a) Ta có:
\( \displaystyle\eqalign{
& \sqrt {{{{{(x - 2)}^4}} \over {{{(3 - x)}^2}}}} + {{{x^2} - 1} \over {x - 3}} \cr
& = {{\sqrt {{{(x - 2)}^4}} } \over {\sqrt {{{(3 - x)}^2}} }} + {{{x^2} - 1} \over {x - 3}} \cr
& = {{{{(x - 2)}^2}} \over {\left| {3 - x} \right|}} + {{{x^2} - 1} \over {x - 3}} \cr} \)
\( \displaystyle\eqalign{
& = {{{x^2} - 4x + 4} \over {3 - x}} + {{{x^2} - 1} \over {x - 3}} \cr
& = {{ - {x^2} + 4x - 4} \over {x - 3}} + {{{x^2} - 1} \over {x - 3}} \cr} \)
\( \displaystyle = {{4x - 5} \over {x - 3}}\) (\(x<3\))
Với \(x = 0,5\) ta có:
\( \displaystyle\eqalign{
& {{4.0,5 - 5} \over {0,5 - 3}} = {{ - 3} \over { - 2,5}} \cr
& = {3 \over {2,5}} = {6 \over 5} = 1,2 \cr} \)
b) Với \(x > -2,\) ta có:
\( \displaystyle\eqalign{
& 4x - \sqrt 8 + {{\sqrt {{x^3} + 2{x^2}} } \over {\sqrt {x + 2} }} \cr
& = 4x - \sqrt 8 + \sqrt {{{{x^3} + 2{x^2}} \over {x + 2}}} \cr} \)
\( \displaystyle\eqalign{
& = 4x - \sqrt 8 + \sqrt {{{{x^2}(x + 2)} \over {x + 2}}} \cr
& = 4x - \sqrt 8 + \sqrt {{x^2}} \cr & = 4x - \sqrt 8 + \left| x \right| \cr} \)
+) Nếu \(x \ge 0 \) thì \( \displaystyle\left| x \right| = x\)
Ta có:
\( \displaystyle\eqalign{
& 4x - \sqrt 8 + \left| x \right| \cr
& = 4x - \sqrt 8 + x = 5x - \sqrt 8 \cr} \)
+) Nếu \(-2 < x < 0\) thì \( \displaystyle\left| x \right| = - x\)
Ta có:
\( \displaystyle4x - \sqrt 8 + \left| x \right|\)\( = 4x - \sqrt 8 - x = 3x - \sqrt 8 \)
Với \(x = - \sqrt 2 <0\) ta có: \( \displaystyle3\left( { - \sqrt 2 } \right) - \sqrt 8\)\( = - 3\sqrt 2 - 2\sqrt 2 = - 5\sqrt 2 \)
