Đáp án:
Bài 1:
a) \(A = \sqrt 5 \)
b) \(x = 5\)
Bài 2:
a) A, B xác định \( \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\x \ne 4\end{array} \right.\).
b) \(B = \dfrac{x}{{x - 4}}\).
c) \(\min P = 8 \Leftrightarrow x = 16\).
Bài 3:
a) \(A\left( {2;0} \right),\,\,B\left( {0; - 4} \right)\).
b) \({C_{ABC}} = 6 + 2\sqrt 5 \).
c) \(m = - 2\) .
Giải thích các bước giải:
Bài 1:
\(\begin{array}{l}a)\,\,A = \sqrt {125} + \sqrt {20} - \sqrt {180} \\\,\,\,\,\,\,A = \sqrt {{5^2}.5} + \sqrt {{2^2}.5} - \sqrt {{5^2}.6} \\\,\,\,\,\,\,A = 5\sqrt 5 + 2\sqrt 5 - 6\sqrt 5 \\\,\,\,\,\,\,A = \left( {5 + 2 - 6} \right)\sqrt 5 \\\,\,\,\,\,\,A = \sqrt 5 \end{array}\)
\(\begin{array}{l}b)\,\,\sqrt {x - 1} + \sqrt {9x - 9} - \sqrt {4x - 4} = 4\,\,\left( {x \ge 1} \right)\\ \Leftrightarrow \sqrt {x - 1} + \sqrt {9\left( {x - 1} \right)} - \sqrt {4\left( {x - 1} \right)} = 4\\ \Leftrightarrow \sqrt {x - 1} + 3\sqrt {x - 1} - 2\sqrt {x - 1} = 4\\ \Leftrightarrow 2\sqrt {x - 1} = 4\\ \Leftrightarrow \sqrt {x - 1} = 2\\ \Leftrightarrow x - 1 = 4\\ \Leftrightarrow x = 5\,\,\,\left( {tm} \right)\end{array}\)
Bài 2:
a) A xác định \( \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\\sqrt x - 2 \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\x \ne 4\end{array} \right.\)
B xác định \( \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\\sqrt x - 2 \ne 0\\4 - x \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\x \ne 4\end{array} \right.\)
b)
\(\begin{array}{l}B = \dfrac{2}{{\sqrt x - 2}} + \dfrac{3}{{\sqrt x + 2}} - \dfrac{{x - 5\sqrt x + 2}}{{4 - x}}\\B = \dfrac{{2\left( {\sqrt x + 2} \right) + 3\left( {\sqrt x - 2} \right) + \left( {x - 5\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\B = \dfrac{{2\sqrt x + 4 + 3\sqrt x - 6 + x - 5\sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\B = \dfrac{x}{{x - 4}}\end{array}\)
c) \(P = A.B = \dfrac{{x - 4}}{{\sqrt x - 2}}.\dfrac{x}{{x - 4}} = \dfrac{x}{{\sqrt x - 2}}\)
\(\,\,\,\,\,\, = \dfrac{{x - 4}}{{\sqrt x - 2}} + \dfrac{4}{{\sqrt x - 2}} = \sqrt x + 2 + \dfrac{4}{{\sqrt x - 2}} = \sqrt x - 2 + \dfrac{4}{{\sqrt x - 2}} + 4\)
Áp dụng BĐT Cô-si cho 2 số dương \(\sqrt x - 2,\,\,\dfrac{4}{{\sqrt x - 2}}\) ta có:
\(\sqrt x - 2 + \dfrac{4}{{\sqrt x - 2}} \ge 2\sqrt {\left( {\sqrt x - 2} \right).\dfrac{4}{{\sqrt x - 2}}} = 4\).
\( \Rightarrow P \ge 4 + 4 = 8\). Vậy \(\min P = 8\) . Dấu “=” xảy ra
\(\begin{array}{l} \Leftrightarrow \sqrt x - 2 = \dfrac{4}{{\sqrt x - 2}} \Leftrightarrow {\left( {\sqrt x - 2} \right)^2} = 4\\ \Leftrightarrow \left[ \begin{array}{l}\sqrt x - 2 = 2\\\sqrt x - 2 = - 2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\sqrt x = 4\\\sqrt x = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 16\,\,\left( {tm} \right)\\x = 0\,\,\left( {ktm} \right)\end{array} \right.\end{array}\)
Bài 3:
a) Cho \(x = 0 \Rightarrow y = 2.0 - 4 = - 4 \Rightarrow d \cap Oy = B\left( {0; - 4} \right)\).
Cho \(y = 0 \Rightarrow 2x - 4 = 0 \Leftrightarrow x = 2 \Rightarrow d \cap Ox = A\left( {2;0} \right)\).
b) Ta có: \(OA = 2,\,\,OB = 4\).
Áp dụng định lí Pytago trong tam giác vuông OAB ta có:
\(AB = \sqrt {O{A^2} + O{B^2}} = \sqrt {{2^2} + {4^2}} = \sqrt {20} = 2\sqrt 5 \).
Vậy chu vi tam giác OAB bằng \(2 + 4 + 2\sqrt 5 = 6 + 2\sqrt 5 \).
c) Để \(\left( {{d_m}} \right):\,\,y = \left( {{m^2} - 2} \right)x + 2m - 2{m^2}\) song song với \(\left( d \right)\) thì:
\(\left\{ \begin{array}{l}{m^2} - 2 = 2\\2m - 2{m^2} \ne - 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{m^2} = 4\\m - {m^2} \ne - 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}m = 2\\m = - 2\end{array} \right.\\m - {m^2} \ne - 2\end{array} \right. \Leftrightarrow m = - 2\).
Vậy \(m = - 2\).
