Thực hiện phép chia đa thức: 

\(x^4-3x^3-7x^2+ax+b=\left(x^2-2x+6\right)\left(x^2-x-15\right)+\left(a-24\right)x+\left(b+90\right)\)

Khi đó: \(\hept{\begin{cases}a-24=3\\b+90=2\end{cases}}\Leftrightarrow\hept{\begin{cases}a=27\\b=-88\end{cases}}\)

\(A=x\left(x+2\right)\left(x+4\right)\left(x+6\right)+8\)

\(=\left(x^2+6x\right)\left(x^2+6x+8\right)+8\)

\(=\left(x^2+6x+4\right)^2-4^2+8\)

\(=\left(x^2+6x+4\right)^2-8\ge-8\)

Dấu \(=\)khi \(x^2+6x+4=0\Leftrightarrow x=-3\pm\sqrt{5}\).

\(B=5+\left(1-x\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=5-\left[\left(x-1\right)\left(x+6\right)\right].\left[\left(x+2\right)\left(x+3\right)\right]\)

\(=5-\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=5-\left(x^2+5x\right)^2+6^2\)

\(=41-\left(x^2+5x\right)^2\le41\)

Dấu \(=\)khi \(x^2+5x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

\(C=\left(x+3\right)^4+\left(x-7\right)^4=\left[\left(x-2\right)+5\right]^4+\left[\left(x-2\right)-5\right]^4\)

\(=2\left(x-2\right)^4+300\left(x-2\right)^2+1250\ge1250\)

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

\(8x^3-\frac{1}{27}\)

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

\(\left(2x-\frac{1}{3}\right)\left(4x^2+\frac{2}{3}x+\frac{1}{9}\right)\)

a) \(A=x^2+3x+3\)

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

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

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

\(\Leftrightarrow x+\frac{3}{2}=0\)

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

b)\(B=x^2+4x+9\)

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

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

Dấu "=" xảy ra \(\Leftrightarrow\left(x+2\right)^2=0\)

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)

c)\(C=x+1-x^2\)

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

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

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

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

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

d)\(D=-4x^2+4x+1\)

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

\(=-\left(2x-1\right)^2+2\le2\)

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

\(\Leftrightarrow2x-1=0\)

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

x² - 4y² + 9 - 6x 

= (x² - 6x + 9) - (2y)²

= (x - 3)² - (2y)² = (x + 2y - 3)(x - 2y - 3)

B= 7800 nha

x=3

a=2

\(A=x^2-6x-7=x^2-6x+9-16=\left(x-3\right)^2-16\ge-16\)

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

\(B=x^2+x-1=x^2+x+\frac{1}{4}-\frac{5}{4}=\left(x+\frac{1}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)

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

\(C=2x^2-5x-9=2\left(x^2-\frac{5}{2}x+\frac{25}{16}\right)-\frac{97}{8}=2\left(x-\frac{5}{4}\right)^2-\frac{97}{8}\ge-\frac{97}{8}\)

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

\(A=x^2+4x+8=x^2+4x+4+4=\left(x+2\right)^2+4\ge4>0\forall x\)

Vậy biểu thức A luôn dương 

\(=\left(x+2\right)^2+4\ge4\)Dấu ''='' xảy ra khi x = -2

Vậy GTNN A là 4 khi x =-2

\(B=4x^2+2x+1=\left(2x\right)^2+2.2x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)

\(=\left(2x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)

Vậy biểu thức B luôn dương 

\(=\left(2x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)Dấu ''='' xảy ra khi x = -1/4 

Vậy GTNN B là 3/4 khi x = 1/4