\(M=5x^2-2x+7=5\left(x^2-\dfrac{2}{5}x+\dfrac{1}{25}\right)+\dfrac{34}{5}\)
\(=5\left(x-\dfrac{1}{5}\right)^2+\dfrac{34}{5}\ge\dfrac{34}{5}\)
\(minM=\dfrac{34}{5}\Leftrightarrow x=\dfrac{1}{5}\)
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\(M=5x^2-2x+7\)
=> \(M=5x^2-5\dfrac{2}{5}x+5\dfrac{1}{25}+\dfrac{34}{5}\)
=> \(M=5\left(x-\dfrac{1}{5}\right)^2+\dfrac{34}{5}\)
mà \(5\left(x-\dfrac{1}{5}\right)^2\)≥0 => \(5\left(x-\dfrac{1}{5}\right)^2+\dfrac{34}{5}\)≥\(\dfrac{34}{5}\)
vậy Min M = 34/5 dấu = xảy ra khi x=1/5
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