Đáp án:
\(\begin{array}{l}
B3:\\
a)\dfrac{3}{2} \ge x\\
b)x \le 0\\
c)x \ge 0\\
d)\forall x\\
e)x > - 3\\
f)x \in \emptyset \\
g)x > 1\\
h)\forall x\\
i)x = 1\\
k)\forall x\\
l)\left[ \begin{array}{l}
x \ge 5\\
x \le 3
\end{array} \right.\\
m)\left\{ \begin{array}{l}
x \ge 2\\
x \ne 5
\end{array} \right.\\
n)5 > x \ge - 2\\
o)\left[ \begin{array}{l}
x \ge 1\\
x < - 2
\end{array} \right.\\
B4:\\
a)3\sqrt 2 - 4\\
b)2 + \sqrt 5 \\
c)4 + \sqrt 2 \\
d)\sqrt 5 - 1\\
e)2 + \sqrt 3 \\
f)3 - \sqrt 3
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B3:\\
a)DK: - 2x + 3 \ge 0 \to \dfrac{3}{2} \ge x\\
b)DK: - 5x \ge 0 \to x \le 0\\
c)DK:x \ge 0\\
d)DK:{x^2} + 1 \ge 0\left( {ld} \right)\forall x\\
KL:\forall x\\
e)DK:x + 3 > 0\\
\to x > - 3\\
f)DK:{x^2} + 6 < 0\left( {vô lý} \right)\\
Do:{x^2} + 6 > 0\forall x\\
KL:x \in \emptyset \\
g)DK:x - 1 > 0\\
\to x > 1\\
h)DK:{x^2} - 2x + 1 \ge 0\\
\to {\left( {x - 1} \right)^2} \ge 0\left( {ld} \right)\forall x\\
KL:\forall x\\
i)DK: - {x^2} - 2x - 1 \ge 0\\
\to - {\left( {x - 1} \right)^2} \ge 0\\
\to {\left( {x - 1} \right)^2} \le 0\\
\to x - 1 = 0\\
\to x = 1\\
k)DK:4{x^2} - 12x + 9 \ge 0\\
\to {\left( {2x - 3} \right)^2} \ge 0\left( {ld} \right)\forall x\\
l)DK:{x^2} - 8x + 15 \ge 0\\
\to \left( {x - 3} \right)\left( {x - 5} \right) \ge 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 3 \ge 0\\
x - 5 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 3 \le 0\\
x - 5 \le 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
x \ge 5\\
x \le 3
\end{array} \right.\\
m)DK:\left\{ \begin{array}{l}
x - 2 \ge 0\\
x - 5 \ne 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x \ge 2\\
x \ne 5
\end{array} \right.\\
n)DK:\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 2 \ge 0\\
5 - x > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 2 \le 0\\
5 - x < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
5 > x \ge - 2\\
\left\{ \begin{array}{l}
x \le - 2\\
x > 5
\end{array} \right.\left( l \right)
\end{array} \right.\\
o)DK:\left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 1 \ge 0\\
x + 2 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 1 \le 0\\
x + 2 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
x \ge 1\\
x < - 2
\end{array} \right.\\
B4:\\
a)\sqrt {{{\left( {4 - 3\sqrt 2 } \right)}^2}} = \left| {4 - 3\sqrt 2 } \right| = 3\sqrt 2 - 4\\
b)\sqrt {{{\left( {2 + \sqrt 5 } \right)}^2}} = \left| {2 + \sqrt 5 } \right| = 2 + \sqrt 5 \\
c)\sqrt {{{\left( {4 + \sqrt 2 } \right)}^2}} = \left| {4 + \sqrt 2 } \right| = 4 + \sqrt 2 \\
d)\sqrt {6 - 2\sqrt 5 } = \sqrt {5 - 2\sqrt 5 .1 + 1} \\
= \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} = \sqrt 5 - 1\\
e)\sqrt {7 + 4\sqrt 3 } = \sqrt {4 + 2.2\sqrt 3 + 3} \\
= \sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} = 2 + \sqrt 3 \\
f)\sqrt {12 - 6\sqrt 3 } = \sqrt {9 - 2.3.\sqrt 3 + 3} \\
= \sqrt {{{\left( {3 - \sqrt 3 } \right)}^2}} = 3 - \sqrt 3
\end{array}\)
