Bài 1.35 trang 12 SBT Đại số 10 Nâng cao

Đề bài

Cho \(A = \left\{ {x \in R|{1 \over {\left| {x - 2} \right|}} > 2} \right\}\) và \(B = \left\{ {x \in R|\left| {x - 1} \right| < 1} \right\}.\)

Hãy tìm \(A \cup B\) và \(A \cap B\)

Lời giải chi tiết

Ta có:

\(\begin{array}{l}\dfrac{1}{{\left| {x - 2} \right|}} > 2 \Leftrightarrow \left\{ \begin{array}{l}\left| {x - 2} \right| > 0\\1 > 2\left| {x - 2} \right|\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \ne 2\\\left| {x - 2} \right| < \dfrac{1}{2}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne 2\\x - 2 >  - \dfrac{1}{2}\\x - 2 < \dfrac{1}{2}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x \ne 2\\x > \dfrac{3}{2}\\x < \dfrac{5}{2}\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x \ne 2\\\dfrac{3}{2} < x < \dfrac{5}{2}\end{array} \right.\\ \Rightarrow A = \left( {\dfrac{3}{2};\dfrac{5}{2}} \right)\backslash \left\{ 2 \right\}\end{array}\)

\(\begin{array}{l}\left| {x - 1} \right| < 1 \Leftrightarrow  - 1 < x - 1 < 1\\ \Leftrightarrow 0 < x < 2\\ \Rightarrow B = \left( {0;2} \right)\end{array}\)

Vậy

\(\begin{array}{l}A \cup B = \left( {0;\dfrac{5}{2}} \right)\backslash \left\{ 2 \right\}\\A \cap B = \left( {\dfrac{3}{2};2} \right)\end{array}\)

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