{a22​=a1​.a3​a32​=a2​.a4​​\Rightarrow{a2a3=a1a2a3a4=a2a3{a2a3=a1a2a3a4=a2a3⇒{a3​a2​​=a2​a1​​a4​a3​​=a3​a2​​​\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}⇒a2​a1​​=a3​a2​​=a4​a3​​

\Rightarrow\frac{a_1^3}{a_2^3}=\frac{a_2^3}{a_3^3}=\frac{a_3^3}{a_4^3}=\frac{a_1}{a_2}.\frac{a_2}{a_3}=\frac{a_3}{a_4}=\frac{a_1}{a_4}\left(1\right)⇒a23​a13​​=a33​a23​​=a43​a33​​=a2​a1​​.a3​a2​​=a4​a3​​=a4​a1​​(1)

Áp dụng tính chất của dãy tỉ số = nhau ta có:

\frac{a_1^3}{a_2^3}=\frac{a_2^3}{a_3^3}=\frac{a_3^3}{a_4^3}=\frac{a_1^3+a_2^3+a_3^3}{a_2^3+a_3^3+a_4^3}\left(2\right)a23​a13​​=a33​a23​​=a43​a33​​=a23​+a33​+a43​a13​+a23​+a33​​(2)

Từ (1) và (2) \Rightarrow\frac{a_1^3+a_2^3+a_3^3}{a_2^3+a_3^3+a_4^3}=\frac{a_1}{a_4}\left(đpcm\right)⇒a23​+a33​+a43​a13​+a23​+a33​​=a4​a1​​(đpcm)

Ta có : \(f\left(1\right)=-\frac{1}{2}.1=-\frac{1}{2}\)

\(f\left(-2\right)=-\frac{1}{2}\left(-2\right)=1\)

\(f\left(-1\right)=-\frac{1}{2}\left(-1\right)=\frac{1}{2}\)

\(f\left(0\right)=-\frac{1}{2}0=0\)

Ta có : \(y=-\frac{1}{2}x=-1\Leftrightarrow x=2\)

\(y=-\frac{1}{2}x=0\Leftrightarrow x=0\)

\(y=-\frac{1}{2}x=2\Leftrightarrow x=-4\)

\(A=x^2+1\ge1\)

Dấu ''='' xảy ra <=> x = 0 

Vậy GTNN A là 1 <=> x = 0