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\(a,x^2+3x+9\)
\(=x^2+2\cdot x\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+\dfrac{27}{4}\)
\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{27}{4}\)
Ta thấy: \(\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\dfrac{3}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}\forall x\)
Dấu \("="\) xảy ra \(\Leftrightarrow x+\dfrac{3}{2}=0\Leftrightarrow x=-\dfrac{3}{2}\)
\(b,2x^2-5x+10\)
\(=2x^2-5x+\dfrac{25}{8}+\dfrac{55}{8}\)
\(=2\left(x^2-2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}\right)+\dfrac{55}{8}\)
\(=2\left(x-\dfrac{5}{4}\right)^2+\dfrac{55}{8}\)
Ta có: \(2\left(x-\dfrac{5}{4}\right)^2\ge0\forall x\)
\(\Rightarrow2\left(x-\dfrac{5}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\forall x\)
Dấu \("="\) xảy ra \(\Leftrightarrow x-\dfrac{5}{4}=0\Leftrightarrow x=\dfrac{5}{4}\)
#\(Toru\)
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Lời giải:
$A=5x^2+y^2+4xy-2x-2y+2020$
$=(4x^2+y^2+4xy)+x^2-2x-2y+2020$
$=(2x+y)^2-2(2x+y)+x^2+2x+2020$
$=(2x+y)^2-2(2x+y)+1+(x^2+2x+1)+2018$
$=(2x+y-1)^2+(x+1)^2+2018\geq 2018$
Vậy GTNN của $A$ là $2018$. Giá trị này đạt tại $2x+y-1=0$ và $x+1=0$
Hay $x=-1; y=3$
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A = 2x2 - 5x + 2
= 2( x2 - 5/2x + 25/16 ) - 9/8
= 2( x - 5/4 )2 - 9/8 ≥ -9/8 ∀ x
Đẳng thức xảy ra <=> x = 5/4
=> MinA = -9/8, đạt được khi x = 5/4
\(A=2x^2-5x+2\)
\(=2\left(x^2-\frac{5}{2}x+1\right)\)
\(=2\left(x^2-2x\frac{5}{4}+\frac{25}{16}\right)-\frac{9}{8}\)
\(=2\left(x-\frac{5}{4}\right)^2-\frac{9}{8}\ge-\frac{9}{8}\forall x\)
Dấu"=" xảy ra khi \(x-\frac{5}{4}=0\Rightarrow x=\frac{5}{4}\)
Vậy \(Min_A=-\frac{9}{8}\Leftrightarrow x=\frac{5}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(A=-2x^2-5x+3\)
\(=-2\left(x^2+\dfrac{5}{2}x-\dfrac{3}{2}\right)\)
\(=-2\left(x^2+2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}-\dfrac{49}{16}\right)\)
\(=-2\left(x+\dfrac{5}{4}\right)^2+\dfrac{49}{8}\)
Ta có: \(\left(x+\dfrac{5}{4}\right)^2\ge0\forall x\)
\(\Rightarrow-2\left(x+\dfrac{5}{4}\right)^2\le0\forall x\)
\(\Rightarrow-2\left(x+\dfrac{5}{4}\right)^2+\dfrac{49}{8}\le\dfrac{49}{8}\forall x\)
Dấu '=' xảy ra khi \(x+\dfrac{5}{4}=0\)
hay \(x=-\dfrac{5}{4}\)
Vậy: Giá trị lớn nhất của biểu thức \(A=-2x^2-5x+3\) là \(\dfrac{49}{8}\) khi \(x=-\dfrac{5}{4}\)
\(A=5x^2-2x=5\left(x^2-\frac{2}{5}x+\frac{1}{25}\right)-\frac{1}{5}\)
\(=5\left(x-\frac{1}{5}\right)^2-\frac{1}{5}\ge-\frac{1}{5}\forall x\)
Dấu "=" xảy ra khi \(\left(x-\frac{1}{5}\right)^2=0\Leftrightarrow x=\frac{1}{5}\)
Vậy \(A_{min}=-\frac{1}{5}\Leftrightarrow x=\frac{1}{5}\)