a, \(A=-x^2-2x+3=-\left(x^2+2x-3\right)=-\left(x^2+2x+1-4\right)\)

\(=-\left(x+1\right)^2+4\le4\)

Dấu ''='' xảy ra khi x = -1 

Vậy GTLN là 4 khi x = -1 

b, \(B=-4x^2+4x-3=-\left(4x^2-4x+3\right)=-\left(4x^2-4x+1+2\right)\)

\(=-\left(2x-1\right)^2-2\le-2\)

Dấu ''='' xảy ra khi x = 1/2 

Vậy GTLN B là -2 khi x = 1/2 

c, \(C=-x^2+6x-15=-\left(x^2-2x+15\right)=-\left(x^2-2x+1+14\right)\)

\(=-\left(x-1\right)^2-14\le-14\)

Vâỵ GTLN C là -14 khi x = 1

Bài 8 : 

b, \(B=x^2-6x+11=x^2-6x+9+2=\left(x-3\right)^2+2\ge2\)

Dấu ''='' xảy ra khi x = 3

Vậy GTNN B là 2 khi x = 3 

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

Dấu ''='' xảy ra khi x = 1/2 

Vậy ...

c, \(x^2-12x+2=x^2-12x+36-34=\left(x-6\right)^2-34\ge-34\)

Dấu ''='' xảy ra khi x = 6

Vậy ...

\(B1,a,A=x^2-6x+11\)

               \(=\left(x^2-6x+9\right)+2\)

                \(=\left(x-3\right)^2+2\ge2\)

Dấu "=" <=> x=3

Vậy ..........

\(b,B=x^2-20x+101\)

        \(=\left(x^2-20x+100\right)+1\)

         \(=\left(x-10\right)^2+1\ge1\)

Dấu "=" <=> x = 10

Vậy .

\(2,a,A=4x-x^2+3\)

            \(=7-\left(x^2-4x+4\right)\)'

             \(=7-\left(x-2\right)^2\le7\)

Dấu ''='' <=> x = 2

Vậy .

\(b,B=-x^2+6x-11\)

       \(=-2-\left(x^2-6x+9\right)\)

        \(=-2-\left(x-3\right)^2\le-2\)

Dấu ""=" <=> x = 3

Vậy..

ko biét

B = -x² + 6x - 11

B = -x² + 6x - 9 - 2

B= - ( x² - 6x + 9 ) - 2

B = - ( x - 3 )² - 2 <= - 2

Dấu "=" xra khi x - 3 = 0 <=> x = 3

Vậy B max = -2 khi x = 3

\(B=-x^2+6x-11=-\left(x^2-6x+11\right).\)

\(=-\left(x^2-6x+9+2\right)=\left[\left(x-3\right)^2+2\right]\)

\(=-\left(x-3\right)^2-2\le-2\)

Để B lớn nhất \(\Rightarrow-\left(x-3\right)^2-2=-2\Rightarrow x=3\)

\(\Rightarrow MAX=-2\) tại x = 3 

\(A=x^2-6x+11\)

\(=x^2-2x.3+3^2+2\)

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

\(\Rightarrow A\ge2\)

\(\Rightarrow MinA=2\)

\(Khi\)\(\left(x-3\right)^2=0\Rightarrow x-3=0\Rightarrow x=3\)

Chúc bn học giỏi nhoa!!!

ban kia lam dung roi do

k tui nha

thanks

A = x² - 2.x.3 + 9 - 9 + 11

A = (x - 3)² + 2

Ta có:

(x -3)² \(\ge\) 0 \(\forall x\in R\)

=> (x - 3)² + 2 \(\ge\)2 \(\forall x\in R\)

=> GTLN x² - 6x + 11 là 2 khi và chỉ khi x = 3

\(x^2-6x+11=\left(x^2-2.3x+3^2\right)+2=\left(x+3\right)^2+2\ge2\)

Vậy Max(A)=2

Dấu "=" xảy ra khi và chỉ khi x=-3

\(A=\left(x^2-4x+4\right)+4=\left(x-2\right)^2+4\ge4\)

\(minA=4\Leftrightarrow x=2\)

\(B=\left(4x^2-12x+9\right)+2=\left(2x-3\right)^2+2\ge2\)

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

\(C=3\left(x^2+2x+1\right)-8=3\left(x+1\right)^2-8\ge-8\)

\(minC=-8\Leftrightarrow x=-1\)

\(D=-\left(x^2-2x+1\right)-4=-\left(x-1\right)^2-4\le-4\)

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

\(E=-\left(4x^2-6x+\dfrac{9}{4}\right)-\dfrac{11}{4}=-\left(2x-\dfrac{3}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\)

\(maxA=-\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{4}\)

\(F=-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{55}{8}=-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{55}{8}\le-\dfrac{55}{8}\)

\(maxF=-\dfrac{55}{8}\Leftrightarrow x=\dfrac{1}{4}\)

\(G=\left(x^2-4xy+4y^2\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-2y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(maxG=\dfrac{3}{4}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-\dfrac{1}{2}\end{matrix}\right.\)

\(H=-\left(x^2-2x+1\right)-\left(y^2+4y+4\right)+16=-\left(x-1\right)^2-\left(y+2\right)^2+16\le16\)

\(maxH=16\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

\(\frac{16}{x^4-6x^2+11}=\frac{16}{\left(x^4-6x^2+9\right)+2}\)

\(=\frac{16}{\left(x^2-3\right)^2+2}\le\frac{16}{2}=8\)

Vậy GTLN là 8 đạt được khi 

\(x^2=3\Leftrightarrow\orbr{\begin{cases}x=-\sqrt{3}\\x=\sqrt{3}\end{cases}}\)

a: \(A=\dfrac{x^4+x^2+11x^2+11}{x^4+x^2+5x^2+5}=\dfrac{\left(x^2+11\right)\left(x^2+1\right)}{\left(x^2+5\right)\left(x^2+1\right)}=\dfrac{x^2+11}{x^2+5}\)

b: \(A=\dfrac{x^2+5+6}{x^2+5}=1+\dfrac{6}{x^2+5}< =1+\dfrac{6}{5}=\dfrac{11}{5}\)

Dấu = xảy ra khi x=0

1. Câu hỏi của Quỳnh Như - Toán lớp 8 - Học toán với OnlineMath

Em tham khảo câu 1 tại link này.

Em cảm ơn cô nhiều

\(a.A=x^2-6x+11\)

     \(=x^2-2.x.3+3^2+2\)

     \(=\left(x-3\right)^2+2\)

      \(\Rightarrow\left(x-3\right)^2\ge0\)

      \(\Rightarrow\left(x-3\right)^2+2\ge2\)

      \(\Rightarrow A\ge2\)

\(A_{min}\Leftrightarrow A=2\)

Dấu "=" xảy ra: \(\left(x-3\right)^2=0\)

                      \(x-3=0\)

                      \(x=3\)

Vậy \(A\)nhỏ nhất khi \(x=3\)