1) \(=\left(2z+3\right)\left(4z^2-6z+9\right)\)

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

3) \(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)

4) \(=\left(2x+1\right)^2\)

5) \(=\left(x-10\right)^2\)

6) \(=\left(y^2-7\right)^2\)

7) \(=\left(5x-4y\right)\left(25x^2+20xy+16y^2\right)\)

Cảm ơn bạn nhiều nha 😁😁😁😁

1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)

2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

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

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

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

6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)

7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)

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

9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)

10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)

11) \(=\left(x+2\right)^3\)

12) \(=\left(x+3\right)^3\)

1. x³ + 8 = (x + 2 )(x² - x + 1)

2. 27 - 8y³ = ( 3 - 2y ) ( 9 + 6y + 4y² )

3. y⁶ + 1 = (y²)³ + 1 = ( y² + 1) ( y⁴ - y² +1 )

4.64x³ - \(\dfrac{1}{8}\)y³ = ( 4x - \(\dfrac{1}{2}\)y ) ( 16x² + 2xy + \(\dfrac{1}{4}\)y²)

5. 125x⁶ - 27y⁹ = (5x²)³ - (3y³)³

= ( 5x² - 3y³)(25x⁴ +15x²y³ + 9y⁶)

a: \(=\left(x+1+5\right)\left(x+1-5\right)=\left(x+6\right)\left(x-4\right)\)

b: =(1-2x)(1+2x)

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

d: =(x+3)^3

e: \(=\left(2x-y\right)^3\)

f: =(x+2y)(x^2-2xy+4y^2)

1) \(A=36x^2+12x+1=\left(6x+1\right)^2\ge0\)

\(minA=0\Leftrightarrow x=-\dfrac{1}{6}\)

2) \(B=9x^2+6x+1=\left(3x+1\right)^2\ge0\)

\(minB=0\Leftrightarrow x=-\dfrac{1}{3}\)

4) \(D=x^2-4x+y^2-8y+6=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\) 

\(minD=-14\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

3) \(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\ge-36\)

\(minC\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

5) \(E=\left(x-8\right)^2+\left(x+7\right)^2=2x^2-2x+113=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{225}{2}\ge\dfrac{225}{2}\)

\(minE=\dfrac{225}{2}\Leftrightarrow x=\dfrac{1}{2}\)

e: Ta có: \(E=\left(x-8\right)^2+\left(x+7\right)^2\)

\(=x^2-16x+64+x^2+14x+49\)

\(=2x^2-2x+113\)

\(=2\left(x^2-x+\dfrac{113}{2}\right)\)

\(=2\left(x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{225}{4}\right)\)

\(=2\left(x-\dfrac{1}{2}\right)^2+\dfrac{225}{2}\ge\dfrac{225}{2}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

\(x^3+8y^3\)

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

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

\(8y^3-125\)

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

\(=\left(2y-5\right)\left(4y^2+10y+25\right)\)

\(a^6-b^3\)

\(=\left(a^2\right)^3-b^3\)

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

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

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

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

\(x^{32}-1\)

\(=\left(x^{16}\right)^2-1^2\)

\(=\left(x^{16}-1\right)\left(x^{16}+1\right)\)

\(=\left(x^8-1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)

\(=\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)

\(=\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)

n *o biets