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bn ơi câu a t chưa làm chưa biết nhưng câu b chắc chắn có Max tại x=-3 nhé ! Nếu bn chỉ tìm ra Min là chưa đủ
\(\int\limits^2_1\frac{8x+5}{6x^2+7x+2}dx=\int\limits^2_1\frac{8x+5}{6\left(x+\frac{1}{2}\right)\left(x+\frac{2}{3}\right)}dx=\frac{1}{6}\int\limits^2_1(\frac{2}{x+\frac{2}{3}}+\frac{6}{x+\frac{1}{2}})dx\:\)
\(=\frac{1}{6}\left(2ln\left|x+\frac{1}{2}\right|+6ln\left|x+\frac{2}{3}\right|\right)\)\(|^2_1\)
=\(\frac{1}{3}ln\left(\left|x+\frac{1}{2}\right|\right)+ln\left(\left|x+\frac{2}{3}\right|\right)\)\(|^2_1\)
= \(\frac{1}{3}ln\frac{5}{2}+ln\frac{8}{3}-\frac{1}{3}ln\frac{3}{2}-ln\frac{5}{3}=\frac{1}{3}ln5-\frac{1}{3}ln3+ln8-ln3=3ln2-\frac{4}{3}ln3+\frac{1}{3}ln5\)
\(\Rightarrow\)a=3,b=\(\frac{-4}{3}\),c=\(\frac{1}{3}\)
P=2
ĐK;x>0
<=> \(\frac{1}{2}\)log2x-log2x-log52>1
<=>\(\frac{1}{2}\)log2x>1+log52
<=> log2x>\(\frac{1+log_{ }^{ }}{2}\)( ví a=2>0)
<=>x>2\(\frac{1+log_{ }^{ }}{2}\)
ĐKXĐ: ...
\(\Leftrightarrow log_3\left(2x-1\right)-log_3\left(x-1\right)^2=3\left(x^2-2x+1\right)-2x+1+1\)
\(\Leftrightarrow log_3\left(2x-1\right)+2x-1=log_3\left(x-1\right)^2+1+3\left(x-1\right)^2\)
\(\Leftrightarrow log_3\left(2x-1\right)+2x-1=log_33\left(x-1\right)^2+3\left(x-1\right)^2\)
Xét hàm \(f\left(t\right)=log_3t+t\) với \(t>0\)
\(f'\left(t\right)=\frac{1}{t.ln3}+1>0\Rightarrow f\left(t\right)\) đồng biến
\(\Rightarrow f\left(2x-1\right)=f\left(3\left(x-1\right)^2\right)\Leftrightarrow2x-1=3\left(x-1\right)^2\)
\(\Leftrightarrow3x^2-8x+4=0\)
\(\Leftrightarrow...\)
14.
\(log_aa^2b^4=log_aa^2+log_ab^4=2+4log_ab=2+4p\)
15.
\(\frac{1}{2}log_ab+\frac{1}{2}log_ba=1\)
\(\Leftrightarrow log_ab+\frac{1}{log_ab}=2\)
\(\Leftrightarrow log_a^2b-2log_ab+1=0\)
\(\Leftrightarrow\left(log_ab-1\right)^2=0\)
\(\Rightarrow log_ab=1\Rightarrow a=b\)
16.
\(2^a=3\Rightarrow log_32^a=1\Rightarrow log_32=\frac{1}{a}\)
\(log_3\sqrt[3]{16}=log_32^{\frac{4}{3}}=\frac{4}{3}log_32=\frac{4}{3a}\)
11.
\(\Leftrightarrow1>\left(2+\sqrt{3}\right)^x\left(2+\sqrt{3}\right)^{x+2}\)
\(\Leftrightarrow\left(2+\sqrt{3}\right)^{2x+2}< 1\)
\(\Leftrightarrow2x+2< 0\Rightarrow x< -1\)
\(\Rightarrow\) có \(-2+2020+1=2019\) nghiệm
12.
\(\Leftrightarrow\left\{{}\begin{matrix}x-2>0\\0< log_3\left(x-2\right)< 1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>2\\1< x-2< 3\end{matrix}\right.\)
\(\Rightarrow3< x< 5\Rightarrow b-a=2\)
13.
\(4^x=t>0\Rightarrow t^2-5t+4\ge0\)
\(\Rightarrow\left[{}\begin{matrix}t\le1\\t\ge4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}4^x\le1\\4^x\ge4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\le0\\x\ge1\end{matrix}\right.\)
