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1.
\(y'=3x^2-3=0\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(y\left(0\right)=5;\) \(y\left(1\right)=3;\) \(y\left(2\right)=7\)
\(\Rightarrow y_{min}=3\)
2.
\(y'=4x^3-8x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-\sqrt{2}\end{matrix}\right.\)
\(f\left(-2\right)=-3\) ; \(y\left(0\right)=-3\) ; \(y\left(-\sqrt{2}\right)=-7\) ; \(y\left(1\right)=-6\)
\(\Rightarrow y_{max}=-3\)
3.
\(y'=\frac{\left(2x+3\right)\left(x-1\right)-x^2-3x}{\left(x-1\right)^2}=\frac{x^2-2x-3}{\left(x-1\right)^2}=0\Rightarrow x=-1\)
\(y_{max}=y\left(-1\right)=1\)
4.
\(y'=\frac{2\left(x^2+2\right)-2x\left(2x+1\right)}{\left(x^2+2\right)^2}=\frac{-2x^2-2x+4}{\left(x^2+2\right)^2}=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
\(y\left(1\right)=1\) ; \(y\left(-2\right)=-\frac{1}{2}\Rightarrow y_{min}+y_{max}=-\frac{1}{2}+1=\frac{1}{2}\)
Bài 2: Mình nghĩ điều kiện sửa thành $a,b\in\mathbb{N}$ thôi thì đúng hơn.
ĐKĐB $\Leftrightarrow \log_2[(2x+1)(y+2)]^{y+2}=8-(2x-2)(y+2)$
$\Leftrightarrow (y+2)\log_2[(2x+1)(y+2)]=8-(2x-2)(y+2)$
$\Leftrightarrow (y+2)[\log_2[(2x+1)(y+2)]+(2x-2)]=8$
$\Leftrightarrow \log_2[(2x+1)(y+2)]+(2x-2)]=\frac{8}{y+2}$
$\Leftrightarrow \log_2(2x+1)+\log_2(y+2)+(2x+1)-3=\frac{8}{y+2}$
$\Leftrightarrow \log_2(2x+1)+(2x+1)=\frac{8}{y+2}+3-\log_2(y+2)=\frac{8}{y+2}+\log_2(\frac{8}{y+2})(*)$
Xét hàm $f(t)=\log_2t+t$ với $t>0$
$f'(t)=\frac{1}{t\ln 2}+1>0$ với mọi $t>0$
Do đó hàm số đồng biến trên TXĐ
$\Rightarrow (*)$ xảy ra khi mà $2x+1=\frac{8}{y+2}$
$\Leftrightarrow 8=(2x+1)(y+2)$
Áp dụng BĐT AM-GM:
$8=(2x+1)(y+2)\leq \left(\frac{2x+1+y+2}{2}\right)^2$
$\Rightarrow 2\sqrt{2}\leq \frac{2x+y+3}{2}$
$\Rightarrow 2x+y\geq 4\sqrt{2}-3$
Vậy $P_{\min}=4\sqrt{2}-3$
$\Rightarrow a=4; b=2; c=-3$
$\Rightarrow a+b+c=3$
Đáp án B.
2.
\(\Leftrightarrow\left(y+2\right)log_2\left(2x+1\right)\left(y+2\right)=8-\left(2x-2\right)\left(y+2\right)\)
\(\Leftrightarrow log_2\left(2x+1\right)\left(y+2\right)=\frac{8}{y+2}-2x+2\)
\(\Leftrightarrow log_2\left(2x+1\right)+log_2\left(y+2\right)=\frac{8}{y+2}-2x+2\)
\(\Leftrightarrow log_2\left(2x+1\right)+\left(2x+1\right)=-log_2\left(y+2\right)+3+\frac{8}{y+2}\)
\(\Leftrightarrow log_2\left(2x+1\right)+\left(2x+1\right)=log_2\left(\frac{8}{y+2}\right)+\frac{8}{y+2}\)
Xét hàm \(f\left(t\right)=log_2t+t\Rightarrow f'\left(t\right)=\frac{1}{t.ln2}+1>0;\forall t>0\)
\(\Rightarrow f\left(t\right)\) đồng biến \(\Rightarrow2x+1=\frac{8}{y+2}\)
\(\Rightarrow2x=\frac{8}{y+2}-1=\frac{6-y}{y+2}\)
\(\Rightarrow P=2x+y=y+\frac{6-y}{y+2}=y+\frac{8}{y+2}-1\)
\(\Rightarrow P=y+2+\frac{8}{y+2}-3\ge2\sqrt{\frac{8\left(y+2\right)}{y+2}}-3=4\sqrt{2}-3\)
\(\Rightarrow\left\{{}\begin{matrix}a=4\\b=2\\c=-3\end{matrix}\right.\) \(\Rightarrow a+b+c=3\)
\(\left(xy-1\right)2^{2xy-1}=\left(x^2+y\right)2^{x^2+y}\)
\(\Leftrightarrow\left(xy-1\right)2^{2\left(xy-1\right)+1}=\left(x^2+y\right)2^{x^2+y}\)
\(\Leftrightarrow2\left(xy-1\right)2^{2\left(xy-1\right)}=\left(x^2+y\right)2^{x^2+y}\)
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
4.
\(xy+y=2\Leftrightarrow xy=2-y\Rightarrow x=\frac{2-y}{y}=\frac{2}{y}-1\)
\(\Rightarrow P=x+y^2=y^2+\frac{2}{y}-1\)
\(\Rightarrow P=y^2+\frac{1}{y}+\frac{1}{y}-1\ge3\sqrt[3]{\frac{y^2}{y.y}}-1=2\)
\(\Rightarrow P_{min}=2\) khi \(x=y=1\)
5.
\(y'=1-\frac{4}{\left(x-3\right)^2}=0\Leftrightarrow\left(x-3\right)^2=4\)
\(\Rightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=5\\x=1< 3\left(l\right)\end{matrix}\right.\)
BBT:
Từ BBT ta có \(y_{min}=y\left(5\right)=7\)
\(\Rightarrow m=7\)
3.
\(y'=-2x^2-6x+m\)
Hàm đã cho nghịch biến trên R khi và chỉ khi \(y'\le0;\forall x\)
\(\Leftrightarrow\Delta'=9+2m\le0\)
\(\Rightarrow m\le-\frac{9}{2}\)
4.
\(y'=x^2-mx-2m-3\)
Hàm đồng biến trên khoảng đã cho khi và chỉ khi \(y'\ge0;\forall x>-2\)
\(\Leftrightarrow x^2-mx-2m-3\ge0\)
\(\Leftrightarrow x^2-3\ge m\left(x+2\right)\Leftrightarrow m\le\frac{x^2-3}{x+2}\)
\(\Leftrightarrow m\le\min\limits_{x>-2}\frac{x^2-3}{x+2}\)
Xét \(g\left(x\right)=\frac{x^2-3}{x+2}\) trên \(\left(-2;+\infty\right)\Rightarrow g'\left(x\right)=\frac{x^2+4x+3}{\left(x+2\right)^2}=0\Rightarrow x=-1\)
\(g\left(-1\right)=-2\Rightarrow m\le-2\)
Đáp án B