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

\(=-\left(x-2\right)^2+9\le9\)Dấu ''='' xày ra khi x = 2

Vậy GTLN của A bằng 9 tại x = 2

b, \(B=-x^2-4y^2+2x-4y+3\)

\(=-\left(x^2-2x+1-1\right)-\left(4y^2+4y+1-1\right)+3\)

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

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

Dấu ''='' xảy ra khi x = 1 ; y = -1/2

Vậy GTNN của B bằng 4 tại x = 1 ; y = -1/2 

\(\left(\frac{x^2-3x}{x^2-9}-1\right):\left(\frac{9-x^2}{x^2+x-6}-\frac{x-3}{2-x}-\frac{x-2}{x+3}\right)\)

\(=\left(\frac{x^2-3x}{x^2-9}-1\right):\left(\frac{9-x^2}{x^2+x-6}+\frac{x-3}{x-2}-\frac{x-2}{x+3}\right)\)

\(=\left(\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-1\right):\left(\frac{\left(3-x\right)\left(3+x\right)}{x^2+x-6}+\frac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}-\frac{\left(x-2\right)^2}{\left(x+3\right)\left(x-2\right)}\right)\)

\(=\left(\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-1\right):\left(\frac{\left(3-x\right)\left(3+x\right)}{x^2+x-6}+\frac{\left(x-3\right)\left(x+3\right)-\left(x-2\right)^2}{x^2+x-6}\right)\)

\(=\left(\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{9-x^2+x^2-9-x^2+4x-4}{x^2+x-6}\)

\(=\frac{x\left(x-3\right)-\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}:\frac{-\left(x-2\right)^2}{x^2+x-6}\)

\(=\frac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}:\frac{-\left(x-2\right)^2}{x^2+x-6}\)

\(=\frac{3}{x+3}.\frac{x^2+x-6}{-\left(x-2\right)^2}\)

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

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

HT

a)

\(C=x^2+x-2\)

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

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

Mà\(\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\)

Vậy \(C_{Min}=-\frac{9}{4}\)khi và chỉ khi\(\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x=-\frac{1}{2}\)

b)

\(D=x^2+y^2+x-6y+5\)

\(=x^2+2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+y^2-2.y.3+3^2+5-\left(\frac{1}{2}\right)^2-3^2\)

\(=\left(x+\frac{1}{2}\right)^2+\left(y-3\right)^2-\frac{17}{4}\)

Mà\(\left(x+\frac{1}{2}\right)^2+\left(y-3\right)^2\ge0\Rightarrow\left(x+\frac{1}{2}\right)^2+\left(y-3\right)^2-\frac{17}{4}\ge-\frac{17}{4}\)

Vậy \(D_{Min}=-\frac{17}{4}\)khi và chỉ khi \(\hept{\begin{cases}\left(x+\frac{1}{2}\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=3\end{cases}}\)

c)

\(E=x^2+10y^2-6xy-10y+26\)

\(=x^2-2.x.3y+\left(3y\right)^2+y^2-2.y.5+5^2+26-5^2\)

\(=\left(x-3y\right)^2+\left(y-5\right)^2+1\)

Mà\(\left(x-3y\right)^2+\left(y-5\right)^2\ge0\Rightarrow\left(x-3y\right)^2+\left(y-5\right)^2+1\ge1\)

Vậy \(E_{Min}=1\)khi và chỉ khi\(\hept{\begin{cases}\left(x-3y\right)^2=0\\\left(y-5\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=15\\y=5\end{cases}}}\)

a, \(E=4x^2+6x+5=4\left(x^2+\frac{2.3}{4}x+\frac{9}{16}-\frac{9}{16}\right)+5\)

\(=4\left(x+\frac{3}{4}\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\forall x\)

Vậy ta có đpcm 

b, \(F=2x^2-3x+7=2\left(x^2-\frac{2.3}{4}x+\frac{9}{16}-\frac{9}{16}\right)+7\)

\(=2\left(x-\frac{3}{4}\right)^2+\frac{47}{8}\ge\frac{47}{8}>0\forall x\)

Vậy ta có đpcm 

c, \(K=5x^2-4x+1=5\left(x^2-\frac{2.2}{5}x+\frac{4}{25}-\frac{4}{25}\right)+1\)

\(=5\left(x-\frac{2}{5}\right)^2+\frac{1}{5}\ge\frac{1}{5}>0\forall x\)

Vậy ta có đpcm 

d, \(Q=3x^2+2x+5=3\left(x^2+\frac{2}{3}x+\frac{1}{9}-\frac{1}{9}\right)+5\)

\(=3\left(x+\frac{1}{3}\right)^2+\frac{14}{3}\ge\frac{14}{3}>0\forall x\)

Vậy ta có đpcm 

= 5 

hehe, easy game :)) (ko biết đúng hay sai đâu, mk sủa thế cho vui)

Hok tốt

đáp án :

5

hk tốt 

Áp dụng BĐT Cauchy Schwarz dạng Engel 

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\left(1+1+1\right)^2}{a+b+c}=\frac{9}{a+b+c}\)

Dấu ''='' xảy ra khi a = b = c