Đáp án:
B1:
7) \(\dfrac{1}{2} > x\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
1)DK: - 2x + 3 \ge 0 \to \dfrac{3}{2} \ge x\\
2)DK:x \ne 0\\
3)DK:x + 3 > 0 \to x > - 3\\
4)DK:{x^2} + 6 < 0\left( {vô lý} \right)\\
\left( {Do:{x^2} + 6 > 0\forall x} \right)\\
\to x \in \emptyset \\
5)DK:3x + 4 \ge 0 \to x \ge - \dfrac{4}{3}\\
6)DK:{x^2} + 1 \ge 0\left( {ld} \right)\forall x\\
7)DK:1 - 2x > 0\\
\to \dfrac{1}{2} > x\\
8)DK:3x + 5 < 0 \to x < - \dfrac{5}{3}\\
B2:\\
1)2\sqrt 3 + 5\sqrt 3 - 4\sqrt 3 = 3\sqrt 3 \\
4)3.2\sqrt 3 - 4.3\sqrt 3 + 5.4\sqrt 3 = 14\sqrt 3 \\
7)3.2\sqrt 5 - 2.3\sqrt 5 + 4\sqrt 5 = 4\sqrt 5 \\
2)5\sqrt 5 + 2\sqrt 5 - 3.3\sqrt 5 = - 2\sqrt 5 \\
5)2\sqrt 3 + 5\sqrt 3 - 3\sqrt 3 = 4\sqrt 3 \\
8)2 + 2\sqrt 2 - 2\sqrt 2 = 2\\
3)2.4\sqrt 2 + 4.2\sqrt 2 - 5.3\sqrt 2 = \sqrt 2 \\
6)2.3\sqrt 2 - 7\sqrt 2 + 9\sqrt 2 = 8\sqrt 2 \\
9)\dfrac{{\sqrt 5 + 1 - \sqrt 5 + 1}}{{5 - 1}} = \dfrac{2}{4} = \dfrac{1}{2}\\
10)\dfrac{{\sqrt 5 + 2 + \sqrt 5 - 2}}{{5 - 4}} = 2\sqrt 5
\end{array}\)
