gọi g là trọng tâm tam giác abc chứng minh ag+bc/2>bg
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(Cho14x^4+4x^3-6x+16=0\)
\(x\left(14x^3+4x^2\right)-6x+16=0\)
\(14x^3+4x^2-6x+16=0\)
\(x\left(14x^2+4x\right)-6x+16=0\)
\(x\left(14x^2+4x+6\right)+16=0\)
\(14x^2+4x+6+16=0\)
\(14x^2+4x+22=0\)
\(14\left(x^2+14+8\right)-10=0\)
\(x^2+22-10=0\)
\(x^2+12=0\)
\(x^2=-12\)
\(x=\pm\sqrt{12}\)
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
hộ caiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii
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\(\frac{AG+BC}{2}>BG\)