Không gian mẫu: \(\dfrac{52!}{\left(4!\right)^{13}}\)
Do đó xác suất: \(P=\dfrac{1}{\dfrac{52!}{\left(4!\right)^{13}}}=\dfrac{\left(4!\right)^{13}}{52!}=...\)
1.
ĐKXĐ: \(-3\le x\le1\)
\(2\left(x+3\right)-m\sqrt{x+3}+5\left(1-x\right)+2m\sqrt{1-x}=4\sqrt{\left(x+3\right)\left(1-x\right)}\)
\(\Leftrightarrow m\left(2\sqrt{1-x}-\sqrt{x+3}\right)=3x-11+4\sqrt{\left(x+3\right)\left(1-x\right)}\)
Đặt \(2\sqrt{1-x}-\sqrt{x+3}=t\Rightarrow t\in\left[-2;4\right]\)
\(t^2=7-3x-4\sqrt{\left(1-x\right)\left(x+3\right)}\)
\(\Rightarrow3x-11+4\sqrt{\left(1-x\right)\left(x+3\right)}=-4-t^2\)
Do đó pt trở thành: \(m.t=-t^2-4\)
- Với \(t=0\) ko phải nghiệm
- Với \(t\ne0\Rightarrow m=\dfrac{-t^2-4}{t}\)
Xét \(f\left(t\right)=\dfrac{-t^2-4}{t}\) với \(t\in\left[-2;4\right]\)
\(f^2\left(t\right)=\dfrac{\left(t^2+4\right)^2}{t^2}\ge4\Rightarrow\left[{}\begin{matrix}f\left(t\right)\le-2\\f\left(t\right)\ge2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}m\le-2\\m\ge2\end{matrix}\right.\)
2.
ĐKXĐ: \(-1\le x\le1\)
Đặt \(\sqrt{1+x^2}-\sqrt{1-x^2}=t\) \(\Rightarrow t\in\left[0;\sqrt{2}\right]\)
\(t^2=2-2\sqrt{1-x^4}\Rightarrow2\sqrt{1-x^4}=2-t^2\)
Pt trở thành:
\(m\left(t+2\right)=2-t^2+t\)
\(\Leftrightarrow m=\dfrac{-t^2+t+2}{t+2}\)
Xét \(f\left(t\right)=\dfrac{-t^2+t+2}{t+2}\) trên \(\left[0;\sqrt{2}\right]\)
\(f\left(t\right)=\dfrac{-t^2}{t+2}+1\le1\) ;
\(f\left(t\right)-\left(\sqrt{2}-1\right)=\dfrac{-t^2+t+2-\left(\sqrt{2}-1\right)\left(t+2\right)}{t+2}=\dfrac{\left(\sqrt{2}-t\right)\left(t+2\sqrt{2}-2\right)}{t+2}\ge0\)
\(\Rightarrow f\left(t\right)\ge\sqrt{2}-1\)
\(\Rightarrow\sqrt{2}-1\le f\left(t\right)\le1\)
\(\Rightarrow\sqrt{2}-1\le m\le1\)
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