Toán tìm GTLN a,M= √3- √x-1 b,N=6 √x -x-1 c,P=1/x- √x -1 15/09/2021 By Maria tìm GTLN a,M= √3- √x-1 b,N=6 √x -x-1 c,P=1/x- √x -1
Đáp án: \(\begin{array}{l} a)\,\,\,Max\,\,M = \sqrt 3 \,\,\,khi\,\,\,x = 1.\\ b)\,\,Max\,\,N = 8\,\,\,khi\,\,\,x = 9.\\ c)\,\,Ma\,xP = – \frac{4}{5}\,\,\,khi\,\,\,x = \frac{1}{4}. \end{array}\) Giải thích các bước giải: \(\begin{array}{l} a)\,\,M = \sqrt 3 – \sqrt {x – 1} \\ DK:\,\,\,x – 1 \ge 0 \Leftrightarrow x \ge 1\\ Do\,\,\sqrt {x – 1} \ge 0\,\,\forall x \ge 1\\ \Rightarrow – \sqrt {x – 1} \le 0\,\,\forall x \ge 1\\ \Rightarrow \sqrt 3 – \sqrt {x – 1} \le \sqrt 3 \,\,\forall x \ge 1\\ Dau\,\, = \,\,\,xay\,\,ra \Leftrightarrow x – 1 = 0 \Leftrightarrow x = 1\,\,\,\left( {tm} \right).\\ Vay\,\,Max\,\,M = \sqrt 3 \,\,\,khi\,\,\,x = 1.\\ b)\,\,N = 6\sqrt x – x – 1 = – \left( {x – 6\sqrt x + 9} \right) + 9 – 1\,\,\,\,\left( {x \ge 0} \right)\\ = – {\left( {\sqrt x – 3} \right)^2} + 8\\ Vi\,\,\,{\left( {\sqrt x – 3} \right)^2} \ge 0\,\,\forall x \ge 0\\ \Rightarrow – {\left( {\sqrt x – 3} \right)^2} \le 0\,\,\forall x \ge 0\\ \Rightarrow – {\left( {\sqrt x – 3} \right)^2} + 8 \le 8\,\,\forall x \ge 0\\ Dau\,\, = \,\,xay\,\,ra \Leftrightarrow \sqrt x – 3 = 0 \Leftrightarrow x = 9\,\,\,\left( {tm} \right)\\ Vay\,\,\,Max\,\,N = 8\,\,\,khi\,\,\,x = 9.\\ c)\,\,P = \frac{1}{{x – \sqrt x – 1}}\,\,\,\,\left( {x \ge 0} \right)\\ Ta\,\,co:\,\,x – \sqrt x – 1 = x – 2.\frac{1}{2}\sqrt x + \frac{1}{4} – \frac{1}{4} – 1\\ = {\left( {\sqrt x – \frac{1}{2}} \right)^2} – \frac{5}{4}\\ Vi\,\,{\left( {\sqrt x – \frac{1}{2}} \right)^2} \ge 0\,\,\,\forall x \ge 0 \Rightarrow {\left( {\sqrt x – \frac{1}{2}} \right)^2} – \frac{5}{4} \ge – \frac{5}{4}\,\,\,\forall x \ge 0\\ \Rightarrow \frac{1}{{{{\left( {\sqrt x – \frac{1}{2}} \right)}^2} – \frac{5}{4}}} \le \frac{1}{{ – \frac{5}{4}}} = – \frac{4}{5}\\ Dau\,\, = \,\,xay\,\,ra \Leftrightarrow \sqrt x – \frac{1}{2} = 0 \Leftrightarrow x = \frac{1}{4}.\\ Vay\,\,Ma\,xP = – \frac{4}{5}\,\,\,khi\,\,\,x = \frac{1}{4}. \end{array}\) Trả lời
Đáp án:
\(\begin{array}{l}
a)\,\,\,Max\,\,M = \sqrt 3 \,\,\,khi\,\,\,x = 1.\\
b)\,\,Max\,\,N = 8\,\,\,khi\,\,\,x = 9.\\
c)\,\,Ma\,xP = – \frac{4}{5}\,\,\,khi\,\,\,x = \frac{1}{4}.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\,\,M = \sqrt 3 – \sqrt {x – 1} \\
DK:\,\,\,x – 1 \ge 0 \Leftrightarrow x \ge 1\\
Do\,\,\sqrt {x – 1} \ge 0\,\,\forall x \ge 1\\
\Rightarrow – \sqrt {x – 1} \le 0\,\,\forall x \ge 1\\
\Rightarrow \sqrt 3 – \sqrt {x – 1} \le \sqrt 3 \,\,\forall x \ge 1\\
Dau\,\, = \,\,\,xay\,\,ra \Leftrightarrow x – 1 = 0 \Leftrightarrow x = 1\,\,\,\left( {tm} \right).\\
Vay\,\,Max\,\,M = \sqrt 3 \,\,\,khi\,\,\,x = 1.\\
b)\,\,N = 6\sqrt x – x – 1 = – \left( {x – 6\sqrt x + 9} \right) + 9 – 1\,\,\,\,\left( {x \ge 0} \right)\\
= – {\left( {\sqrt x – 3} \right)^2} + 8\\
Vi\,\,\,{\left( {\sqrt x – 3} \right)^2} \ge 0\,\,\forall x \ge 0\\
\Rightarrow – {\left( {\sqrt x – 3} \right)^2} \le 0\,\,\forall x \ge 0\\
\Rightarrow – {\left( {\sqrt x – 3} \right)^2} + 8 \le 8\,\,\forall x \ge 0\\
Dau\,\, = \,\,xay\,\,ra \Leftrightarrow \sqrt x – 3 = 0 \Leftrightarrow x = 9\,\,\,\left( {tm} \right)\\
Vay\,\,\,Max\,\,N = 8\,\,\,khi\,\,\,x = 9.\\
c)\,\,P = \frac{1}{{x – \sqrt x – 1}}\,\,\,\,\left( {x \ge 0} \right)\\
Ta\,\,co:\,\,x – \sqrt x – 1 = x – 2.\frac{1}{2}\sqrt x + \frac{1}{4} – \frac{1}{4} – 1\\
= {\left( {\sqrt x – \frac{1}{2}} \right)^2} – \frac{5}{4}\\
Vi\,\,{\left( {\sqrt x – \frac{1}{2}} \right)^2} \ge 0\,\,\,\forall x \ge 0 \Rightarrow {\left( {\sqrt x – \frac{1}{2}} \right)^2} – \frac{5}{4} \ge – \frac{5}{4}\,\,\,\forall x \ge 0\\
\Rightarrow \frac{1}{{{{\left( {\sqrt x – \frac{1}{2}} \right)}^2} – \frac{5}{4}}} \le \frac{1}{{ – \frac{5}{4}}} = – \frac{4}{5}\\
Dau\,\, = \,\,xay\,\,ra \Leftrightarrow \sqrt x – \frac{1}{2} = 0 \Leftrightarrow x = \frac{1}{4}.\\
Vay\,\,Ma\,xP = – \frac{4}{5}\,\,\,khi\,\,\,x = \frac{1}{4}.
\end{array}\)