\(B=9x-3x^2\)

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

\(=-3\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}\right)\)

\(=-3\left(x-\frac{3}{2}\right)^2+\frac{27}{4}\)

Vì \(-3\left(x-\frac{3}{2}\right)^2\ge0;\forall x\)

\(\Rightarrow-3\left(x-\frac{3}{2}\right)^2+\frac{27}{4}\ge0+\frac{27}{4};\forall x\)

V đang làm bài thì lỡ tay nhấn gửi làm tiếp nhé

Hay\(B\ge\frac{27}{4};\forall x\)

Dấu"="xảy ra \(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=0\)

                       \(\Leftrightarrow x=\frac{3}{2}\)

Vậy \(B_{min}=\frac{27}{4}\Leftrightarrow x=\frac{3}{2}\)

Tìm max chứ

\(B=9x-3x^2=-3\left(x^2-3x\right)\)

\(=-3\left(x^2-3x+\frac{9}{4}-\frac{9}{4}\right)\)

\(=-3\left[\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\right]\)

\(=-3\left[\left(x-\frac{3}{2}\right)^2\right]+\frac{27}{4}\le\frac{27}{4}\)

\(A=\frac{3\left(x^2+x+1\right)+6x}{x^2+x+1}=3+\frac{6x}{x^2+x+1};\left(x-1\right)^2\ge0< =>x^2+x+1\ge3x;\)

=> \(A\le3+\frac{6x}{3x}=5\). Max A =5 khi x=1

\(B=\frac{7\left(x^2+x+2\right)+7-7x}{x^2+x+2}=7-\frac{7\left(x-1\right)}{x^2+x+2};\)\(\left(x-3\right)^2\ge0< =>x^2+x+2\ge7\left(x-1\right)\)

=> \(B\ge7-\frac{7\left(x-1\right)}{7\left(x-1\right)}=6\)MinB = 6 khi x =3

\(A=x^2-6x+10=\left(x-3\right)^2+1\ge1\)

\(\Rightarrow A_{min}=1\Leftrightarrow x=3\)

\(B=4x^2-4x+25=\left(2x-1\right)^2+24\ge24\)

\(\Rightarrow B_{min}=24\Leftrightarrow x=\frac{1}{2}\)

\(C=3x^2+9x+12=3\left(x+\frac{3}{2}\right)^2+\frac{21}{4}\ge\frac{21}{4}\)

\(\Rightarrow C_{min}=\frac{21}{4}\Leftrightarrow x=\frac{-3}{2}\)

\(D=9x^2+3x+\frac{1}{x}+1420=9x^2-6x+1+9x+\frac{1}{x}+1419\)

\(D=\left(3x-1\right)^2+9x+\frac{1}{x}+1419\)

Áp dụng BĐT cauchy :\(9x+\frac{1}{x}\ge2\sqrt{9x.\frac{1}{x}}=6\)

\(\Rightarrow D\ge\left(3x-1\right)^2+1419+6\ge1425\)

dấu = xảy ra khi \(\left\{\begin{matrix}x=\frac{1}{3}\\9x=\frac{1}{x}\end{matrix}\right.\Leftrightarrow x=\frac{1}{3}}\)

min=1425 khi x=1/3 dg ban ko giai dc sr nhe :)

a/ x² + 3x + 1

\(=x^2+2.\frac{3}{2}.x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+1\)

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

       Vậy MinA = -5/4 khi x + 3/2 = 0 => x = -3/2

b/ 9x² + 3x + 1

\(=\left(3x\right)^2+2.3x.\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+1\)

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

         Vậy MinB = 3/4 khi 3x + 1/2 = 0 => 3x = -1/2 => x = -1/6

c/ -x² + 2x - 1 = -(x² - 2x + 1) = -(x - 1)² \(\le0\)

          Vậy MaxC = 0 khi x - 1 = 0 => x = 1

a.\(=x^2+2.\frac{3}{2}x+\frac{9}{4}-\frac{5}{4}=\left(x+\frac{3}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)

dấu = xảy ra khi x=-3/2

b,c tt

mai ktra rồi mk cần gấp lắm

1 biểu thức làm lun cả Min và Max lun ak?

(nếu có)