Đáp án:
B6:
b. A=2
Giải thích các bước giải:
\(\begin{array}{l}
B5:\\
a.A = x\sqrt 2 - \sqrt {2{x^2} + 2\sqrt 2 x + 1} \\
= x\sqrt 2 - \sqrt {{{\left( {x\sqrt 2 + 1} \right)}^2}} \\
= x\sqrt 2 - \left| {x\sqrt 2 + 1} \right|\\
\to \left[ \begin{array}{l}
A = x\sqrt 2 - x\sqrt 2 - 1\left( {DK:x \ge - \dfrac{1}{{\sqrt 2 }}} \right)\\
A = x\sqrt 2 + x\sqrt 2 + 1\left( {DK:x < - \dfrac{1}{{\sqrt 2 }}} \right)
\end{array} \right.\\
\left[ \begin{array}{l}
A = - 1\\
A = 2x\sqrt 2 + 1
\end{array} \right.\\
b.A = - 3\\
\to 2x\sqrt 2 + 1 = - 3\\
\to 2x\sqrt 2 = - 4\\
\to x = - \dfrac{4}{{2\sqrt 2 }} = - \sqrt 2 \left( {TM} \right)\\
B6:\\
a.DK:{x^2} - 1 \ge 0 \to \left[ \begin{array}{l}
1 \ge x\\
x \ge - 1
\end{array} \right.\\
b.A = \sqrt {{x^2} - 1 + 2\sqrt {{x^2} - 1} .1 + 1} - \sqrt {{x^2} - 1 - 2\sqrt {{x^2} - 1} .1 + 1} \\
= \sqrt {{{\left( {\sqrt {{x^2} - 1} + 1} \right)}^2}} - \sqrt {{{\left( {\sqrt {{x^2} - 1} - 1} \right)}^2}} \\
= \left| {\sqrt {{x^2} - 1} + 1} \right| - \left| {\sqrt {{x^2} - 1} - 1} \right|\\
\to \left[ \begin{array}{l}
A = \sqrt {{x^2} - 1} + 1 - \sqrt {{x^2} - 1} + 1\left( {DK:\sqrt {{x^2} - 1} - 1 \ge 0 \to \left[ \begin{array}{l}
x \ge \sqrt 2 \\
x \le - \sqrt 2
\end{array} \right.} \right)\\
A = \sqrt {{x^2} - 1} + 1 + \sqrt {{x^2} - 1} - 1\left( {DK: - \sqrt 2 < x < \sqrt 2 } \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
A = 2\\
A = 2\sqrt {{x^2} - 1}
\end{array} \right.\\
b.\left| x \right| \ge \sqrt 2 \\
\to \left[ \begin{array}{l}
x \ge \sqrt 2 \\
x \le - \sqrt 2
\end{array} \right.\\
\to A = 2
\end{array}\)
