`M=-9x^2+6x-3`

`M=-(9x^2-6x+3)`

`M=-(9x^2-6x+1+2)`

`M=-(3x-1)^2-2`

Vì `-(3x-1)^2 <= 0 AA x`

`<=>-(3x-1)^2-2 <= -2 AA x`

  Hay `M <= -2 AA x`

Dấu "`=`" xảy ra `<=>(3x-1)^2=0<=>3x-1=0<=>x=1/3`

Vậy `GTLN` của `M` là `-2` khi `x=1/3`

\(M=-9x^2+6x-3\)

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

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

\(M=-\left(3x-1\right)^2-2\)

\(\Rightarrow Max_M=-2\) khi \(3x-1=0\)

                                 \(\Leftrightarrow x=\dfrac{1}{3}\)

\(A=\frac{5x^2+4x-1}{x^2}=\frac{9x^2-\left(4x^2-4x+1\right)}{x^2}=9-\frac{\left(2x-1\right)^2}{x^2}\le9\)

Dấu \(=\)khi \(2x-1=0\Leftrightarrow x=\frac{1}{2}\).

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

Dấu \(=\)khi \(x+2=0\Leftrightarrow x=-2\).

a) \(A=6x-x^2-11=-\left(x^2-6x+9\right)-2=-\left(x-3\right)^2-2\le-2\)

Dấu \(=\)khi \(x-3=0\Leftrightarrow x=3\).

b) \(B=x^2-5x-2=x^2-2.\frac{5}{2}x+\left(\frac{5}{2}\right)^2-\frac{33}{4}=\left(x-\frac{5}{2}\right)^2-\frac{33}{4}\ge-\frac{33}{44}\)

Dấu \(=\)khi \(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\).

khó hiểu quá 

bn giải giúp mình đi

a) \(6x-x^2-11\)

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

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

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

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

Mà: \(\left(x-3\right)^2\ge0\)

\(\Rightarrow-\left(x-3\right)^2\le0\)

\(\Rightarrow-\left(x-3\right)^2-2\le0-2\)

\(\Rightarrow A\le-2\)

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

Vậy giá trị lớn nhất của biểu thức \(6x-x^2-11=-2\) khi \(x=3\)

b) \(x^2-5x-2\)

\(=\left(x^2-2.\frac{5}{2}x+\frac{25}{4}\right)-\frac{33}{4}\)

\(=\left(x-\frac{5}{2}\right)^2-\frac{33}{4}\)

Mà: \(\left(x-\frac{5}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-\frac{5}{2}\right)^2-\frac{33}{4}\ge\frac{-33}{4}\forall x\)

Dấu '' = '' xảy ra khi: \(x-\frac{5}{2}=0\Rightarrow x=\frac{5}{2}\)

Vậy giá trị nhỏ nhất của biểu thức \(x^2-5x-2=\frac{-33}{4}\)  khi \(x=\frac{5}{2}\)

2) \(P=\frac{4}{2x^2+2xy+y^2+5x+20}=\frac{4}{\left(x^2+2xy+y^2\right)+\left(x^2+5x+\frac{25}{4}\right)+\frac{75}{4}}\)

\(=\frac{4}{\left(x+y\right)^2+\left(x+\frac{5}{2}\right)^2+\frac{75}{4}}\)

Để P đạt GTLN 

=> Mẫu thức đạt GTNN

mà \(\left(x+y\right)^2+\left(x+\frac{5}{2}\right)^2+\frac{75}{4}\ge\frac{75}{4}\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+y=0\\x+\frac{5}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{5}{2}\\y=\frac{5}{2}\end{cases}}\)

Thay x = -5/2 và y = 5/2 vào P 

Khi đó P = \(\frac{4}{\left(-\frac{5}{2}+\frac{5}{2}\right)^2+\left(-\frac{5}{2}+\frac{5}{2}\right)^2+\frac{75}{4}}=\frac{4}{\frac{75}{4}}=\frac{16}{75}\)

Vậy Max P = 16/75 <=> x = -5/2 ; y = 5/2

1) Ta có P = x² + 2xy + 3y² + 5y + 10

= (x² + 2xy + y²) + (2y² + 5y + 10) 

= \(\left(x+y\right)^2+2\left(y^2+\frac{5}{2}y+5\right)=\left(x+y\right)^2+2\left(y^2+\frac{5}{2}y+\frac{25}{16}+\frac{55}{16}\right)\)

= \(\left(x+y\right)^2+2\left(y+\frac{5}{4}\right)^2+\frac{55}{8}\ge\frac{55}{8}\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+y=0\\y+\frac{5}{4}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{4}\\y=-\frac{5}{4}\end{cases}}\)

Vạy Min P = 55/8 <=> x = 5/4 ; y = -5/4 

Ta có: \(B=-\left(2x^2-5x+8\right)\)

 \(\Rightarrow B=-\left[2x^2-2.2x.\frac{5}{4}+\left(\frac{5}{4}\right)^2\right]+\frac{27}{4}\)

\(\Rightarrow B=-\left(2x-\frac{5}{4}\right)^2+\frac{27}{4}\)

\(\Rightarrow B=27-\left(2x-\frac{5}{4}\right)^2\)

Vì \(\left(2x-\frac{5}{4}\right)^2\ge0\Rightarrow B\le\frac{27}{4}\)

Dấu "=" xảy ra khi \(2x-\frac{5}{4}=0\Rightarrow x=\frac{5}{8}\)

Vậy B_max=\(\frac{27}{4}\) khi \(x=\frac{5}{8}\)

-B = 2x^2 - 5x + 8 = 2.(x^2 - 5/2 x + 25/16 ) + 39/8 = 2.(x-5/4)^2 + 39/8 >= 39/8

=> B <= -39/8

Dấu "=" xảy ra <=> x-5/4 = 0 <=> x=5/4

Vậy Max B = -39/8 <=> x=5/4

tham khảo

Dấu "=" xảy ra khi 

k có j ngoài 

"tham khảo

Dấu "=" xảy ra khi 

bạn ạ

:")

1/

a, \(A=4x^2-4x+5=4x^2-4x+1+4=\left(2x-1\right)^2+4\ge4\)

Dấu "=" xảy ra khi x=1/2

Vậy Amin=4 khi x=1/2

b, \(B=3x^2+6x-1=3\left(x^2+2x+1\right)-4=3\left(x+1\right)^2-4\ge-4\)

Dấu "=" xảy ra khi x=-1

Vậy Bmin = -4 khi x=-1

2/

a, \(A=10+6x-x^2=-\left(x^2-6x+9\right)+19=-\left(x-3\right)^2+19\le19\)

Dấu "=" xảy ra khi x=3

Vậy Amax = 19 khi x=3

b, \(B=7-5x-2x^2=-2\left(x^2-\frac{5}{2}x+\frac{25}{16}\right)+\frac{31}{8}=-2\left(x-\frac{5}{4}\right)^2+\frac{31}{8}\le\frac{31}{8}\)

Dấu "=" xảy ra khi x=5/4

Vậy Bmax = 31/8 khi x=5/4