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

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) $(3x+5)^2\\=(3x)^2+2.3x.5+5^2\\=9x^2+30x+25$

b) $(6x+\dfrac{1}{3})^2\\=(6x)^2+2.6x.\dfrac{1}{3}+(\dfrac{1}{3})^2\\=36x^2+4x+\dfrac{1}{9}$

c) $(5x-4y)^2\\=(5x)^2-2.5x.4y+(4y)^2\\=25x^2-40xy+16y^2$

d) $(5x-3)(5x+3)\\=(5x)^2-(3)^2\\=25x^2-9$

a) \(\left(2x+1\right)^3\)

\(=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1+1\)

\(=8x^3+12x^2+6x+1\)

b) \(\left(x-3\right)^3\)

\(=x^3-3.x^2.3+3.x.3^2-3^3\)

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

Bài 2: 

a: \(x^3+15x^2+75x+125=\left(x+5\right)^3\)

b: \(1-15y+75y^2-125y^3=\left(1-5y\right)^3\)

c: \(8x^3+4x^2y+\dfrac{3}{2}xy^2+8y^3=\left(2x+2y\right)^3\)

a) \(=x^3+27-54-x^3=-27\)

b) \(=8x^3+y^3\)

a.x^3-1^3

b.x^3-5^3

c)(2x)^3+3^3

d)x^3+1/2^3

\(a,=x^2+x+\dfrac{1}{4}\\ b,=4x^2+2x+\dfrac{1}{4}\\ c,=x^2-2+\dfrac{1}{x^2}\\ d,=4x^2+\dfrac{8}{3}x+\dfrac{4}{9}x^2\\ e,=a^2-1\\ f,=25x^4-4\)

\(a,\left(x+\dfrac{1}{2}\right)^2=x^2+x+\dfrac{1}{4}\)

\(b,\left(2x+\dfrac{1}{2}\right)^2=4x^2+2x+\dfrac{1}{4}\)

\(c,\left(x-\dfrac{1}{x}\right)^2=x^2-2+\dfrac{1}{x^2}\)

\(d,\left(\dfrac{2x+2}{3x}\right)^2=\dfrac{\left(2x+2\right)^2}{9x^2}=\dfrac{4x^2+8x+4}{9x^2}\)

\(e,\left(a-1\right).\left(a+1\right)=a^2-1\)

\(f,\left(5x^2-2\right).\left(5x^2+2\right)=25x^4-4\)

a. \(\left(x+\dfrac{1}{2}\right)^2=x^2+x+\dfrac{1}{4}\)

b. \(\left(2x+\dfrac{1}{2}\right)^2=4x^2+2x+\dfrac{1}{4}\)

c. \(\left(x-\dfrac{1}{x}\right)^2=x^2-2+\dfrac{1}{x^2}\)

d. \(\left(2x+\dfrac{2}{3x}\right)^2=4x^2+\dfrac{8}{3}+\dfrac{4}{9x^2}\)

e. (a - 1)(a + 1) = a² - 1

f. (5x² - 2)(5x² + 2) = 25x⁴ - 4 

g. (2a - 3)(2a + 3) = 4a² - 9

a) \(=4x^2-12x+9\)

b) \(=4x^2+2x+\dfrac{1}{4}\)

c) \(=4x^2-\dfrac{4}{3}x+\dfrac{1}{9}\)

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

e) \(=\left(3-\dfrac{x}{2}\right)\left(9+\dfrac{3x}{2}+\dfrac{x^2}{4}\right)\)

f) \(=\left(125-4x\right)\left(125^2+500x+16x^2\right)\)

\(1,=-\left(y^2+12y+36\right)=-y^2-12y-36\)

\(2,=-\left(16-8y+y^2\right)=-16+8y-y^2\)

\(3,=-\left(\dfrac{4}{9}+\dfrac{4}{3}x+x^2\right)=-\dfrac{4}{9}-\dfrac{4}{3}x-x^2\)

\(4,=-\left(x^2-3x+\dfrac{9}{4}\right)=-x^2+3x-\dfrac{9}{4}\)

\(5,-\left(2+3y\right)^2=-\left(4+12y+9y^2\right)=-4-12y-9y^2\)

.... mấy ý còn lại bn tự lm nhé, tương tự thhooi

1) \(-\left(y+6\right)^2=-y^2-12y-36\)

2) \(-\left(4-y\right)^2=-y^2+8y-16\)

3) \(-\left(x+\dfrac{2}{3}\right)^2=-x^2-\dfrac{4}{3}x-\dfrac{4}{9}\)

4) \(-\left(x-\dfrac{3}{2}\right)^2=-x^2+3x-\dfrac{9}{4}\)

5) \(-\left(3y+2\right)^2=-9y^2-12y-4\)

6) \(-\left(2y-3\right)^2=-4y^2+12y-9\)

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

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