Tìm min(GTNN) của :
A = 2x2 + 6x
B = x2 - 2x + y2 - 4y + 6
C = x2 - 2xy + 6y2 - 12x + 2y + 45
D = (x - 1) (x + 2) (x + 3) (x + 6)
a, -x2 + 2x + 3
b, x2 - 2x + 4y2 - 4y + 8 c, -x2 - y2 + xy + 2x + 2y + 4 d, x2 + 5y2 - 4xy - 2y + 2015 e, 2x2 + y2 + 6x + 2y + 2xy + 2018A= -x2+2x+3
=>A= -(x2-2x+3)
=>A= -(x2-2.x.1+1+3-1)
=>A=-[(x-1)2+2]
=>A= -(x+1)2-2
Vì -(x+1)2 ≤0=> A≤-2
Dấu "=" xảy ra khi
-(x+1)2=0 => x=-1
Vây A lớn nhất= -2 khi x= -1
B=x2-2x+4y2-4y+8
=> B= (x2-2x+1)+(4y2-4y+1)+6
=> B=(x-1)2+(2y+1)2+6
=> B lớn nhất=6 khi x=1 và y=-1/2
Bài 1: Tìm GTNN của biểu thức sau:
a) A= 2x2 + x
b) B = x2 + 2x + y2- 4y + 6
c) C = 4x2 + 4x + 9y2 - 6y - 5
d) D = (2 + x)( x + 4) - ( x - 1)( x + 3 )2
b) Ta có: \(B=x^2+2x+y^2-4y+6\)
\(=x^2+2x+1+y^2-4y+4+1\)
\(=\left(x+1\right)^2+\left(y-2\right)^2+1\ge1\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
Vậy: \(B_{min}=1\) khi (x,y)=(-1;2)
c) Ta có: \(C=4x^2+4x+9y^2-6y-5\)
\(=4x^2+4x+1+9y^2-6y+1-7\)
\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(C_{min}=-7\) khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)
\(A=2x^2+x=2\left(x^2+\dfrac{1}{2}x\right)=2\left(x^2+2.\dfrac{1}{4}x+\dfrac{1}{16}-\dfrac{1}{16}\right)\)
\(=2\left[\left(x+\dfrac{1}{4}\right)^2-\dfrac{1}{16}\right]\ge-\dfrac{1}{8}\) dấu"=' xảy ra<=>x=\(-\dfrac{1}{4}\)
\(B=x^2+2x+y^2-4y+6\)
\(=x^2+2x+1+y^2-4y+4+1=\left(x+1\right)^2+\left(y-2\right)^2+1\)
\(\ge1\) dấu"=" xảy ra<=>x=-1;y=2
\(C=4x^2+4x+9y^2-6y-5\)
\(=4x^2+4x+1+9y^2-6y+1-7\)
\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\)
dấu"=" xảy ra<=>x=\(-\dfrac{1}{2},y=\dfrac{1}{3}\)
\(D=\left(2+x\right)\left(x+4\right)-\left(x-1\right)\left(x+3\right)^2\)
=\(x^2+6x+8-\left(x-1\right)\left(x+3\right)^2\)
\(=\left(x+3\right)^2-1-\left(x-1\right)\left(x+3\right)^2\)
\(=\left(x+3\right)^2\left(2-x\right)-1\ge-1\)
dấu"=" xảy ra\(< =>\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
a) 3x(x+1)-x(3x+2)
b) 2x(x2-5x+6)+(x-1)(x+3)
c) (x2-xy+y2)-(x2+2xy+y2)
d) (2/5xy+x-y)-(3x+4y)-2/5xy
e) 2xy(x2-4xy+4y2)
f) (x+y)(xy+5)
g) (x3-2x2-x+2):(x-1)
h) (2x2+3x-2):(2x-1)
Tìm min(GTNN) của :
A = 2x2 + 6x
B = x2 - 2x + y2 - 4y + 6
C = x2 - 2xy + 6y2 - 12x + 2y + 45
D = (x - 1) (x + 2) (x + 3) (x + 6)
A = 2x2 + 6x = 2( x2 + 3x + 9/4 ) - 9/2 = 2( x + 3/2 )2 - 9/2 ≥ -9/2 ∀ x
Dấu "=" xảy ra khi x = -3/2
=> MinA = -9/2 <=> x = -3/2
B = x2 - 2x + y2 - 4y + 6 = ( x2 - 2x + 1 ) + ( y2 - 4y + 4 ) + 1 = ( x - 1 )2 + ( y - 2 )2 + 1 ≥ 1 ∀ x, y
Dấu "=" xảy ra khi x = 1 ; y = 2
=> MinB = 1 <=> x = 1 ; y = 2
C = x2 - 2xy + 6y2 - 12x + 2y + 45
= ( x2 - 2xy + y2 - 12x + 12y + 36 ) + ( 5y2 - 10y + 5 ) + 4
= [ ( x2 - 2xy + y2 ) - ( 12x - 12y ) + 36 ] + 5( y2 - 2y + 1 ) + 4
= [ ( x - y )2 - 2( x - y ).6 + 62 ] + 5( y - 1 )2 + 4
= ( x - y - 6 )2 + 5( y - 1 )2 + 4 ≥ 4 ∀ x, y
Dấu "=" xảy ra khi x = 7 ; y = 1
=> MinC = 4 <=> x = 7 ; y = 1
D = ( x - 1 )( x + 2 )( x + 3 )( x + 6 )
= [ ( x - 1 )( x + 6 ) ][ ( x + 2 )( x + 3 ) ]
= ( x2 + 5x - 6 )( x2 + 5x + 6 )
= ( x2 + 5x )2 - 36 ≥ -36 ∀ x
Dấu "=" xảy ra <=> x2 + 5x = 0
<=> x( x + 5 ) = 0
<=> x = 0 hoặc x = -5
=> MinD = -36 <=> x = 0 hoặc x = -5
Tìm min(GTNN) của :
A = 2x2 + 6x
B = x2 - 2x + y2 - 4y + 6
C = x2 - 2xy + 6y2 - 12x + 2y + 45
D = (x - 1) (x + 2) (x + 3) (x + 6)
1. Tìm x,y:
a) (x+2)2 + (x-3)2 = 2x ( x+ 7)
b) x3- 3x2 + 3x - 126 = 0
c) x2 + y2 - 2x + 4y + 5 = 0
d) 2x2 - 2xy + y2 + 4x + 4 = 0
\(a.\left(x^2+4x+4\right)+\left(x^2-6x+9\right)=2x^2+14x\)
\(x^2+4x+4+x^2-6x+9-2x^2-14x=0\)
\(-18x+13=0\)
\(x=\dfrac{13}{18}\)
Vậy \(S=\left\{\dfrac{13}{18}\right\}\)
\(b.\left(x-1\right)^3-125=0\)
\(\left(x-1\right)^3=125\)
\(x-1=5\)
\(x=6\)
Vậy \(S=\left\{6\right\}\)
\(c.\left(x-1\right)^2+\left(y +2\right)^2=0\)
\(Do\left(x-1\right)^2\ge0\forall x;\left(y+2\right)^2\ge0\forall y\)
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)
Mà \(\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy \(S=\left\{1;-2\right\}\)
\(d.x^2-4x+4+x^2-2xy+y^2=0\)
\(\left(x-2\right)^2+\left(x-y\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x-y\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
Vậy \(S=\left\{2;2\right\}\)
Tìm Min/Max
A= y2-4y+9
B=x2-x+1
C=2x2-6x
A = y^2 - 4y + 9 = y^2 - 4y + 4 + 5
= ( y - 2 )^2 + 5 >= 5
Dấu ''='' xảy ra khi y = 2
Vậy GTNN A là 5 khi y = 2
B = x^2 - x + 1 = x^2 - x + 1/4 + 3/4 = ( x - 1/2 )^2 + 3/4 >= 3/4
Dấu ''='' xảy ra khi x = 1/2
Vậy GTNN B là 3/4 khi x = 1/2
C = 2x^2 - 6x = 2 ( x^2 - 3x + 9 / 4 - 9/4 )
= 2 ( x - 3/2 )^2 - 9/2 >= -9/2
Dấu ''='' xảy ra khi x = 3/2
Vậy GTNN C là -9/2 khi x = 3/2
a) Ta có: \(A=y^2-4y+9\)
\(=y^2-4y+4+5\)
\(=\left(y-2\right)^2+5\ge5\forall y\)
Dấu '=' xảy ra khi y=2
b) Ta có: \(B=x^2-x+1\)
\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
cho x,y∈ R ; x≠y
tìm min P=x2-6xy+6y2/x2-2xy+y2
\(P=\dfrac{x^2-6xy+6y^2}{x^2-2xy+y^2}=\dfrac{-3\left(x^2-2xy+y^2\right)+4x^2-12xy+9y^2}{x^2-2xy+y^2}\)
\(=-3+\left(\dfrac{2x-3y}{x-y}\right)^2\ge-3\)
\(P_{min}=-3\) khi \(2x=3y\)
tìm gtnn (gtln) của
a) 4x2+12x+1 b) 4x2-3x+10
c)2x2+5x+10 d) x-x2+2
e) 2x-2x2 f) 4x2+2y2+4xy+4y+5
a) \(4x^2+12x+1=\left(4x^2+12x+9\right)-8=\left(2x+3\right)^2-8\ge-8\)
\(ĐTXR\Leftrightarrow x=-\dfrac{3}{2}\)
b) \(4x^2-3x+10=\left(4x^2-3x+\dfrac{9}{16}\right)+\dfrac{151}{16}=\left(2x-\dfrac{3}{4}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\)
\(ĐTXR\Leftrightarrow x=\dfrac{3}{8}\)
c) \(2x^2+5x+10=\left(2x^2+5x+\dfrac{25}{8}\right)+\dfrac{55}{8}=\left(\sqrt{2}x+\dfrac{5\sqrt{2}}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\)
\(ĐTXR\Leftrightarrow x=-\dfrac{5}{4}\)
d) \(x-x^2+2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{9}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\)
\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)
e) \(2x-2x^2=-2\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{2}=-2\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\le\dfrac{1}{2}\)
\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)
f) \(4x^2+2y^2+4xy+4y+5=\left(4x^2+4xy+y^2\right)+\left(y^2+4y+4\right)+1=\left(2x+y\right)^2+\left(y+2\right)^2+1\ge1\)
\(ĐTXR\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
a: Ta có: \(4x^2+12x+1\)
\(=4x^2+12x+9-8\)
\(=\left(2x+3\right)^2-8\ge-8\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)
b: Ta có: \(4x^2-3x+10\)
\(=4\left(x^2-\dfrac{3}{4}x+\dfrac{5}{2}\right)\)
\(=4\left(x^2-2\cdot x\cdot\dfrac{3}{8}+\dfrac{9}{64}+\dfrac{151}{64}\right)\)
\(=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{8}\)
c: Ta có: \(2x^2+5x+10\)
\(=2\left(x^2+\dfrac{5}{2}x+5\right)\)
\(=2\left(x^2+2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{55}{16}\right)\)
\(=2\left(x+\dfrac{5}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{5}{4}\)