c: OS=3*2=6(cm)

=>R₁=6/2=3(cm)

Diện tích hình tròn ngoại tiếp tứ giác SEOF là \(3^2\cdot3,14=28,26\left(cm^2\right)\)

2: \(P=\left(\dfrac{\sqrt{x}+1}{x-\sqrt{x}}+\dfrac{\sqrt{x}-2}{x-1}\right):\dfrac{1}{x-1}\)

\(=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{x-1}{1}\)

\(=\dfrac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x-1}{1}\)

\(=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}}{\sqrt{x}}=\dfrac{2x+1}{\sqrt{x}}\)

\(C=\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}+\dfrac{4}{x-4};\left(x\ge0;x\ne4\right)\)

\(=\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{2}{\sqrt{x}-2}\)

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\(x^2-\left(2m+1\right)x+m^2+1=0\)

\(\text{Δ}=\left(2m+1\right)^2-4\cdot1\cdot\left(m^2+1\right)\)

\(=4m^2+4m+1-4m^2-4=4m-3\)

Để phương trình có hai nghiệm phân biệt thì Δ>0

=>4m-3>0

=>4m>3

=>\(m>\dfrac{3}{4}\)

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2m+1\\x_1x_2=\dfrac{c}{a}=m^2+1\end{matrix}\right.\)

\(A=\left(2x_1-x_2\right)\left(x_1-2x_2\right)\)

\(=2x_1^2+2x_2^2-5x_1x_2\)

\(=2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]-5x_1x_2\)

\(=2\left[\left(2m+1\right)^2-2\left(m^2+1\right)\right]-5\left(m^2+1\right)\)

\(=2\left(2m+1\right)^2-9\left(m^2+1\right)\)

\(=2\left(4m^2+4m+1\right)-9m^2-9\)

\(=8m^2+8m+2-9m^2-9\)

\(=-m^2+8m-7\)

\(=-\left(m^2-8m+7\right)\)

\(=-\left(m^2-8m+16-9\right)\)

\(=-\left(m-4\right)^2+9< =9\forall m\)

Dấu '=' xảy ra khi m=4