gọi M là trung điểm BC

=> \(\frac{AG+BC}{2}=AM+BM\)

Vì theo tính chất trọng tâm AG=2AM=BG

trong tam giác BMG thì tổng của 2 cạnh luôn lớn hơn cạnh còn lại 

=>AM+BM>BG hay\(\frac{AG+BC}{2}>BG\)

\(Cho14x^4+4x^3-6x+16=0\)

\(x\left(14x^3+4x^2\right)-6x+16=0\)

\(14x^3+4x^2-6x+16=0\)

\(x\left(14x^2+4x\right)-6x+16=0\)

\(x\left(14x^2+4x+6\right)+16=0\)

\(14x^2+4x+6+16=0\)

\(14x^2+4x+22=0\)

\(14\left(x^2+14+8\right)-10=0\)

\(x^2+22-10=0\)

\(x^2+12=0\)

\(x^2=-12\)

\(x=\pm\sqrt{12}\)

Để B có nghiệm

=> B = 0

=> 2x⁴ - 8x² = 0

=> 2x²(x² - 4) = 0

=> \(\orbr{\begin{cases}2x^2=0\\x^2-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)

Vậy x \(\in\left\{0;2;-2\right\}\)là nghiệm của đa thức B 

thanks nhìu

help meeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

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