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DL
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DL
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TN
14 tháng 6 2017
\(B=4x^2+5y^2-4xy+3x-y\)
\(\Leftrightarrow\left(4x^2-4xy+3x\right)+5y^2-y\)
\(\Leftrightarrow\left[4x^2-4x\left(y-\dfrac{3}{4}\right)+\left(y-\dfrac{3}{4}\right)^2\right]+5y^2-y-y^2+\dfrac{3}{2}y-\dfrac{9}{16}\)\(\Leftrightarrow\left(2x-y+\dfrac{3}{4}\right)^2+\left(4y^2-\dfrac{1}{2}y+\dfrac{1}{64}\right)-\dfrac{37}{64}\)
\(\Leftrightarrow\left(2x-y+\dfrac{3}{4}\right)^2+\left(2y-\dfrac{1}{8}\right)^2-\dfrac{37}{64}\ge\dfrac{-37}{64}\)
Vậy Min B = \(\dfrac{-37}{64}\) khi \(\left[{}\begin{matrix}\left(2x-y+\dfrac{3}{4}\right)^2=0\\\left(2y-\dfrac{1}{8}\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-y+\dfrac{3}{4}=0\\2y-\dfrac{1}{8}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-y+\dfrac{3}{4}=0\\2y=\dfrac{1}{8}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-\dfrac{1}{16}+\dfrac{3}{4}=0\\y=\dfrac{1}{16}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-11}{32}\\y=\dfrac{1}{16}\end{matrix}\right.\)
TN
14 tháng 6 2017
\(C=9y^2+2x^2-6y-6xy+5x-1\)
\(=\left(9y^2+6y-6xy\right)+2x^2+5x-1\)
\(=\left[9y^2+6y\left(1-x\right)+\left(1-x\right)^2\right]+2x^2+5x-1-1+2x-x^2\)\(=\left(3y-x+1\right)^2+\left(x^2+3x+\dfrac{9}{4}\right)-\dfrac{17}{4}\)
\(=\left(3y-x+1\right)^2+\left(x+\dfrac{3}{2}\right)^2-\dfrac{17}{4}\)
Vậy Min C = \(\dfrac{-17}{4}\) khi \(\left[{}\begin{matrix}\left(3y-x+1\right)^2=0\\\left(x+\dfrac{3}{2}\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3y-x+1=0\\x+\dfrac{3}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3y-\left(\dfrac{-3}{2}\right)+1=0\\x=\dfrac{-3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=\dfrac{-5}{6}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
20 tháng 7 2022
a: \(B=\dfrac{10x}{\left(x+4\right)\left(x-1\right)}-\dfrac{2x-3}{x+4}-\dfrac{x+1}{x-1}\)
\(=\dfrac{10x-\left(2x^2-2x-3x+3\right)-\left(x^2+5x+4\right)}{\left(x+4\right)\left(x-1\right)}\)
\(=\dfrac{10x-2x^2+5x-3-x^2-5x-4}{\left(x+4\right)\left(x-1\right)}\)
\(=\dfrac{-3x^2+10x-7}{\left(x+4\right)\left(x-1\right)}\)
\(=\dfrac{-\left(3x^2-10x+7\right)}{\left(x-1\right)\left(x+4\right)}=-\dfrac{\left(x-1\right)\left(3x-7\right)}{\left(x-1\right)\left(x+4\right)}\)
\(=\dfrac{-3x+7}{x+4}\)
b: \(B+3=\dfrac{-3x+7+3x+12}{x+4}=\dfrac{19}{x+4}>0\)
=>B>-3
DL
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