TK
2
K
Khách
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DT
2
21 tháng 5 2017
\(M=4x^2-10x+\frac{9}{2x}+2018\)
\(=4x^2-12x+2x+\frac{9}{2x}+2018\)
\(=\left(4x^2-12x+9\right)+\left(2x+\frac{9}{2x}\right)+2009\)
\(=\left[\left(2x\right)^2-2.2x.3+3^2\right]+\left(2x+\frac{9}{2x}\right)+2009\)
\(=\left(2x-3\right)^2+\left(2x+\frac{9}{2x}\right)+2009\)
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 \(x=\frac{3}{2}\)
DH
0
14 tháng 10 2018
Bo may la binh day k di hieu ashdbfgbgygygggydfsghuyfhdguuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu3
DT
1
HN
18 tháng 9 2021
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)
OF
3
T
19 tháng 8 2023
Có: \(\left|2022-2x+y\right|\ge0\forall x,y\)
\(\left(x-y-2021\right)^2\ge0\forall x,y\)
\(\Rightarrow\left|2022-2x+y\right|+\left(x-y-2021\right)^2\ge0\forall x,y\)
Mặt khác: \(\left|2022-2x+y\right|+\left(x-y-2021\right)^2=0\)
nên \(\left\{{}\begin{matrix}2022-2x+y=0\\x-y-2021=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}-2x+y=-2022\\x-y=2021\end{matrix}\right.\)
\(\Rightarrow-2x+y+x-y=-2022+2021\)
\(\Rightarrow-x=-1\Leftrightarrow x=1\)
Khi đó: \(1-y=2021\) \(\Leftrightarrow y=-2020\)
\(\Rightarrow x+y=1-2020=-2019\)
19 tháng 8 2023
|2022-2x+y|+(x-y-2021)^2=0
=>2022-2x+y=0 và x-y-2021=0
=>x-y=2021 và 2x-y=2022
=>x=1 và y=-2020
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