\(\left(125x^3+1\right):\left(5x+1\right)\)

mà \(125x^3+1=\left(5x\right)^3+1=\left(5x+1\right)\left(25x^2-5x+1\right)\)

\(\Rightarrow\frac{\left(5x+1\right)\left(25x^2-5x+1\right)}{5x+1}=25x^2-5x+1\)

a,(x+2y)³ =x³+3.x².2y+3.x.(2y)²+(2y)³

= x³+6x²y+12xy²+8y³

b, phần b tương tự dấu thay đổi một tí

c, (5x+1)(5x+1)= (5x+1)²

=25x²+10x+1

a)  \(\left(x+2y\right)^3=x^3+6x^2y+12xy^2+8y^3\)

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

c)  \(\left(5x+1\right)\left(5x-1\right)=25x^2-1\)

bạn Đường Quỳnh Giang phần c bị thiếu 10x rồi

=2x-1

-11 - 2i

(125x3 + 1) : (5x + 1)

= [(5x)3 + 1] : (5x + 1)

= (5x + 1)[(5x)2 – 5x + 1]] : (5x + 1)

= (5x)2 – 5x + 1

= 25x2 – 5x + 1

\(\left(\dfrac{1}{3}.x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)=\left(\dfrac{1}{3}.x\right)^3+\left(2y\right)^3=\dfrac{1}{27}x^3+8y^3\)

b: \(f\left(x\right)=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3=x^6-\dfrac{1}{27}\)

\(\left(x-1\right)-\left(x-2\right)\left(x+2\right)\) 

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

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

\(=x-1-x^2+2^2\) 

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

\(=x-x^2+3\)

 a/ (x-1)²-(x-2)(x+2)

=(x-1)-(x²-2²)

=(x-1)-x²-2²

=x-x² +(2-1)(2+1)

=x-x²+3