\(1,=x^6+27\\ 2,=8x^3+1\\ 3,=x^6+8\\ 4,=27x^3+8\)

1. (x² + 3)(x⁴ - 3x² + 9)

= x⁶ + 27

2. (2x + 1)(4x² - 2x + 1)

= 8x³ + 1

3. (x² + 2)(x⁴ - 2x² + 4)

= x⁶ + 8

4. (3x + 2)(9x² - 6x + 4)

= 27x³ + 8

1: \(\left(x^2+3\right)\left(x^4-3x^2+9\right)=x^6+27\)

2: \(\left(2x+1\right)\left(4x^2-2x+1\right)=8x^3+1\)

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

4: \(\left(3x+2\right)\left(9x^2-6x+4\right)=27x^3+8\)

\(...\Rightarrow\left(x+3\right)\left(x+3\right)^2-\left(9x^3+6x^2+x\right)+\left(2x+1\right)\left(2x-1\right)^2=28\)

\(\Rightarrow\left(x+3\right)^3-9x^3-6x^2-x+\left(4x^2-1\right)\left(2x-1\right)^{ }=28\)

\(\Rightarrow\left(x+3\right)^3-9x^3-6x^2-x+\left(4x^2-1\right)\left(2x-1\right)^{ }=28\)

\(\Rightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3-4x^2-2x+1=28\)

\(\Rightarrow-x^2+24x+28=28\)

\(\Rightarrow x^2-24x=0\)

\(\Rightarrow x\left(x-24\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-24=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=24\end{matrix}\right.\)

a) \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\Rightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\Rightarrow\left(2x-3\right)\left(7x-2x+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-3=0\\5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

b) \(\left(2x-7\right).\left(x-2\right)\left(x^2-4\right)=0\Rightarrow\left(2x-7\right)\left(x-2\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}2x-7=0\\\left(x-2\right)^2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)

c)\(\left(9x^2-25\right)-\left(6x-10\right)=0\Rightarrow\left(3x-5\right)\left(3x+5\right)-2\left(3x-5\right)=0\Rightarrow\left(3x-5\right)\left(3x+5-2\right)=0\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=1\end{matrix}\right.\)

a: Ta có: \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\)

\(\Leftrightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(5x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)

b: Ta có: \(\left(2x-7\right)\left(x-2\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)^2\cdot\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)

c: Ta có: \(\left(9x^2-25\right)-\left(6x-10\right)=0\)

\(\Leftrightarrow\left(3x-5\right)\left(3x+5-2\right)=0\)

\(\Leftrightarrow\left(3x-5\right)\left(3x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)

\(A=\left(2x-1\right)^2+9\ge9\\ A_{min}=9\Leftrightarrow x=\dfrac{1}{2}\\ B=2\left(x^2-2\cdot\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{1}{8}=2\left(x-\dfrac{3}{4}\right)^2+\dfrac{1}{8}\ge\dfrac{1}{8}\\ B_{min}=\dfrac{1}{8}\Leftrightarrow x=\dfrac{3}{4}\\ C=\left(4x^2+4xy+y^2\right)+2\left(2x+y\right)+1+\left(y^2+4y+4\right)-4\\ C=\left[\left(2x+y\right)^2+2\left(2x+y\right)+1\right]+\left(y+2\right)^2-4\\ C=\left(2x+y+1\right)^2+\left(y+2\right)^2-4\ge-4\\ C_{min}=-4\Leftrightarrow\left\{{}\begin{matrix}2x=-1-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-2\end{matrix}\right.\)

\(D=\left(3x-1-2x\right)^2=\left(x-1\right)^2\ge0\\ D_{min}=0\Leftrightarrow x=1\\ G=\left(9x^2+6xy+y^2\right)+\left(y^2+4y+4\right)+1\\ G=\left(3x+y\right)^2+\left(y+2\right)^2+1\ge1\\ G_{min}=1\Leftrightarrow\left\{{}\begin{matrix}3x=-y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-2\end{matrix}\right.\)

\(H=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(2y^2+4y+2\right)+2\\ H=\left(x-y\right)^2+\left(x+1\right)^2+2\left(y+1\right)^2+2\ge2\\ H_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=-1\\y=-1\end{matrix}\right.\Leftrightarrow x=y=-1\)

Ta luôn có \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz\ge0\\ \Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\\ \Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3xy+3yz+3xz\\ \Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\\ \Leftrightarrow\dfrac{3^2}{3}\ge xy+yz+xz\\ \Leftrightarrow K\le3\\ K_{max}=3\Leftrightarrow x=y=z=1\)

\(a,=9x^2+12x+4-18x-4+10x^2=19x^2-6x\\ b,=8x^3-27-8x^3-36x^2-54x-27=-36x^2-54x-54\)