\(A=-x^2+4x-4-5=-\left(x-2\right)^2-5\)

Vì  \(-\left(x-2\right)^2\le0\Rightarrow-\left(x-2\right)^2-5\le-5\)

Vậy  GTLN  của  A  là  \(-5\)

1)

a)  \(M=\)\(x^2\)\(+\)\(4x\)\(+\)\(9\)

\(=\)\(x^2\)\(+\)\(2x\)\(.\)\(2\)\(+\)\(4\)\(+\)\(5\)

\(=\left(x+2\right)^2\)\(+\)\(5\)\(>;=\)\(5\)

Dấu bằng xảy ra khi x + 2 = 0

                               x      = -2

Vậy GTNN của M bằng 5 khi x = -2

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

\(=\)\(x^2\)\(-\)\(2x\)\(.\)\(10\)\(+\)\(100\)\(+\)\(1\)

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

Dấu bằng xảy ra khi x - 10 = 0

                              x        =   10

Vậy GTNN của N bằng 1 khi x = 10

2)

a)  \(C=\)\(-y^2\)\(+\)\(6y\)\(-\)\(15\)

\(=\)\(-y^2\)\(+\)\(2y\)\(.\)\(3\)\(-\)\(9\)\(-\)\(6\)

\(=\)\(-\left(y-3\right)^2\)\(-\)\(6\)\(< ;=\)\(6\)

Dấu bằng xảy ra khi y - 3 = 0

                               y      = 3

Vậy GTLN của C bằng -6 khi y = 3

b)  \(B=\)\(-x^2\)\(+\)\(9x\)\(-\)\(12\)

\(=\)\(-x^2\)\(+\)\(2x\)\(.\)\(\frac{9}{2}\)\(-\)\(\frac{81}{4}\)\(+\)\(\frac{81}{4}\)\(-\)\(12\)

\(=\)\(-\left(x-\frac{9}{2}\right)^2\)\(+\)\(\frac{33}{4}\)\(< ;=\)\(\frac{33}{4}\)

Dấu bằng xảy ra khi  \(x-\frac{9}{2}=0\)

                                \(x=\frac{9}{2}\)

Vậy GTLN của B bằng  \(\frac{33}{4}\)khi x =  \(\frac{9}{2}\)

a) M = x² + 4x + 9 = x² + 4x + 4 + 5 = (x + 2)² + 5 

Vì : \(\left(x+2\right)^2\ge0\forall x\in R\) 

Nên M = (x + 2)² + 5 \(\ge5\forall x\in R\)

Vậy M_min = 5 khi x = -2

b) N = x² - 20x + 101 = x² - 20x + 100 + 1 = (x - 10)² + 1 

Vì \(\left(x-10\right)^2\ge0\forall x\in R\)

Nên : N = (x - 10)² + 1 \(\ge1\forall x\in R\)

Vậy N_min = 1 khi x = 10

Bài 2 : 

a) C = -y² + 6y - 15 = -(y² - 6y + 15) = -(y² - 6y + 9 + 6) = -(y² - 6y + 9) - 6 = -(y - 3)² - 6

Vì \(-\left(y-3\right)^2\le0\forall x\in R\)

 Nên : C = -(y - 3)² - 6 \(\le-6\forall x\in R\)

Vậy C_min = -6 khi y = 3 

b) B = -x² + 9x - 12 = -(x² - 9x + 12) = -(x² - 9x +  \(\frac{81}{4}-\frac{33}{4}\)) = \(-\left(x-\frac{9}{2}\right)^2+\frac{33}{4}\)

Vì \(-\left(x-\frac{9}{2}\right)^2\le0\forall x\in R\)

Nên :  B = \(-\left(x-\frac{9}{2}\right)^2+\frac{33}{4}\) \(\le\frac{33}{4}\forall x\in R\)

Vậy B_min = \(\frac{33}{4}\) khi \(x=\frac{9}{2}\)

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

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 \(x+2=0\Leftrightarrow x=-2\).

\(A=\frac{1}{x^2-4x+9}=\frac{1}{x^2-4x+4+5}=\frac{1}{\left(x-2\right)^2+5}\le\frac{1}{5}\forall x\in R\)

Vậy GTLN của A bằng 1/5 khi x = 2.

Để \(\frac{1}{x^2-4x+9}\)đạt GTLN

\(\Leftrightarrow x^2-4x+9\)đạt giá trị nhỏ nhất

Ta có : \(x^2-4x+9\)

\(=\left(x-2\right)^2+5\ge5\)

Dấu " = " xảy ra : \(\Leftrightarrow x=2\)

Vậy \(max_A=\frac{1}{5}\Leftrightarrow x=2\)

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

\(\Leftrightarrow A=\frac{1}{\left(x-2\right)^2+5}\le\frac{1}{5}\)

\(maxA=\frac{1}{5}\)dấu " = " xảy ra khi \(x=2\)

Vậy : ................

\(\left\{{}\begin{matrix}4x^2+9y^2=9\\A=x-2y+3\end{matrix}\right.\)

Áp dụng bất đẳng thức Bunhiacopxki cho các cặp số \(\left(\dfrac{1}{2};2x\right);\left(-\dfrac{2}{3};3y\right)\)

\(x-2y=\dfrac{1}{2}.x+\left(-\dfrac{2}{3}\right).3y\)

\(\Rightarrow\left[\dfrac{1}{2}.2x+\left(-\dfrac{2}{3}\right).3y\right]^2\le\left(\dfrac{1}{4}+\dfrac{4}{9}\right)\left(4x^2+9y^2\right)=\dfrac{25}{36}.9\)

\(\Rightarrow x-2y\le\dfrac{5}{6}.3=\dfrac{5}{2}\)

\(\Rightarrow A=x-2y+3\le\dfrac{5}{2}+3\)

\(\Rightarrow A=x-2y+3\le\dfrac{11}{2}\)

Dấu "=" xảy ra khi và chỉ khi

\(\dfrac{\dfrac{1}{2}}{2x}=\dfrac{-\dfrac{2}{3}}{3y}\)

\(\Rightarrow\dfrac{2x}{\dfrac{1}{2}}=\dfrac{3y}{-\dfrac{2}{3}}\)

\(\Rightarrow\dfrac{4x^2}{\dfrac{1}{4}}=\dfrac{9y^2}{\dfrac{4}{9}}=\dfrac{4x^2+9y^2}{\dfrac{1}{4}+\dfrac{4}{9}}=\dfrac{9}{\dfrac{25}{36}}=\dfrac{9.36}{25}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{9.36}{25}.\dfrac{1}{16}\\y^2=\dfrac{9.36}{25}.\dfrac{4}{36}=\dfrac{9.4}{25}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3.6}{5}.\dfrac{1}{4}=\dfrac{9}{10}\\y=\dfrac{3.2}{5}=\dfrac{6}{5}\end{matrix}\right.\)

Vậy \(GTLN\left(A\right)=\dfrac{11}{2}\left(tạix=\dfrac{9}{10};y=\dfrac{6}{5}\right)\)

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