Câu 2:

ĐKXĐ: x<>0

\(B=\dfrac{-x^2-x-1}{x^2}\)

\(=-1-\dfrac{1}{x}-\dfrac{1}{x^2}\)

\(=-\left(\dfrac{1}{x^2}+\dfrac{1}{x}+1\right)\)

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

\(=-\left(\dfrac{1}{x}+\dfrac{1}{2}\right)^2-\dfrac{3}{4}< =-\dfrac{3}{4}\forall x< >0\)

Dấu '=' xảy ra khi 1/x+1/2=0

=>1/x=-1/2

=>x=-2

a: =-x^2+6x-4

=-(x^2-6x+4)

=-(x^2-6x+9-5)

=-(x-3)^2+5<=5

Dấu = xảy ra khi x=3

b: =3(x^2-5/3x+7/3)

=3(x^2-2*x*5/6+25/36+59/36)

=3(x-5/6)^2+59/12>=59/12

Dấu = xảy ra khi x=5/6

c: \(=-\left(x-3\right)^2+2\left|x-3\right|\)

\(=-\left[\left(\left|x-3\right|\right)^2-2\left|x-3\right|+1-1\right]\)

\(=-\left(\left|x-3\right|-1\right)^2+1< =1\)

Dấu = xảy ra khi x=4 hoặc x=2

Biểu thức nào em?

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

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

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

\(\left|2x-1\right|+3\ge3\Leftrightarrow\dfrac{3+\left|2x-1\right|}{14}\ge\dfrac{3}{14}\)

Dấu \("="\Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)

\(\dfrac{-4x^2+4x}{15}=\dfrac{-4x^2+4x-1+1}{15}=\dfrac{-\left(2x-1\right)^2+1}{15}\)

Ta có \(-\left(2x-1\right)^2+1\le1\Leftrightarrow\dfrac{-\left(2x-1\right)^2+1}{15}\le\dfrac{1}{15}\)

Dấu \("="\Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)

a: =x^2-6x+9+x^2-6x+9

=2(x-3)^2>=0

Dấu = xảy ra khi x=3

b: =-(x^2+4x+y^2-2y)

=-(x^2+4x+4+y^2-2y+1-5)

=-(x+2)^2-(y-1)^2+5<=5

Dấu = xảy ra khi x=-2 và y=1