\(a^3+3a=b^3+3b=2=>a^2+3a-b^2-3b=0\)

\(=>\left(a-b\right)\left(a+b\right)+3\left(a-b\right)=0\)

\(=>\left(a-b\right)\left(a+b+3\right)=0\)

\(=>\orbr{\begin{cases}a-b=0\\a+b+3=0\end{cases}=>\orbr{\begin{cases}a+b=2a=2b\\a+b=-3\end{cases}}}\)

Vì \(x=2018\Rightarrow x+1=2019\)

Thay x+1=2019 vào biểu thức A  ta được :

\(A=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-...-\left(x+1\right)x+x+1\)

\(=x^6-x^6-x^5+x^5+x^4-...-x^2-x+x+1\)

\(=1\)

\(A=x^6-2019x^5+2018x^4-2019x^3+2019x^2-2019x+2019\)

\(=x^6-2018x^5-x^5+2018x^4+x^4-2018x^3-x^3+2018x^2+x^2\)

\(-2018x-x+2019\)

\(=x^5\left(x-2018\right)-x^4\left(x-2018\right)-x^3\left(x-2018\right)+x^2\left(x-2018\right)\)

\(+x\left(x-2018\right)-\left(x-2018\right)+1\)

= 1

\(4x^2+81\)

\(=\left(2x\right)^2+9^2\)????

Liệu đề có sai không vậy , mk nghĩ là \(4x^2-81\)thì đúng hơn

\(4x^2-81\)

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

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

_Hắc phong_

Mình nghĩ phải là \(4x^2-81\)

x³ - 6x² - x + 30

= (x + 2).x² - 6x² - x + 30/x + 2

= x² - 8x + 15

= (x + 2)(x - 3)(x - 5)

\(x^3-6x^2-x+30\)

\(=\left(x^3-8x^2+15x\right)+\left(2x^2-16x+30\right)\)

\(=x\left(x^2-8x+15\right)+2\left(x^2-9x+15\right)\)

\(=\left(x^2-8x+15\right)\left(x+2\right)\)

\(=\left(x^2-3x-5x+15\right)\left(x+2\right)\)

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

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

a)\(2x^2-12x=-18\)

\(\Leftrightarrow2x^2-12x+18=0\)

\(\Leftrightarrow2\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow2\left(x-3\right)^2=0\)

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

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

b) \(\left(4x^2-4x+1\right)-x^2=0\)

\(\Leftrightarrow\left(2x-1\right)^2-x^2=0\)

\(\Leftrightarrow\left(2x-1-x\right)\left(2x-1+x\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\3x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{3}\end{cases}}}\)

_Minh ngụy_

\(x^2-ay-y^2-ax\)

\(=\left(x^2-y^2\right)-\left(ax+ay\right)\)

\(=\left(x-y\right)\left(x+y\right)-a\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-a\right)\)

_Minh ngụy_

Ta có:

VT = (x² + y²)(a² + b²)

= x²a² + x²b² + y²a² + y²b²

= (a²x² + b²y² + 2axby) + (a²y² - 2aybx + b²x²)

= (ax + by)² + (ay - bx)²

=> VT = VP => đpcm

a) \(27x^3+8^3\)

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

\(=\left(3x+2\right)\left[\left(3x\right)^2+6x+2^2\right]\)

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

b) \(8x^3-y^3\)

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

c) \(x^2+4xy+4y^2\)

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

\(27x^3+8\)

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

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

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

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

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

\(x^2+4xy+4y^2\)

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

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

_Minh ngụy_