Tim giá trị nhỏ nhất của 2x^2+5x+8
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Ta có: A = 2x2 - 5x + 3 = 2(x2 - 5/2x + 25/16) - 1/8 = 2(x - 5/4)2 - 1/8 \(\le\)-1/8 \(\forall\)x
Dấu "=" xảy ra <=> x - 5/4 = 0 <=> x = 5/4
Vậy MinA = -1/8 <=> x = 5/4
\(A=2x^2-5x+3=2\left(x^2-\frac{5}{2}x+\frac{3}{2}\right)\)
\(=2\left(x^2-\frac{5}{2}x+\frac{25}{16}-\frac{1}{16}\right)\)
\(=2\left[\left(x-\frac{5}{4}\right)^2-\frac{1}{16}\right]\)
\(=2\left[\left(x-\frac{5}{4}\right)^2\right]-\frac{1}{8}\ge\frac{-1}{8}\)
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F = 5x2 + 2y2 + 4xy - 2x + 4y + 8
F = ( 4x2 + 4xy + y2 ) + ( x2 - 2x + 1 ) + ( y2 + 4y + 4 ) + 3
F = ( 2x + y )2 + ( x - 1 )2 + ( y + 2 )2 + 3
\(\hept{\begin{cases}\left(2x+y\right)^2\\\left(x-1\right)^2\\\left(y+2\right)^2\end{cases}}\ge0\forall x,y\Rightarrow\left(2x+y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3\ge3\forall x,y\)
Đẳng thức xảy ra <=> \(\hept{\begin{cases}2x+y=0\\x-1=0\\y+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-2\end{cases}}\)
Vậy MinF = 3 <=> x = 1 , y = -2
G = 5x2 + 5y2 + 8xy + 2y + 2020
= x2 + ( 4x2 + 8xy + 4y2 ) + ( y2 + 2y + 1 ) + 2019
= x2 + ( 2x + 2y )2 + ( y + 1 )2 + 2019
\(\hept{\begin{cases}x^2\\\left(2x+2y\right)^2\\\left(y+1\right)^2\end{cases}}\ge0\forall x,y\Rightarrow x^2+\left(2x+2y\right)^2+\left(y+1\right)^2+2019\ge2019\forall x,y\)
Tuy nhiên đẳng thức không xảy ra :P
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\(B=2x^2-5x+3\)
\(=2\left(x^2-\frac{5}{2}x+\frac{3}{2}\right)\)
\(=2\left(x^2-\frac{5}{2}x+\frac{25}{16}-\frac{1}{16}\right)\)
\(=2\left[\left(x-\frac{5}{4}\right)^2-\frac{1}{16}\right]\)
\(=2\left[\left(x-\frac{5}{4}\right)^2\right]-\frac{1}{32}\ge\frac{-1}{32}\)
\(B=2x^2-5x+3\)
\(=2\left(x^2-\frac{5}{2}x+\frac{3}{2}\right)\)
\(=2\left(x^2-\frac{5}{4}\cdot2x+\left(\frac{5}{4}\right)^2-\left(\frac{5}{4}\right)^2+\frac{3}{2}\right)\)
\(=2\left[\left(x-\frac{5}{4}\right)^2-\frac{25}{16}+\frac{3}{2}\right]\)
\(=2\left[\left(x-\frac{5}{4}\right)^2-\frac{1}{16}\right]\)
\(=2\left(x-\frac{5}{4}\right)^2-\frac{1}{8}\)
có\(2\left(x-\frac{5}{4}\right)^2\ge0\)
\(\Rightarrow\left(x-\frac{5}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\)
\(\Rightarrow GTNNB=-\frac{1}{8}\)
với \(\left(x-\frac{5}{4}\right)^2=0;x=\frac{5}{4}\)
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\(P=5x^2+y^2-2x(y+8)+2023\\=5x^2+y^2-2xy-16x+2023\\=(x^2-2xy+y^2)+(4x^2-16x+16)+2007\\=(x-y)^2+4(x^2-4x+4)+2007\\=(x-y)^2+4(x-2)^2+2007\)
Ta thấy: \(\left(x-y\right)^2\ge0\forall x;y\)
\(4\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow (x-y)^2+4(x-2)^2\ge0\forall x;y\\\Rightarrow P=(x-y)^2+4(x-2)^2+2007\ge2007\forall x;y\)
Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}x-y=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=2\end{matrix}\right.\Leftrightarrow x=y=2\)
Vậy \(Min_P=2007\) khi \(x=y=2\).
\(\text{#}Toru\)
\(P=5x^2+y^2-2x\left(y+8\right)+2023\)
\(=x^2-2xy+y^2+4x^2-16x+2023\)
\(=\left(x-y\right)^2+4x^2-16x+16+2007\)
\(=\left(x-y\right)^2+\left(2x-4\right)^2+2007>=2007\)
Dấu = xảy ra khi x-y=0 và 2x-4=0
=>x=y=2
Theo đề bài ta có :
\(2x^2+5x+8=2\left(x+1,25\right)^2-3,125+8\)
= \(2\left(x+1,25\right)^2+4,875\)
=> \(Max=4,875\Leftrightarrow x=-1,25\)
Quyên Min nha không phải Max