rút gọn :
(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}})^{2}
\(\frac{(\sqrt{x+2\sqrt{x-1}}+ \sqrt{x-2\sqrt{x-1}})^{2}}{(\sqrt{x+\sqrt{2x-1}}+ \sqrt{x-\sqrt{2x-1}})^{2}}\)
rút gọn : (\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}})^{2} \(\frac{(\sqrt{x+2\sqrt{x-1}}+ \sqrt{x-2\sqrt{x-1}})^{2}}{(\sqrt{x+\sqrt{2x-1}}+ \sqrt{x-\sq
By Faith
Đáp án:
Giải thích các bước giải:
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
*)\\
A = \sqrt {x + 4\sqrt {x – 4} } + \sqrt {x – 4\sqrt {x – 4} } \,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 4} \right)\\
= \sqrt {\left( {x – 4} \right) + 4\sqrt {x – 4} + 4} + \sqrt {\left( {x – 4} \right) – 4\sqrt {x – 4} + 4} \\
= \sqrt {{{\left( {\sqrt {x – 4} + 2} \right)}^2}} + \sqrt {{{\left( {\sqrt {x – 4} – 2} \right)}^2}} \\
= \sqrt {x – 4} + 2 + \left| {\sqrt {x – 4} – 2} \right|\\
TH1:\,\,\,\,\sqrt {x – 4} – 2 \ge 0 \Leftrightarrow \sqrt {x – 4} \ge 2 \Leftrightarrow x \ge 8\\
\Rightarrow A = \sqrt {x – 4} + 2 + \sqrt {x – 4} – 2 = 2\sqrt {x – 4} \\
TH2:\,\,\,\,\,\sqrt {x – 4} – 2 < 0 \Leftrightarrow \sqrt {x – 4} < 2 \Leftrightarrow 4 \le x < 8\\
\Rightarrow A = \sqrt {x – 4} + 2 + 2 – \sqrt {x – 4} = 4\\
*)\\
B = \dfrac{{{{\left( {\sqrt {x + 2\sqrt {x – 1} } + \sqrt {x – 2\sqrt {x – 1} } } \right)}^2}}}{{{{\left( {\sqrt {x + \sqrt {2x – 1} } + \sqrt {x – \sqrt {2x – 1} } } \right)}^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 1} \right)\\
= \dfrac{{{{\left( {\sqrt {\left( {x – 1} \right) + 2\sqrt {x – 1} + 1} + \sqrt {\left( {x – 1} \right) – 2\sqrt {x – 1} + 1} } \right)}^2}}}{{\dfrac{1}{2}.{{\left[ {\sqrt 2 .\left( {\sqrt {x + \sqrt {2x – 1} } + \sqrt {x – \sqrt {2x – 1} } } \right)} \right]}^2}}}\\
= \dfrac{{{{\left( {\sqrt {{{\left( {\sqrt {x – 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x – 1} – 1} \right)}^2}} } \right)}^2}}}{{\dfrac{1}{2}.{{\left( {\sqrt {2x + 2\sqrt {2x – 1} } + \sqrt {2x – 2\sqrt {2x – 1} } } \right)}^2}}}\\
= \dfrac{{2.{{\left( {\sqrt {x – 1} + 1 + \left| {\sqrt {x – 1} – 1} \right|} \right)}^2}}}{{{{\left( {\sqrt {{{\left( {\sqrt {2x – 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {2x – 1} – 1} \right)}^2}} } \right)}^2}}}\\
= \dfrac{{2.{{\left( {\sqrt {x – 1} + 1 + \left| {\sqrt {x – 1} – 1} \right|} \right)}^2}}}{{{{\left( {\sqrt {2x – 1} + 1 + \sqrt {2x – 1} – 1} \right)}^2}}}\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 1 \Rightarrow \sqrt {2x – 1} – 1 \ge 0} \right)\\
= \dfrac{{2.{{\left( {\sqrt {x – 1} + 1 + \left| {\sqrt {x – 1} – 1} \right|} \right)}^2}}}{{{{\left( {2\sqrt {2x – 1} } \right)}^2}}}\\
= \dfrac{{{{\left( {\sqrt {x – 1} + 1 + \left| {\sqrt {x – 1} – 1} \right|} \right)}^2}}}{{2.\left( {2x – 1} \right)}}\\
TH1:\,\,\,\,\,\sqrt {x – 1} – 1 \ge 0 \Leftrightarrow x \ge 2\\
B = \dfrac{{{{\left( {\sqrt {x – 1} + 1 + \sqrt {x – 1} – 1} \right)}^2}}}{{2.\left( {2x – 1} \right)}} = \dfrac{{4.\left( {x – 1} \right)}}{{2.\left( {2x – 1} \right)}} = \dfrac{{2x – 2}}{{2x – 1}}\\
TH1:\,\,\,\sqrt {x – 1} – 1 < 0 \Leftrightarrow 1 \le x < 2\\
B = {\dfrac{{\left( {\sqrt {x – 1} + 1 + 1 – \sqrt {x – 1} } \right)}}{{2.\left( {2x – 1} \right)}}^2} = \dfrac{2}{{2x – 1}}
\end{array}\)