1. a. \(A=8a-8a^2+3=-8\left(a-\frac{1}{2}\right)^2+5\)

Vì \(\left(a-\frac{1}{2}\right)^2\ge0\forall a\)\(\Rightarrow-8\left(a-\frac{1}{2}\right)^2+5\le5\)

Dấu "=" xảy ra \(\Leftrightarrow-8\left(a-\frac{1}{2}\right)^2=0\Leftrightarrow a-\frac{1}{2}=0\Leftrightarrow a=\frac{1}{2}\)

Vậy Amax = 5 <=> a = 1/2

b. \(B=b-\frac{9b^2}{25}=-\frac{9}{25}\left(b-\frac{25}{18}\right)^2+\frac{25}{36}\)

Vì \(\left(b-\frac{25}{18}\right)^2\ge0\forall b\)\(\Rightarrow-\frac{9}{25}\left(b-\frac{25}{18}\right)^2+\frac{25}{36}\le\frac{25}{36}\)

Dấu "=" xảy ra \(\Leftrightarrow-\frac{9}{25}\left(b-\frac{25}{18}\right)^2=0\Leftrightarrow b-\frac{25}{18}=0\Leftrightarrow b=\frac{25}{18}\)

Vậy Bmax = 25/36 <=> b = 25/18

a,\(A=8a-8a^2+3\)

       \(=-8\left(a^2-a\right)+3\)

       \(=-8\left(a^2-2a\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\right)+3\)

       \(=-8\left[\left(a-\frac{1}{2}\right)^2-\frac{1}{4}\right]+3\)

       \(=-8\left(a-\frac{1}{2}\right)^2+2+3\)

       \(=-8\left(a-\frac{1}{2}\right)^2+5\le5\forall a\) 

Dấu"=" xảy ra khi \(\left(a-\frac{1}{2}\right)^2=0\Rightarrow a=\frac{1}{2}\)

Vậy \(Max_A=5\)khi\(a=\frac{1}{2}\)

bài 2:

b,\(D=d^2+10e^2-6de-10e+26\)

\(=d^2-23de+\left(3e\right)^2+e^2-2.5e+5^2+1\)

\(=\left(d-3e\right)^2+\left(e-5\right)^2+1\ge1\forall d,e\)

Dấu"=" xảy ra khi\(\orbr{\begin{cases}\left(d-3e\right)^2=0\\\left(e-5\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}d=15\\e=5\end{cases}}}\)

vậy \(D_{min}=1\)khi \(d=15;e=5\)

c,:\(E=4x^4+12x^2+11\)

\(=\left(2x^2\right)^2+2.2x^2.3+3^2+2\)

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

còn 1 đoạn nx bạn tự lm tiếp,lm giống như D

Ta có: A = 12x - 4x² + 3 = -(4x² - 12x - 3) = -4(x² - 3x + 9/4) + 12 = -4(x + 3/2)² + 12

Ta luôn có: -4(x + 3/2)² \(\le\) 0 \(\forall\)x

=> -4(x + 3/2)² + 12 \(\le\) 12 \(\forall\)x

hay A \(\le\) 12 \(\forall\)x

Dấu "=" xảy ra <=> (x + 3/2)² = 0 <=> x + 3/2 = 0 <=> x = -3/2

Vậy A_max = 12 tại x = -3/2

Ta có:

\(4x^2+12x+100=\left(2x+3\right)^2+91\)

\(\Rightarrow B=\frac{-9}{\left(2x+3\right)^2+91}\)

Vì \(\left(2x+3\right)^2\ge0;\forall x\)

\(\Rightarrow\left(2x+3\right)^2+91\ge0+91;\forall x\)

\(\Rightarrow\frac{9}{\left(2x+3\right)^2+91}\le\frac{9}{91};\forall x\)

\(\Rightarrow\frac{-9}{\left(2x+3\right)^2+91}\ge\frac{-9}{91};\forall x\)

Dấu '"=" xảy ra \(\Leftrightarrow2x+3=0\)

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

Vậy MIN \(B=\frac{-9}{91}\)\(\Leftrightarrow x=\frac{-3}{2}\)

TL:

\(B=\frac{-9}{\left(2x+6\right)^2+64}\) 

 Để B_min \(\Rightarrow\left(2x+6\right)^2+64\) nhỏ nhất

Mà \(\left(2x+6\right)^2+64\ge64\forall x\in R\) 

dấu "=" xảy ra <=> \(\left(2x+6\right)^2=0\Leftrightarrow2x+6=0\Leftrightarrow2x=-6\Leftrightarrow x=-3\) 

=>B_min =\(\frac{-9}{64}\) tại x=-3

Vậy.......

\(B=12x-8y-4x^2-y^2+1\)

\(=-\left(4x^2-12x+y^2+8y-1\right)\)

\(=-\left[\left(4x^2-12x+9\right)+\left(y^2+8y+16\right)-24\right]\)

\(=\left[\left(2x-3\right)^2+\left(y+4\right)^2-24\right]\)

\(=-\left(2x-3\right)^2-\left(y+4\right)^2+24\)

\(\Rightarrow B_{max}=24\Leftrightarrow-\left(2x-3\right)^2-\left(y+4\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}2x-3=0\\y+4=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{3}{2}\\y=-4\end{cases}}}\)

Ta có:  B = 12x - 8y - 4x² - y² + 1 = (-4x² + 12x - 9) - (y² + 8y + 16) + 26 = -4(x² - 3x + 9/4) - (y + 4)² + 26 = -4(x - 3/2)² - (y + 4)² + 26

Ta luôn có: -4(x - 3/2)² \(\le\) 0 \(\forall\) x (vì  4(x - 3/2)² \(\ge\)0 \(\forall\)x)

             -(y + 4)² \(\le\) 0 \(\forall\)y  (vì (y + 4)² \(\ge\)0 \(\forall\) y)

=> -4(x - 3/2)² - (y + 4)² + 26 \(\le\) 26 \(\forall\)x,y

hay B \(\le\) 26 \(\forall\)x, y

Dấu "=" xảy ra khi : \(\hept{\begin{cases}\left(x-\frac{3}{2}\right)^2=0\\\left(y+4\right)^2=0\end{cases}}\) <=> \(\hept{\begin{cases}x-\frac{3}{2}=0\\y+4=0\end{cases}}\) <=> \(\hept{\begin{cases}x=\frac{3}{2}\\y=-4\end{cases}}\)

Vậy B_max = 26 tại x = 3/2 và y = -4

\(1,a,A=x^2-6x+25\)

\(=x^2-2.x.3+9-9+25\)

\(=\left(x-3\right)^2+16\)

Ta có :

\(\left(x-3\right)^2\ge0\)Với mọi x

\(\Rightarrow\left(x-3\right)^2+16\ge16\)

Hay \(A\ge16\)

\(\Rightarrow A_{min}=16\)

\(\Leftrightarrow x=3\)

\(b,B=4x^2+4x-2\)

\(B=4x^2+4x+1-3\)

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

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

Ta có : 

\(\left(2x+1\right)^2\ge0\)với mọi x

\(\Rightarrow\left(2x+1\right)^2-3\ge-3\)

\(\Leftrightarrow B\ge-3\)

\(\Rightarrow B_{min}=-3\)

\(\Leftrightarrow x=-\frac{1}{2}\)

a) \(4x^2+12x+10=\left(2x+3\right)^2+1\ge1\)

Dấu "="\(\Leftrightarrow x=-2\)

b) \(B=\left(3x-1\right)^2+4\ge4\)

Dấu "="\(\Leftrightarrow x=\frac{1}{3}\)

a, \(A=4x^2+12x+10\)

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

Dấu"=" xảy ra<=> \(\left(2x+1\right)^2=0\)

                     \(\Leftrightarrow x=\frac{-1}{2}\)

\(b,B=9x^2-6x+5\)

      \(=\left(3x-1\right)^2+4\ge4\forall x\)

Dấu"=" xảy ra<=> \(\left(3x-1\right)^2=0\)

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

Mình đang cần gấp . Đảm bảo k trả đầy đủ + kb :'>

2.    \(Q=\left(x-3\right)\left(4x+5\right)+2019\)

        \(Q=4x^2+5x-12x-15+2019\)   

        \(Q=4x^2-7x+2004\)  

        \(Q=\left(2x\right)^2-2.2x.\frac{7}{4}+\frac{49}{16}+2019-\frac{49}{16}\) 

        \(Q=\left(2x-\frac{7}{4}\right)^2+\frac{32255}{16}\)  

        \(Do\) \(\left(2x-\frac{7}{4}\right)^2\ge0\forall x\) \(Nên\) \(\left(2x-\frac{7}{4}\right)^2+\frac{32255}{16}\ge\frac{32255}{16}\)  

        \(\Rightarrow Q\ge\frac{32255}{16}\) 

         \(Vậy\) \(MinQ=\frac{32255}{16}\Leftrightarrow x=\frac{7}{8}\)

3. \(T=4\left(a^3+b^3\right)-6\left(a^2+b^2\right)\)  

   \(T=4\left(a+b\right)\left(a^2-ab+b^2\right)-6a^2-6b^2\) 

   \(T=4\left(a^2-ab+b^2\right)-6a^2-6b^2\)  (do a+b=1)

   \(T=4a^2-4ab+4a^2-6a^2-6b^2\) 

   \(T=-2a^2-4ab-2b^2\)

   \(T=-2\left(a^2+2ab+b^2\right)\) 

   \(T=-2\left(a+b\right)^2\)

   \(T=-2.1^2=-2.1=-2\) (do a+b=1)

\(A=x^2-3x+5\)

\(=x^2-3x+\frac{9}{4}+\frac{11}{4}\)

\(=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)

\(\left(x-\frac{3}{2}\right)^2\ge0\Rightarrow A\ge\frac{11}{4}\)

Dấu "=" xảy ra khi \(x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)

Vậy Min A = \(\frac{11}{4}\Leftrightarrow x=\frac{3}{2}\)

a) \(A=x^2-3x+5\)

\("="\Leftrightarrow x=\frac{11}{4}\Rightarrow x=\frac{3}{2};\frac{11}{4}\)

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

\("="\Leftrightarrow x=5\Rightarrow x=0;5\)

c) \(C=4x-x^2+3\)

\("="\Leftrightarrow x=7\Rightarrow x=2;7\)

d) \(D=x^4+x^2+2\)

\("="\Leftrightarrow x=2\Rightarrow x=0;2\)

\(A=x^2+y^2+4x-6y+25\)

\(=x^2+4x+4+y^2-6y+9+12\)

\(=\left(x+2\right)^2+\left(y-3\right)^2+12\ge12\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}x+2=0\\ y-3=0\end{cases}\Rightarrow\begin{cases}x=-2\\ y=3\end{cases}\)

A=x2+y2+4x−6y+25

\(= x^{2} + 4 x + 4 + y^{2} - 6 y + 9 + 12\)

\(= \left(\left(\right. x + 2 \left.\right)\right)^{2} + \left(\left(\right. y - 3 \left.\right)\right)^{2} + 12 \geq 12 \forall x , y\)

Dấu '=' xảy ra khi \(\left{\right. x + 2 = 0 \\ y - 3 = 0 \Rightarrow \left{\right. x = - 2 \\ y = 3\)