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a: Hàm số nghịch biến trên R
b: \(\dfrac{f\left(x_1\right)-f\left(x_2\right)}{x_1-x_2}=\dfrac{x_1^2-4x_1+5-x_2^2+4x_2-5}{x_1-x_2}\)
\(=x_1+x_2-4\)
Trường hợp 1: x<=2
\(\Leftrightarrow x_1+x_2-4< =0\)
Vậy: Hàm số nghịch biến khi x<=2
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ:
a. \(\left\{{}\begin{matrix}x+1\ge0\\x^2-3x+2\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ne\left\{1;2\right\}\end{matrix}\right.\)
Hay \(D=[-1;1)\cup\left(1;2\right)\cup\left(2;+\infty\right)\)
b. \(\left\{{}\begin{matrix}x+2\ge0\\1-x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\x\ne1\end{matrix}\right.\)
c. \(\left\{{}\begin{matrix}1+x\ge0\\1-x\ge0\end{matrix}\right.\) \(\Leftrightarrow-1\le x\le1\)
d. \(\left\{{}\begin{matrix}1-4x^2\ge0\\x^2-2x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{1}{2}\le x\le\dfrac{1}{2}\\x\ne0\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Điều kiện: \(\left\{{}\begin{matrix}5x-10\ge0\\8-x\ge0\end{matrix}\right.\)\(\Leftrightarrow2\le x\le8\)
Bình phương hai vế ta được: \(5x-10=\left(8-x\right)^2\)
\(\Leftrightarrow5x-10=64-16x+x^2\)
\(\Leftrightarrow x^2-21x+74=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{21+\sqrt{145}}{2}\left(loai\right)\\x=\dfrac{21-\sqrt{145}}{2}\left(nhan\right)\end{matrix}\right.\)
Vậy nghiệm của phương trình là \(x=\dfrac{21-\sqrt{145}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{2x+3}{x-1}\ge1\Leftrightarrow\dfrac{2x+3}{x-1}-1\ge0\)
\(\Leftrightarrow\dfrac{x+4}{x-1}\ge0\)
\(\Rightarrow\left[{}\begin{matrix}x\le-4\\x>1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tham khảo tại link sau:
https://hoc24.vn/cau-hoi/ai-giup-em-cau-2-voi-a.3401576227354
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left\{{}\begin{matrix}3x_1+x_2=-2\\x_1+x_2=2m-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x_1=-2m\\x_1+x_2=2m-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=-m\\x_2=2m-2+m=3m-2\end{matrix}\right.\)
\(x_1\cdot x_2=m^2-3m\)
\(\Leftrightarrow-3m^2+2m-m^2+3m=0\)
\(\Leftrightarrow-4m^2+5m=0\)
\(\Leftrightarrow m\left(4m-5\right)=0\)
=>m=0 hoặc m=5/4
Câu 11: A
Câu 10: D
Câu 9: C