Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(M=4x^2-10x+\frac{9}{2x}+2018\)
\(=4x^2-12x+2x+\frac{9}{2x}+2018\)
\(=\left(4x^2-12x+9\right)+\left(2x+\frac{9}{2x}\right)+2009\)
\(=\left[\left(2x\right)^2-2.2x.3+3^2\right]+\left(2x+\frac{9}{2x}\right)+2009\)
\(=\left(2x-3\right)^2+\left(2x+\frac{9}{2x}\right)+2009\)
Ta có : \(2x+\frac{9}{2x}\ge2\sqrt{2x\cdot\frac{9}{2x}}=2.\sqrt{9}=6\)
\(\Rightarrow M\ge\left(2x-3\right)^2+6+2009\ge2015\)
Dấu "=" xảy ra <=> \(x=\frac{3}{2}\)
Vậy GTNN của M là \(2015\) tại \(x=\frac{3}{2}\)
Bo may la binh day k di hieu ashdbfgbgygygggydfsghuyfhdguuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu3
Ta có:
\(y=3x\left(k^2-k+1\right)=-2\)
\(\Rightarrow3\left(-3\right)\left(k^2-k+1\right)=-2\)
\(\Rightarrow-9\left(k^2-k+1\right)=-2\)
\(\Rightarrow k^2-k+1=\frac{2}{9}\)
\(\Rightarrow\left(k-\frac{1}{2}\right)^2=\frac{2}{9}-\frac{3}{4}=\frac{35}{36}\) (Vô nghiệm)
ok cảm ơn
công nhận nó dễ thật