Nội dung bài giảng
Rút gọn các biểu thức:
a) \(\sqrt {{{x - 2\sqrt x + 1} \over {x + 2\sqrt x + 1}}} \) (x ≥ 0);
b) \({{x - 1} \over {\sqrt y - 1}}\sqrt {{{{{(y - 2\sqrt y + 1)}^2}} \over {{{(x - 1)}^4}}}} \) (x ≠1, y ≠ 1 và y ≥ 0).
Gợi ý làm bài
a) Vì x ≥ 0 nên \(x = {\left( {\sqrt x } \right)^2}\)
Ta có:
\(\eqalign{
& \sqrt {{{x - 2\sqrt x + 1} \over {x + 2\sqrt x + 1}}} \cr
& = \sqrt {{{{{\left( {\sqrt x } \right)}^2} - 2\sqrt x + 1} \over {{{\left( {\sqrt x } \right)}^2} + 2\sqrt x + 1}}} \cr
& = \sqrt {{{{{\left( {\sqrt x - 1} \right)}^2}} \over {{{\left( {\sqrt x + 1} \right)}^2}}}} \cr} \)
\( = {{\sqrt {{{\left( {\sqrt x - 1} \right)}^2}} } \over {\sqrt {{{\left( {\sqrt x + 1} \right)}^2}} }} = {{\left| {\sqrt x - 1} \right|} \over {\left| {\sqrt x + 1} \right|}} = {{\left| {\sqrt x - 1} \right|} \over {\sqrt x + 1}}\)
- Nếu \(\sqrt x - 1 \ge 0 \Leftrightarrow x \ge 1\) thì \(\left| {\sqrt x - 1} \right| = \sqrt x - 1\)
Ta có: \({{\left| {\sqrt x - 1} \right|} \over {\sqrt x + 1}} = {{\sqrt x - 1} \over {\sqrt x + 1}}\) (với x ≥ 1)
- Nếu \(\sqrt x - 1 < 0 \Leftrightarrow x < 1\) thì \(\left| {\sqrt x - 1} \right| = 1 - \sqrt x \)
Ta có: \({{\left| {\sqrt x - 1} \right|} \over {\sqrt x + 1}} = {{1 - \sqrt x } \over {\sqrt x + 1}}\) (với 0 ≤ x < 1)
b) Vì y ≥ 0 nên \(y = {\left( {\sqrt y } \right)^2}\)
Ta có:
\(\eqalign{
& {{x - 1} \over {\sqrt y - 1}}\sqrt {{{{{\left( {y - 2\sqrt y + 1} \right)}^2}} \over {{{(x - 1)}^4}}}} \cr
& = {{x - 1} \over {\sqrt y - 1}}{{\sqrt {{{\left( {y - 2\sqrt y + 1} \right)}^2}} } \over {\sqrt {{{(x - 1)}^4}} }} \cr} \)
\(\eqalign{
& = {{x - 1} \over {\sqrt y - 1}}{{\left| {y - 2\sqrt y + 1} \right|} \over {{{(x - 1)}^2}}} \cr
& = {{\left| {{{\left( {\sqrt y } \right)}^2} - 2\sqrt y + 1} \right|} \over {\left( {\sqrt y - 1} \right)(x - 1)}} = {{\left| {{{\left( {\sqrt y - 1} \right)}^2}} \right|} \over {\left( {\sqrt y - 1} \right)(x - 1)}} \cr} \)
\( = {{{{\left( {\sqrt y - 1} \right)}^2}} \over {\left( {\sqrt y - 1} \right)(x - 1)}} = {{\sqrt y - 1} \over {x - 1}}\) (x ≠ 1, y ≠ 1, y ≥ 0)