c: Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\left(y-\dfrac{1}{3}\right)^2\ge0\forall y\)

Do đó: \(\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2\ge0\forall x,y\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-\dfrac{1}{3}\right)^2-10\ge-10\forall x,y\)

Dấu '=' xảy ra khi x=-1 và \(y=\dfrac{1}{3}\)

GTNN là gì z.tui ko  hiểu nên ko giải được!

GTNN là giá trị nhỏ nhất

A = x² + 5x + 7 

   = ( x² + 5x + 25/4 ) + 3/4

   = ( x + 5/2 )² + 3/4

\(\left(x+\frac{5}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\frac{5}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

Đẳng thức xảy ra <=> x + 5/2 = 0 => x = -5/2

=> MinA = 3/4 <=> x = -5/2

B = 6x - x² - 5

   = -( x² - 6x + 9 ) + 4

   = -( x - 3 )² + 4

\(-\left(x-3\right)^2\le0\forall x\Rightarrow-\left(x-3\right)^2+4\le4\)

Đẳng thức xảy ra <=> x - 3 = 0 => x = 3

=> MaxB = 4 <=> x = 3

C = ( 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

\(\left(x^2+5x\right)^2\ge0\forall x\Rightarrow\left(x^2+5x\right)^2-36\ge-36\)

Đẳng thức xảy ra <=> x² + 5x = 0

                             <=> x( x + 5 ) = 0

                             <=> x = 0 hoặc x = -5

=> MinC = -36 <=> x = 0 hoặc x = -5

Thank bn.😊😉

1:

a: =x^2-7x+49/4-5/4

=(x-7/2)^2-5/4>=-5/4

Dấu = xảy ra khi x=7/2

b: =x^2+x+1/4-13/4

=(x+1/2)^2-13/4>=-13/4

Dấu = xảy ra khi x=-1/2

e: =x^2-x+1/4+3/4=(x-1/2)^2+3/4>=3/4

Dấu = xảy ra khi x=1/2

f: x^2-4x+7

=x^2-4x+4+3

=(x-2)^2+3>=3

Dấu = xảy ra khi x=2

2:

a: A=2x^2+4x+9

=2x^2+4x+2+7

=2(x^2+2x+1)+7

=2(x+1)^2+7>=7

Dấu = xảy ra khi x=-1

b: x^2+2x+4

=x^2+2x+1+3

=(x+1)^2+3>=3

Dấu = xảy ra khi x=-1

rất tiếc em mới học lớp 6

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jnymrjd,5

Bài 5:

a) \(A=x^2-4x+9=\left(x^2-4x+4\right)+5=\left(x-2\right)^2+5\ge5\)

\(minA=5\Leftrightarrow x=2\)

b) \(B=x^2-x+1=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(minB=\dfrac{3}{4}\Leftrightarrow x=\dfrac{1}{2}\)

c) \(C=2x^2-6x=2\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\ge-\dfrac{9}{2}\)

\(minC=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{3}{2}\)

Bài 4:

a) \(M=4x-x^2+3=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

\(maxM=7\Leftrightarrow x=2\)

b) \(N=x-x^2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

\(maxN=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)

c) \(P=2x-2x^2-5=-2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{9}{2}=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\le-\dfrac{9}{2}\)

\(maxP=-\dfrac{9}{2}\Leftrightarrow x=\dfrac{1}{2}\)