a) Ta có: \(A=x^2-5x+7\)

\(=x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}+\dfrac{3}{4}\)

\(=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{2}\)

b) Ta có: \(B=2x^2-8x+15\)

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

\(=2\left(x^2-4x+4+\dfrac{7}{2}\right)\)

\(=2\left(x-2\right)^2+7\ge7\forall x\)

Dấu '=' xảy ra khi x=2

a. `A=x^2-5x+7`

`=x^2-2.x. 5/2 + (5/2)^2 +3/4`

`=(x-5/2)^2 + 3/4`

`=> A_(min) =3/4 <=> x-5/2 =0<=>x=5/2`

b) `B=2x^2-8x+15`

`=[(\sqrt2x)^2 -2.\sqrt2 x . 2\sqrt2 +(2\sqrt2)^2] +7`

`=(\sqrt2x-2\sqrt2)^2+7`

`=> B_(min)=7 <=> x=2`.

a) \(A=x^2-5x+7\)

\(=x^2-2.\dfrac{5}{2}x+\left(\dfrac{5}{2}\right)^2+\dfrac{3}{4}\)

\(=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\)

Mặt khác, ta có \(\left(x-\dfrac{5}{2}\right)^2\ge0\forall x\)  \(\Rightarrow\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)

Dấu "=" xảy ra khi \(\left(x-\dfrac{5}{2}\right)^2=0\Leftrightarrow x-\dfrac{5}{2}=0\Leftrightarrow x=\dfrac{5}{2}\) 

Vậy \(A_{min}=\dfrac{3}{4}\) khi \(x=\dfrac{5}{2}\)

b) \(B=2x^2-8x+15\)

\(=4x^2-2.2x.2+2^2+11\)

\(=\left(2x-2\right)^2+11\)

Vì \(\left(2x-2\right)^2\ge0\forall x\) nên \(\left(2x-2\right)^2+11\ge11\forall x\)

Dấu "=" xảy ra khi  \(\left(2x-2\right)^2=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)

Vậy \(B_{min}=11\) khi \(x=1\)

1) \(A=x^2+8x+15=\left(x^2+8x+16\right)-1=\left(x+4\right)^2-1\ge-1\)

\(minA=-1\Leftrightarrow x=-4\)

2) \(B=7x-x^2-5=-\left(x^2-7x+\dfrac{49}{4}\right)+\dfrac{29}{4}=-\left(x-\dfrac{7}{2}\right)^2+\dfrac{29}{4}\le\dfrac{29}{4}\)

\(maxB=\dfrac{29}{4}\Leftrightarrow x=\dfrac{7}{2}\)

Ta có: \(A=x^2+8x+15\)

\(=x^2+8x+16-1\)

\(=\left(x+4\right)^2-1\ge-1\forall x\)

Dấu '=' xảy ra khi x=-4

Lớp 8 nhé, mình chọn nhầm

a) = 3(x²-2x+1) +1-3

GTNN = -2

B) tt

I zì:vv

a) Ta có: \(A=4x^2+4x+11=4x^2+4x+1=10=\left(2x+1\right)^2+10\ge10\forall x\)

Vậy Min_A=10 khi \(x=-\dfrac{1}{2}\)

b) Ta có: \(B=5-8x-x^2=-\left(x^2+8x-5\right)=-\left(x^2+8x+16-21\right)\)

\(=-\left(x+4\right)^2+21\le21\forall x\)

Vậy Max_B=21 khi x=-4

A= 5-8x-x²

=-x²-8x+21-16

=21-(x²+8x+16)

=21-(x+4)²\(\ge\)21-0=21

Dấu = khi x=-4

Vậy Amax=21 khi x=-4

B= x²+x+1

\(=x^2+\frac{x}{2}+\frac{x}{2}+\frac{1}{4}+\frac{3}{4}\)

\(=x\left(x+\frac{1}{2}\right)+\frac{1}{2}\left(x+\frac{1}{2}\right)+\frac{3}{4}\)

\(=\left(x+\frac{1}{2}\right)\left(x+\frac{1}{2}\right)+\frac{3}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge0+\frac{3}{4}=\frac{3}{4}\)

Dấu = khi x=-1/2

Vậy Bmin=3/4 khi x=-1/2

bạn xem lại đề bài 1 là GTNN hay GTLN nha

Toán lớp 1 cái gì,xạo.Toán trung học thì có.

Lớp 1 mà làm được cái này thì...THIÊN TÀI