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\(P=xy-3\left(x+y\right)+9\)
Đặt \(x+y=a\Rightarrow1< a\le\sqrt{2}\)
\(a^2=x^2+y^2+2xy=1+2xy\Rightarrow xy=\frac{a^2-1}{2}\)
\(P=\frac{a^2-1}{2}-3a+9\Rightarrow2P=a^2-6a+17\)
\(2P=a^2-6a-2+6\sqrt{2}+19-6\sqrt{2}\)
\(2P=\left(a+\sqrt{2}\right)\left(a-\sqrt{2}\right)-6\left(a-\sqrt{2}\right)+19-6\sqrt{2}\)
\(2P=\left(\sqrt{2}-a\right)\left(6-\sqrt{2}-a\right)+19-6\sqrt{2}\ge19-6\sqrt{2}\)
\(\Rightarrow P\ge\frac{19-6\sqrt{2}}{2}\)
Dấu "=" xảy ra khi \(a=\sqrt{2}\) hay \(x=y=\frac{\sqrt{2}}{2}\)
\(\left\{{}\begin{matrix}x^2=2log_a\left(ab\right)=2\left(1+log_ab\right)\\y^2=2log_b\left(ab\right)=2\left(1+log_ba\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2log_ab=x^2-2\\2log_ba=y^2-2\end{matrix}\right.\) \(\Rightarrow\left(x^2-2\right)\left(y^2-2\right)=4\)
\(\Leftrightarrow y^2-2=\frac{4}{x^2-2}\Rightarrow y^2=\frac{2x^2}{x^2-2}\) (\(x\ge\sqrt{2}\))
\(\Rightarrow P=f\left(x\right)=8x+\frac{x\sqrt{2}}{\sqrt{x^2-2}}=0\)
\(\Rightarrow f'\left(x\right)=8-\frac{2\sqrt{2}x}{\left(x^2-2\right)^2\sqrt{\frac{x^2}{x^2-2}}}=0\)
\(\Leftrightarrow\left(x^2-2\right)^3=\frac{1}{8}\Leftrightarrow x^2-2=\frac{1}{2}\Rightarrow x=\frac{\sqrt{10}}{2}\)
\(\Rightarrow P_{min}=P\left(\frac{\sqrt{10}}{2}\right)=5\sqrt{10}\Rightarrow\left\{{}\begin{matrix}m=0\\n=5\\m=10\end{matrix}\right.\) \(\Rightarrow m+n+p=15\)
Câu 2. Đặt A=x2+y2+1
Nhập \(2^A=\left(A-2x+1\right)4^x\) vào máy tính Casio. Cho x=0.01, tìm A
Máy sẽ giải ra, A=1.02=1+2x
\(\Leftrightarrow x^2+y^2+1=1+2x\)
\(\Leftrightarrow x^2+y^2-2x=1\)
\(\Leftrightarrow\left(x-1\right)^2+y^2=1\) (C)
Có (C) là đường tròn tâm (1,0) bán kính R=1
Lại có: P=\(\frac{8x+4}{2x-y+1}\)
\(\Leftrightarrow x\left(2P-8\right)-yP+P-4=0\) (Q)
Có (Q) là phương trình đường thẳng.
Để x,y có nghiệm thì đường thẳng và đường tròn giao nhau nghĩa là d(I,(Q))\(\le R\)
\(\Leftrightarrow\frac{\left|x\left(2P-8\right)-yP+P-4\right|}{\sqrt{\left(2P-8\right)^2+P^2}}\le1\)
\(\Leftrightarrow\frac{\left|2P-8+P-4\right|}{\sqrt{\left(2P-8\right)^2+1}}\le1\)
\(\Leftrightarrow\left(3P-12\right)^2\le5P^2-32P+64\)
\(\Leftrightarrow4P^2-40P+80\le0\)
\(\Leftrightarrow5-\sqrt{5}\le P\le5+\sqrt{5}\)
Vậy GTNN của P gần số 3 nhất. Chọn C
\(\left(x+y\right)xy=x^2+y^2-xy\)
\(\Leftrightarrow\left(x+y\right)xy=\left(x+y\right)^2-3xy\)
Đặt \(x+y=t\Rightarrow xy=\frac{t^2}{t+3}\)
Lại có \(\left(x+y\right)^2\ge4xy\Rightarrow t^2\ge\frac{4t^2}{t+3}\)
\(\Leftrightarrow t^2\left(\frac{t-1}{t+3}\right)\ge0\Rightarrow\left[{}\begin{matrix}t\ge1\\t< -3\end{matrix}\right.\)
\(A=\frac{x^3+y^3}{\left(xy\right)^3}=\frac{\left(x+y\right)\left(x^2+y^2-xy\right)}{\left(xy\right)^3}=\frac{\left(x+y\right)\left(x+y\right)xy}{\left(xy\right)^3}=\left(\frac{x+y}{xy}\right)^2\)
\(A=\left(\frac{t\left(t+3\right)}{t^2}\right)^2=\left(\frac{t+3}{t}\right)^2=\left(1+\frac{3}{t}\right)^2\)
\(\Rightarrow y'=-\frac{6\left(t+3\right)}{t^3}< 0\) \(\forall t\ge1;t< -3\)
\(\lim\limits_{x\rightarrow-\infty}\left(1+\frac{3}{t}\right)^2=1\Rightarrow A_{max}=A\left(1\right)=16\)
\(\Rightarrow M=16\) khi \(x=y=\frac{1}{2}\)
Lời giải:
Từ điều kiện đb \(\ln x+\ln y\geq \ln (x^2+y)\Leftrightarrow \ln (xy)\geq \ln (x^2+y)\)
\(\Leftrightarrow xy\geq x^2+y\Leftrightarrow y(x-1)\geq x^2\)
\(\bullet\)Nếu \(x\geq 1\Rightarrow y\geq \frac{x^2}{x-1}\)
Khi đó \(P=x+y\geq x+\frac{x^2}{x-1}=2x+1+\frac{1}{x-1}=2(x-1)+\frac{1}{x-1}+3\)
Áp dụng định lý AM-GM:
\(P\geq 2\sqrt{2(x-1).\frac{1}{x-1}}+3=2\sqrt{2}+3\) hay \(P_{\min}=2\sqrt{2}+3\)
\(\bullet \)Nếu \(x<1\Rightarrow \ln x<0\) kéo theo \(\ln x+\ln y<\ln y\)
Mà \(\ln(x^2+y)\geq \ln (0+y)=\ln y\) nên \(\ln x+\ln y<\ln (x^2+y)\) (không thỏa mãn đkđb) (loại)
Vậy \(P_{\min}=2\sqrt{2}+3\)
Đáp án B
\(2018^{2\left(x^2-y+1\right)}=\frac{2x+y}{x^2+2x+1}\)
\(\Leftrightarrow2\left(x^2-y+1\right)=log_{2018}\left(\frac{2x+y}{x^2+2x+1}\right)\)
\(\Leftrightarrow2\left(x^2+2x+1-2x-y\right)=log_{2018}\left(2x+y\right)-log_{2018}\left(x^2+2x+1\right)\)
\(\Leftrightarrow2\left(x^2+2x+1\right)+log_{2018}\left(x^2+2x+1\right)=log_{2018}\left(2x+y\right)+2\left(2x+y\right)\)
Đặt \(f\left(u\right)=log_{2018}u+2u\)
\(\begin{matrix}x^2+2x+1>0\\2x+y>0\end{matrix}\Rightarrow u>0\)
\(f'\left(u\right)=\frac{1}{u.ln2018}+2>0\)
Suy ra hàm số đồng biến
\(\Leftrightarrow f\left(x^2+2x+1\right)=f\left(2x+y\right)\)\(\Leftrightarrow x^2+2x+1=2x+y\) (tính chất hàm đồng biến)
\(\Leftrightarrow y=x^2+1\)
\(P=2y-3x=2x^2-3x+2\)
\(P=2\left(x-\frac{3}{4}\right)^2+\frac{7}{8}\)
\(P_{min}=\frac{7}{8}\) khi \(x=\frac{3}{4}\)
\(\left(xy-1\right)2^{2xy-1}=\left(x^2+y\right)2^{x^2+y}\)
\(\Leftrightarrow\left(xy-1\right)2^{2\left(xy-1\right)+1}=\left(x^2+y\right)2^{x^2+y}\)
\(\Leftrightarrow2\left(xy-1\right)2^{2\left(xy-1\right)}=\left(x^2+y\right)2^{x^2+y}\)
Do vế phải luôn dương \(\Rightarrow VT>0\Rightarrow xy-1>0\) (1)
Xét hàm \(f\left(t\right)=t.2^t\) với \(t>0\Rightarrow f'\left(t\right)=2^t+t.2^t.ln2>0\)
\(\Rightarrow f\left(t\right)\) đồng biến \(\Rightarrow f\left(t_1\right)=f\left(t_2\right)\Leftrightarrow t_1=t_2\)
\(\Rightarrow2\left(xy-1\right)=x^2+y\Rightarrow2xy-y=x^2+2\) (thay \(x=\dfrac{1}{2}\) thấy ko phải nghiệm)
\(\Rightarrow y=\dfrac{x^2+2}{2x-1}\) (2)
Thay (2) vào (1): \(xy-1>0\Rightarrow x.\left(\dfrac{x^2+2}{2x-1}\right)-1>0\Rightarrow\dfrac{x^3+2x}{2x-1}-1>0\)
\(\Rightarrow\dfrac{x^3+1}{2x-1}>0\Rightarrow2x-1>0\) (do \(x>0\Rightarrow x^3+1>0\))
Vậy \(y=\dfrac{x^2+2}{2x-1}=\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{9}{4\left(2x-1\right)}=\dfrac{2x-1}{4}+\dfrac{9}{4\left(2x-1\right)}+\dfrac{1}{2}\)
\(\Rightarrow y\ge2\sqrt{\dfrac{\left(2x-1\right)}{4}.\dfrac{9}{4\left(2x-1\right)}}+\dfrac{1}{2}=2\)
\(\Rightarrow y_{min}=2\) khi \(\dfrac{2x-1}{4}=\dfrac{9}{4\left(2x-1\right)}\Rightarrow x=2\)
Đáp án B
\(5^{x+3y}+5^{xy+1}+xy+1+x+3y=\frac{1}{5^{xy+1}}+\frac{1}{5^{x+3y}}\)
\(\Leftrightarrow5^{x+3y}-5^{-x-3y}+x+3y=5^{-xy-1}-5^{-\left(-xy-1\right)}+\left(-xy-1\right)\)
Xét hàm \(f\left(t\right)=5^t-\frac{1}{5^t}+t\Rightarrow f'\left(t\right)=5^t.ln5+\frac{ln5}{5^t}+1>0\)
\(\Rightarrow f\left(t\right)\) đồng biến
\(\Rightarrow x+3y=-xy-1\)
\(\Rightarrow y\left(x+3\right)=-x-1\)
\(\Rightarrow y=\frac{-x-1}{x+3}\)
\(\Rightarrow T=f\left(x\right)=x-\frac{2x+2}{x+3}+1\)
\(f'\left(x\right)=\frac{\left(x+1\right)\left(x+5\right)}{\left(x+3\right)^2}>0;\forall x\ge0\)
\(\Rightarrow f\left(x\right)_{min}=f\left(0\right)=\frac{1}{3}\Rightarrow m=\frac{1}{3}\)