64.
Đặt \(y=cosx\in\left[-1;1\right]\)
\(ln\left(m+ln\left(m+y\right)\right)=y\)
\(\Rightarrow e^y=m+ln\left(m+y\right)\)
Đặt \(ln\left(m+y\right)=z\)
\(\Rightarrow\left\{{}\begin{matrix}e^y=m+z\\e^z=m+y\end{matrix}\right.\)
\(\Rightarrow e^y+y=e^z+z\)
Hàm \(f\left(t\right)=e^t+t\) có \(f'\left(t\right)=e^t+1>0;\forall t\) nên đồng biến
\(\Rightarrow y=z\)
\(\Rightarrow ln\left(m+y\right)=y\Rightarrow m+y=e^y\)
\(\Rightarrow m=e^y-y\)
Xét \(f\left(y\right)=e^y-y\) với \(y\in\left[-1;1\right]\)
\(f'\left(y\right)=e^y-1=0\Rightarrow y=0\)
\(f\left(0\right)=1\) ; \(f\left(-1\right)=1+\dfrac{1}{e}\) ; \(f\left(1\right)=e-1\)
\(\Rightarrow1\le f\left(y\right)\le e-1\)
\(\Rightarrow\) Pt có nghiệm khi \(1\le m\le e-1\)
65.
ĐKXĐ: \(x\ge1\)
\(4log_3x+2a\sqrt{\dfrac{3}{4}log_3x}+4a-3=0\)
Đặt \(\sqrt{\dfrac{3}{4}log_3x}=t\ge0\Rightarrow log_3x=\dfrac{4}{3}t^2\)
\(\Rightarrow\dfrac{16}{3}t^2+2a.t+4a-3=0\)
\(\Rightarrow a=\dfrac{3-\dfrac{16}{3}t^2}{2t+4}\)
Hàm \(f\left(t\right)=\dfrac{9-16t^2}{6\left(t+2\right)}\) có \(f'\left(t\right)=-\dfrac{16t^2+64t+9}{6\left(t+2\right)^2}< 0;\forall t>0\) nên nghịch biến khi \(t\ge0\)
\(\Rightarrow\) Pt có nghiệm duy nhất khi \(a\le f\left(0\right)=\dfrac{3}{4}\)
Hình như tất cả các đáp án đều sai
66.
Đặt \(log_2x=t\)
\(\Rightarrow t^2+2t+8-4m-4\sqrt{t+m}=0\)
\(\Leftrightarrow t^2+6t+9-4\left(t+m\right)-4\sqrt{t+m}-1=0\)
\(\Leftrightarrow\left(t+3\right)^2-\left(2\sqrt{t+m}+1\right)^2=0\)
\(\Leftrightarrow\left(t+4-2\sqrt{t+m}\right)\left(t+2-2\sqrt{t+m}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2\sqrt{t+m}=t+4\\2\sqrt{t+m}=t+2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4\left(t+m\right)=t^2+8t+16\\t\ge-4\end{matrix}\right.\\\left\{{}\begin{matrix}4\left(t+m\right)=t^2+4t+4\\t\ge-2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4m=t^2+4t+16\\t\ge-4\end{matrix}\right.\\\left\{{}\begin{matrix}4m=t^2+4\\t\ge-2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4m\ge12\\4m\ge4\end{matrix}\right.\) \(\Rightarrow m\ge1\)
\(\Rightarrow\) Có \(2023-1+1=2023\) giá trị nguyên của m
