\(D=\left(3x+2\right)^2-\left(3x+2\right)\left(3x-2\right)-6x\)

\(\Leftrightarrow D=\left(3x+2\right)\left(3x+2-3x+2\right)-6x\)

\(\Leftrightarrow D=4\left(3x+2\right)-6x\)

\(\Leftrightarrow D=12x+8-6x\)

\(\Leftrightarrow D=6x+8\)

Vậy giá trị của D phụ thuộc vào giá trị của biến x

Kiên thưc cần nhơ:

- Cac hằng đẳng thưc đã học, phep nhân đơn thưc - đa thưc.

- Cộng trừ cac đơn thưc đồng thưc

- Cộng trừ cac đơn thưc đồng dạng

\(27x^2\left(x+1\right)-\left(3x+1\right)^3=-8\)

\(\Rightarrow27x^3+27x^2-27x^3-27x^2-9x-1=-8\)

\(\Rightarrow-9x-1=-8\)

\(\Rightarrow-9x=-7\)

\(\Rightarrow x=\frac{7}{9}\)

\(27x^2\left(x+1\right)-\left(3x+1\right)^3\)

\(27x^3+27^2-27x^3-27x^2-9x-1=-8\)

\(-9x-1=-8\)

\(-9x=-7\)

\(x=\frac{7}{9}\)

\(4x^2-81=0\)

\(\Rightarrow\left(2x\right)^2-9^2=0\)

\(\Rightarrow\left(2x-9\right).\left(2x+9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-9=0\\2x+9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{9}{2}\\x=-\frac{9}{2}\end{cases}}}\)

Vậy ...

\(4x^2-81=0\)

\(\Leftrightarrow\left(2x\right)^2-9^2=0\)

\(\Leftrightarrow\left(2x-9\right)\left(2x+9\right)=0\)

\(2x-9=0\)

\(2x=9\)

\(x=\frac{9}{2}\)

\(2x+9=0\)

\(2x=-9\)

\(x=-\frac{9}{2}\)

Ta có:\(\left(ay-bx\right)^2\ge0\Leftrightarrow a^2y^2-2axby+b^2x^2\ge0\)

\(\Leftrightarrow a^2y^2+b^2x^2\ge2abxy\)

\(\Leftrightarrow a^2x^2+a^2y^2+b^2y^2+b^2x^2\ge a^2x^2+2abxy+b^2y^2\)

\(\Leftrightarrow\left(a^2+b^2\right).\left(x^2+y^2\right)\ge\left(ax+by\right)^2\)

Vậy...

\(-\left(4x^2-6x+18\right)=-\left[\left(2x-3\right)^2+9\right]< 0\)

A = ( x - 2 )² - ( 2x + 1 )² 

A = x² - 4x + 4 - 4x² + 4x + 1 

A = - 3x² + 5 

B = ( x - 2y )² - ( x - 2y ) . ( 2y + x ) 

B = x² - 4xy + 4y² - ( 2xy + x² - 4y² - 2xy ) 

B = x² - 4xy + 4y² - 2xy - x² + 4y² + 2xy 

B = 8y² - 4xy 

a) \(\left(x^2-x+2\right)^2+\left(x-2\right)^2\)

\(=\left(x^4-2x^3+5x^2-4x+4\right)+\left(x^2-4x+4\right)\)

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

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

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

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

\(x^4-9x^3+28x^2-36x+16\)

\(=x^4-x^3-8x^3+8x^2+20x^2-20x-16x+16\)

\(=\left(x^4-x^3\right)-\left(8x^3-8x^2\right)+\left(20x^2-20x\right)-\left(16x-16\right)\)

\(=x^3\left(x-1\right)-8x^2\left(x-1\right)+20x\left(x-1\right)-16\left(x-1\right)\)

\(=\left(x-1\right)\left(x^3-8x^2+20x-16\right)\)

\(=\left(x-1\right)\left(x^3-2x^2-6x^2+12x+8x-16\right)\)

\(=\left(x-1\right)[x^2\left(x-2\right)-6x\left(x-2\right)+8\left(x-2\right)]\)

\(=\left(x-1\right)\left(x-2\right)\left(x^2-6x+8\right)\)

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

\(=\left(x-1\right)\left(x-2\right)[x\left(x-4\right)-2\left(x-4\right)]\)

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

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

\(a,5\left(x-2\right)\left(x+2\right)-5.x^{28}\)

\(=\left(5x-10\right)\left(x+2\right)-5x^{28}\)

\(=5x^2+10x-10x-20-5x^{28}\)

\(=5x^2-2x-5x^{28}\)

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

\(=a^2-2ab+b^2+4ab\)

\(=a^2+2ab+b^2\)

a) 5(x - 2)(x + 2) - 5.x²⁸

= 5(x² - 4) - 5.x²⁸

= 5.x² - 5.4 - 5.x²

= 5x² - 20 - 5x²⁸ 

= -5x²⁸ + 5x² - 20

b) (a - b)² + 4ab

= a² - 2ab + b² + 4ab

= a² - 2ab + b²