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Câu 40: -6<2x<=8

=>-3<x<=4

=>A=(-3;4]

=>\(C_{R}A\) =R\A=(-∞;3]\(\cup\) (4;+∞)

|x+1|<=2

=>-2<=x+1<=2

=>-3<=x<=1

=>B=[-3;1]

=>\(C_{R}B\) =R\B=(-∞;-3)\(\cup\) (1;+∞)

\(\left(C_{R}A\right)\) \\(\left(C_{R}B\right)\) =[-3;1]

=>Không có câu nào đúng

Câu 39:

Để A giao B=rỗng thì -m+2>2m+1 hoặc -m+5<=2m-3

=>-3m>-1 hoặc -3m<=-8

=>m<1/3 hoặc m>=8/3

=>Chọn B

8.

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+2x+3}=a>0\\\sqrt{x^2+4x+5}=b>0\end{matrix}\right.\) \(\Rightarrow2a^2-b^2=x^2+1\)

Pt trở thành:

\(\sqrt{2a^2-b^2}+2a=3b\)

\(\Leftrightarrow\sqrt{2a^2-b^2}=3b-2a\)

\(\Rightarrow2a^2-b^2=4a^2-12ab+9b^2\)

\(\Leftrightarrow2a^2-12ab+10b^2=0\Rightarrow\left[{}\begin{matrix}a=b\\a=5b\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+2x+3}=\sqrt{x^2+4x+5}\\\sqrt{x^2+2x+3}=5\sqrt{x^2+4x+5}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+3=x^2+4x+5\\x^2+2x+3=25\left(x^2+4x+5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\24x^2+98x+122=0\left(vn\right)\end{matrix}\right.\)

9.

ĐKXĐ: \(-1\le x\le1\)

Đặt \(\left\{{}\begin{matrix}\sqrt{1+x}=a\ge0\\\sqrt{1-x}=b\ge0\end{matrix}\right.\) \(\Rightarrow a^2+2b^2=3-x=-\left(x-3\right)\)

Pt trở thành:

\(a-2b-3ab=-\left(a^2+2b^2\right)\)

\(\Leftrightarrow a-2b+a^2-3ab+2b^2=0\)

\(\Leftrightarrow a-2b+\left(a-b\right)\left(a-2b\right)=0\)

\(\Leftrightarrow\left(a-2b\right)\left(a-b+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\a+1=b\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{1+x}=2\sqrt{1-x}\\\sqrt{1+x}+1=\sqrt{1-x}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}1+x=4\left(1-x\right)\\x+2+2\sqrt{1+x}=1-x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=3\Rightarrow x=\dfrac{3}{5}\\-1-2x=2\sqrt{1+x}\left(1\right)\end{matrix}\right.\)

Xét (1) \(\Leftrightarrow\left\{{}\begin{matrix}-1-2x\ge0\\\left(-1-2x\right)^2=4\left(1+x\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{1}{2}\\x^2=\dfrac{3}{4}\end{matrix}\right.\) \(\Rightarrow x=-\dfrac{\sqrt{3}}{2}\)

Vậy \(x=\left\{\dfrac{3}{5};-\dfrac{\sqrt{3}}{2}\right\}\)

5.

Tọa độ dỉnh của (P) là: \(I\left(-\dfrac{b}{2a};\dfrac{-\Delta}{4a}\right)\Rightarrow I\left(1;-4m-2\right)\)

Để I thuộc \(y=3x-1\)

\(\Rightarrow-4m-2=3.1-1\)

\(\Rightarrow m=-1\)

6.a.

Với \(a\ne0\)

 \(\left\{{}\begin{matrix}64a+8b+c=0\\-\dfrac{b}{2a}=5\\\dfrac{4ac-b^2}{4a}=12\end{matrix}\right.\)  \(\Leftrightarrow\left\{{}\begin{matrix}64a+8b+c=0\\b=-10a\\4ac-b^2=48a\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}c=-64a-8b=-64a-8\left(-10a\right)=16a\\b=-10a\\4ac-b^2=48a\end{matrix}\right.\)

\(\Rightarrow4a.16a-\left(-10a\right)^2=48a\)

\(\Rightarrow a=-\dfrac{4}{3}\Rightarrow b=\dfrac{40}{3}\Rightarrow c=-\dfrac{64}{3}\)

Hay pt (P): \(y=-\dfrac{4}{3}x^2+\dfrac{40}{3}x-\dfrac{64}{3}\)

b.

Thay tọa độ 3 điểm vào pt (P) ta được:

\(\left\{{}\begin{matrix}c=-1\\a+b+c=-1\\a-b+c=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=1\\b=-1\\c=-1\end{matrix}\right.\)

Pt (P): \(y=x^2-x-1\)

c.

Do (P) đi qua 3 điểm có tọa độ (1;16); (-1;0); (5;0) nên ta có:

\(\left\{{}\begin{matrix}a+b+c=16\\a-b+c=0\\25a+5b+c=0\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}a=-2\\b=8\\c=10\end{matrix}\right.\)

hay pt (P) có dạng: \(y=-2x^2+8x+10\)