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a) Ta có: \(M=x^2-3x+10\)
\(=x^2-2\cdot x\cdot\frac{3}{2}+\frac{9}{4}+\frac{31}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{31}{4}\)
Ta có: \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{31}{4}\ge\frac{31}{4}\forall x\)
Dấu '=' xảy ra khi \(x-\frac{3}{2}=0\)
hay \(x=\frac{3}{2}\)
Vậy: Giá trị nhỏ nhất của biểu thức \(M=x^2-3x+10\) là \(\frac{31}{4}\) khi \(x=\frac{3}{2}\)
b) Ta có: \(N=2x^2+5y^2+4xy+8x-4y-100\)
\(=x^2+8x+16+x^2+4xy+4y^2+y^2-4y+4-120\)
\(=\left(x+4\right)^2+\left(x+2y\right)^2+\left(y-2\right)^2-120\)
Ta có: \(\left(x+4\right)^2\ge0\forall x\)
\(\left(x+2y\right)^2\ge0\forall x,y\)
\(\left(y-2\right)^2\ge0\forall y\)
Do đó: \(\left(x+4\right)^2+\left(x+2y\right)^2+\left(y-2\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x+4\right)^2+\left(x+2y\right)^2+\left(y-2\right)^2-120\ge-120\forall x,y\)
Dấu '=' xảy ra khi
\(\left\{{}\begin{matrix}x+4=0\\x+2y=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\-4+2y=0\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\2y=4\\y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=2\\y=2\end{matrix}\right.\)
Vậy: Giá trị nhỏ nhất của biểu thức \(N=2x^2+5y^2+4xy+8x-4y-100\) là -120 khi x=-4 và y=2
\(A=x-x^2=-\left(x^2-2\times x\times\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\right)=-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\)
\(\left(x-\frac{1}{2}\right)^2\ge0\)
\(\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
\(-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\le\frac{1}{4}\)
Vậy Max A = \(\frac{1}{4}\) khi x = \(\frac{1}{2}\)
***
\(B=5-8x-x^2=-\left(x^2+2\times x\times4+4^2-4^2-5\right)=-\left[\left(x+4\right)^2-21\right]\)
\(\left(x+4\right)^2\ge0\)
\(\left(x+4\right)^2-21\ge-21\)
\(-\left[\left(x+4\right)^2-21\right]\le21\)
Vậy Max B = 21 khi x = - 4
***
\(C=5-x^2+2x-4y^2-4y=-\left(x^2-2\times x\times1+1^2-1^2+\left(2y\right)^2-2\times2y\times1+1^2-1^2-5\right)=-\left[\left(x-1\right)^2+\left(2y-1\right)^2-7\right]\)
\(\left(x-1\right)^2\ge0\)
\(\left(2y-1\right)^2\ge0\)
\(\left(x-1\right)^2+\left(2y-1\right)^2-7\ge-7\)
\(-\left[\left(x-1\right)^2+\left(2y-1\right)^2-7\right]\le7\)
Vậy Max C = 7 khi x = 1 và y = \(\frac{1}{2}\)
\(A=\left(x^2-4x+4\right)+4=\left(x-2\right)^2+4\ge4\)
\(minA=4\Leftrightarrow x=2\)
\(B=\left(4x^2-12x+9\right)+2=\left(2x-3\right)^2+2\ge2\)
\(minB=2\Leftrightarrow x=\dfrac{3}{2}\)
\(C=3\left(x^2+2x+1\right)-8=3\left(x+1\right)^2-8\ge-8\)
\(minC=-8\Leftrightarrow x=-1\)
\(D=-\left(x^2-2x+1\right)-4=-\left(x-1\right)^2-4\le-4\)
\(maxD=-4\Leftrightarrow x=1\)
\(E=-\left(4x^2-6x+\dfrac{9}{4}\right)-\dfrac{11}{4}=-\left(2x-\dfrac{3}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\)
\(maxA=-\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{4}\)
\(F=-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{55}{8}=-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{55}{8}\le-\dfrac{55}{8}\)
\(maxF=-\dfrac{55}{8}\Leftrightarrow x=\dfrac{1}{4}\)
\(G=\left(x^2-4xy+4y^2\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-2y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(maxG=\dfrac{3}{4}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-\dfrac{1}{2}\end{matrix}\right.\)
\(H=-\left(x^2-2x+1\right)-\left(y^2+4y+4\right)+16=-\left(x-1\right)^2-\left(y+2\right)^2+16\le16\)
\(maxH=16\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
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 $x=-1; y=3$
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
a) Từ M = x − 3 2 2 + 31 4 ≥ 31 4 ⇒ M min = 31 4 ⇔ x = 3 2 .
b) Ta có N = ( x + 2 y ) 2 + ( y – 2 ) 2 + ( x + 4 ) 2 – 120 ≥ - 120 .
Tìm được N min = -120 Û x = -4 và y = 2.
a) Đặt \(A=x^2-2x+1\)
Ta có: \(A=x^2-2x+1=\left(x-1\right)^2\)
Vì \(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow A_{min}=0\)
Dấu "=" xảy ra khi: \(x-1=0\)
\(\Leftrightarrow x=1\)
Vậy \(A_{min}=0\)\(\Leftrightarrow\)\(x=1\)
b) Ta có: \(M=x^2-3x+10\)
\(\Leftrightarrow M=\left(x^2-3x+\frac{9}{4}\right)+\frac{31}{4}\)
\(\Leftrightarrow M=\left(x-\frac{3}{2}\right)^2+\frac{31}{4}\)
Vì \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\)\(\Rightarrow\)\(\left(x-\frac{3}{2}\right)^2+\frac{31}{4}\ge\frac{31}{4}\forall x\)
\(\Rightarrow\)\(M_{min}=\frac{31}{4}\)
Dấu "=" xảy ra khi: \(x-\frac{3}{2}=0\)
\(\Leftrightarrow x=\frac{3}{2}\)
Vậy \(M_{min}=\frac{31}{4}\)\(\Leftrightarrow\)\(x=\frac{3}{2}\)
\(A=x^2-8x+16+x^2+4xy+4y^2+y^2+4y+4+2004\)
\(=\left(x-4\right)^2+\left(x+2y\right)^2+\left(y+2\right)^2+2004\ge2004\)
Dấu ''='' xảy ra khi x = 4 ; y = -2