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Rút gọn các biểu thức sau :
a) \(\sqrt {4{a^2}} \) với \(a \ge 0\);
b) \(\sqrt {0,16{{\left( {x – 2} \right)}^2}} \) với \(x \ge 2\);
c) \(\sqrt {25.{{\left( {3 – a} \right)}^2}} + 3\);
d) \(\dfrac{1}{{2\left( {x – 5} \right)}}\sqrt {36.{{\left( {x – 5} \right)}^2}} – 5\) với \(x \ne 5\).
Sử dụng công thức \(\sqrt {A.B} = \sqrt A .\sqrt B \) và \(\sqrt {{A^2}} = \left| A \right| = \left\{ \begin{array}{l}A\;\;khi\;\;A \ge 0\\ – A\;\;khi\;\;A < 0\end{array} \right..\)
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\(a)\;\;\sqrt {4{a^2}} = \left| {2a} \right| = 2a\) vì \(a \ge 0.\)
\(b)\;\;\sqrt {0,16{{\left( {x – 2} \right)}^2}} = \sqrt {0,16} .\sqrt {{{\left( {x – 2} \right)}^2}} = 0,4.\left| {x – 2} \right| = 0,4\left( {x – 2} \right)\) vì \(x \ge 2.\)
\(\begin{array}{l}c)\;\sqrt {25.{{\left( {3 – a} \right)}^2}} + 3 = \sqrt {25} .\sqrt {{{\left( {3 – a} \right)}^2}} + 3 = 5\left| {3 – a} \right| + 3\\ = \left\{ \begin{array}{l}5\left( {3 – a} \right) + 3\;\;\;khi\;\;a \le 3\\5\left( {a – 3} \right) + 3\;\;\;khi\;\;a > 3\end{array} \right. = \left\{ \begin{array}{l} – 5a + 18\;\;khi\;\;a \le 3\\5a – 12\;\;\;\;khi\;\;a > 3\end{array} \right..\end{array}\)
\(\begin{array}{l}d)\;\;\dfrac{1}{{2\left( {x – 5} \right)}}\sqrt {36.{{\left( {x – 5} \right)}^2}} – 5 = \dfrac{1}{{2\left( {x – 5} \right)}}.\sqrt {36} .\sqrt {{{\left( {x – 5} \right)}^2}} \\ = \dfrac{1}{{2\left( {x – 5} \right)}}.6.\left| {x – 5} \right| = \dfrac{3}{{x – 5}}.\left| {x – 5} \right|\\ = \left\{ \begin{array}{l}\dfrac{3}{{x – 5}}.\left( {x – 5} \right)\;\;\;khi\;\;\;x \ge 5\\\dfrac{3}{{x – 5}}\left[ { – \left( {x – 5} \right)} \right]\;\;\;khi\;\;\;x < 5\end{array} \right. = \left\{ \begin{array}{l}3\;\;khi\;\;x \ge 5\\ – 3\;\;khi\;\;x < 5\end{array} \right..\end{array}\)