Đề số 2

5*60/(1/3)=230 bản bị lẻ ở đây

6660 trong 19980 s

\(\left(3-x\right)\left(x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3-x=0\\x+7=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-7\end{matrix}\right.\)

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Lời giải:

Điểm $(a,b)$ biểu diễn $z=a+bi$.

Do $z$ bị giới hạn bởi $y=2; y=-2$ nên $-2\leq b\leq 2$.

$a$ biểu diễn hoành độ, không bị ảnh hưởng gì bởi $y=2; y=-2$ nên $a$ là số thực bất kỳ.

Đáp án D.

Chời, mắt toi bị lé hay sao mà ko thấy câu 2 luôn :v

\(DKXD:\dfrac{x-1}{2x-3}>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\\2x-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\\2x-3< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< 1\end{matrix}\right.\)

\(\Rightarrow D=\left(-\infty;1\right)\cup\left(\dfrac{3}{2};+\infty\right)\)

\(y'=\dfrac{\left(\dfrac{x-1}{2x-3}\right)'}{\dfrac{x-1}{2x-3}.ln10}=\dfrac{\dfrac{2x-3-2x+2}{\left(2x-3\right)^2}}{\dfrac{x-1}{2x-3}.ln10}=\dfrac{-1}{\left(2x-3\right)\left(x-1\right).ln10}=...\)

\(\log_ab=3\Leftrightarrow a^3=b\)

\(\log_ab-\log_ac=\log_a\dfrac{b}{c}=3+2=5\Leftrightarrow a^5=\dfrac{b}{c}\Leftrightarrow c=\dfrac{b}{a^5}=\dfrac{a^3}{a^5}=\dfrac{1}{a^2}\)

\(\log_ax=k\Rightarrow\log_a\dfrac{b}{x}=3-k\Leftrightarrow\log_a\dfrac{b}{a^3b^2\sqrt{c}}=3-k\)

\(\Leftrightarrow\log_a\dfrac{1}{a^3.a^3.\sqrt{\dfrac{1}{a^2}}}=3-k\Leftrightarrow3-k=\log_a\dfrac{1}{a^5}\)

\(\Leftrightarrow\log_a\dfrac{1}{a}=\dfrac{3-k}{5}\Leftrightarrow\dfrac{1}{a}=a^{\dfrac{3-k}{5}}\Leftrightarrow-1=\dfrac{3-k}{5}\Leftrightarrow k=8\Rightarrow\log_ax=8\)

ĐKXĐ: \(x>0\)

a.

\(log^2_{0.2}x-5log_{0.2}x+6=0\)

\(\Leftrightarrow\left(log_{0.2}x-2\right)\left(log_{0.2}x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}log_{0.2}x=2\\log_{0.2}x=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{25}\\x=\dfrac{1}{125}\end{matrix}\right.\)

b.

\(log_{0.2}^2x-5log_{0.2}x+6>0\)

\(\Leftrightarrow\left[{}\begin{matrix}log_{0.2}x>3\\log_{0.2}x< 2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}0< x< \dfrac{1}{125}\\x>\dfrac{1}{25}\end{matrix}\right.\)

B. 

\(f\left(x\right)=x^2-3x+2\int\limits^1_0f\left(x\right).f'\left(x\right)dx\)

\(\Leftrightarrow f\left(x\right)=x^2-3x+f^2\left(x\right)|^1_0\)

\(\Leftrightarrow f\left(x\right)=x^2-3x+f^2\left(1\right)-f^2\left(0\right)=x^2-3x+C\)

Với \(C=f^2\left(1\right)-f^2\left(0\right)\)

\(f\left(1\right)=C-2\) ; \(f\left(0\right)=C\Rightarrow f^2\left(1\right)-f^2\left(0\right)=\left(C-2\right)^2-C^2=C\)

\(\Rightarrow-4C+4=C\Rightarrow C=\dfrac{4}{5}\)

\(\Rightarrow f\left(x\right)=x^2-3x+\dfrac{4}{5}\)

\(\Rightarrow\int\limits^a_0f\left(x\right)dx=\int\limits^a_0\left(x^2-3x+\dfrac{4}{5}\right)dx=\left(\dfrac{1}{3}x^3-\dfrac{3}{2}x^2+\dfrac{4}{5}x\right)|^a_0\)

\(=\dfrac{a^3}{3}-\dfrac{3a^2}{2}+\dfrac{4a}{5}=\dfrac{4a}{5}\)

\(\Rightarrow\dfrac{a^3}{3}-\dfrac{3a^2}{2}=0\Rightarrow a=\dfrac{9}{2}\)

Chọn B