rightarrow \frac{{14}}{x} = \frac{{12}}{{24}} = \frac{1}{2}\\
\Rightarrow x = 2.14 = 28\left( {cm} \right)
\end{array}$
b) Áp dụng t.c đường phân giác :
$\begin{array}{l}
\frac{{AC}}{{AD}} = \frac{{CV}}{{VD}}\\
\Rightarrow \frac{x}{{60}} = \frac{{32}}{{40}} = \frac{4}{5}\\
\Rightarrow x = 60.\frac{4}{5} = 48\left( {cm} \right)\\
B3:\\
a)Dkxd:x \ne a;x \ne - a\\
Pt:\frac{{x + a}}{{a - x}} - \frac{{x - a}}{{a + x}} = \frac{{a\left( {3a + 1} \right)}}{{{a^2} - {x^2}}}\\
\Rightarrow \frac{{\left( {x + a} \right)\left( {x + a} \right) + \left( {a - x} \right)\left( {a - x} \right)}}{{\left( {a - x} \right)\left( {a + x} \right)}} = \frac{{3{a^2} + a}}{{{a^2} - {x^2}}}\\
\Rightarrow {x^2} + 2ax + {a^2} + {a^2} - 2ax + {x^2} = 3{a^2} + a\\
\Rightarrow 2{x^2} + 2{a^2} = 3{a^2} + a\\
\Rightarrow 2{x^2} = {a^2} + a\\
+ a = - 3 \Rightarrow 2{x^2} = 9 - 3 = 6\\
\Rightarrow {x^2} = 3\\
\Rightarrow x = \pm \sqrt 3 \left( {tm} \right)\\
b)a = 1\\
\Rightarrow 2{x^2} = 2\\
\Rightarrow {x^2} = 1\\
\Rightarrow \left[ \begin{array}{l}
x = 1\left( {ktm:x \ne a} \right)\\
x = - 1\left( {ktm:x \ne - a} \right)
\end{array} \right.
\end{array}$
Vậy pt vô nghiệm khi a=1
