a/ \(36+x^2-12x=x^2-2x.6+6^2=\left(x+6\right)^2\)

b/ \(\left(x+2y\right)^2=x^2+2x.2y+\left(2y\right)^2=x^2+4xy+4y^2\)

c/ \(\left(\sqrt{x}-2\sqrt{y}\right)^2=\left(\sqrt{x}\right)^2-2\sqrt{x}.2\sqrt{y}+\left(2\sqrt{y}\right)^2=x-4\sqrt{xy}+4y\)

(x² + 3²)² 

^{=x^4 + 18x^2+81}

(x² + 3²)² 

= (x²)² + 2x² . 3² + (3²)² 

= x⁴ + 18x² + 81

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

c) \(\left(x^2+3^2\right)^2=x^4+18x+81\)

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

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

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

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

c) \(\left(2x+3y^2\right)^2=4x^2+12xy^2+9y^4\)

c) \(\left(4x-2y^2\right)^2=16x^2-16xy^2+4y^4\)

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

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

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

c) \(\left(\dfrac{x}{3}-\dfrac{y}{4}\right)\left(\dfrac{x}{3}+\dfrac{y}{4}\right)=\dfrac{x^2}{9}-\dfrac{y^2}{16}\)

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

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

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

c) \(\left(2x+\dfrac{3}{5}\right)\left(\dfrac{3}{5}-2x\right)=\dfrac{9}{25}-4x^2\)

c) \(\left(\dfrac{1}{2}x-\dfrac{4}{3}\right)\left(\dfrac{4}{4}+\dfrac{1}{2}x\right)=\dfrac{1}{4}x^2-\dfrac{16}{9}\)

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

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

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

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

\(\left(x+5y\right)^2=x^2+10xy+25y^2\)

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

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

a, \(\left(\frac{x}{y}-\frac{2}{3}\right)\left(\frac{x}{y}+\frac{2}{3}\right)=\left(\frac{x}{y}\right)^2-\left(\frac{2}{3}\right)^2\)

b,\(\left(2\sqrt{x}-\frac{2}{3}\right)\left(\frac{2}{3}+2\sqrt{x}\right)=\left(2\sqrt{x}-\frac{2}{3}\right)\left(2\sqrt{x}+\frac{2}{3}\right)\)

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

Trả lời:

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

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

a: \(\left(2x^2+3y\right)^3\)

\(=8x^6+3\cdot4x^4\cdot3y+3\cdot2x^2\cdot9y^2+27y^3\)

\(=8x^6+36x^4y+54x^2y^2+27y^3\)

b: \(\left(2a^2b+\dfrac{1}{3}ab^2\right)^2\)

\(=4a^4b^2+2\cdot2a^2b\cdot\dfrac{1}{3}ab^2+\dfrac{1}{9}a^2b^4\)

\(=4a^4b^2+\dfrac{4}{3}a^3b^3+\dfrac{1}{9}a^2b^4\)

\(9\left(x-3y\right)^2-25\left(2x+y\right)^2\)

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

\(=\left(3x-9y\right)^2-\left(10x+5y\right)^2\)

\(=\left[3x-9y+10x+5y\right]\left[3x-9y-\left(10x+5y\right)\right]\)

\(=\left(13x-4y\right)\left(-7x-14y\right)\)

\(=-7\left(x+2y\right)\left(13x-4y\right)\)

9(x - 3y)² - 25(2x + y)²

= 3².(x - 3y)² - 5².(2x + y)²

= (3x - 9y)² - (10x + 5y)²

= (3x - 9y - 10x - 5y)(3x - 9y + 10x + 5y)

= (-7x - 14y)(13x - 4y)

= -7(x + 2y)(13x - 4y)

\(27a^3-b^3+9ab^2-27a^2b\)

\(=\left(3a\right)^3-3\cdot\left(3a\right)^2b+3\cdot3a\cdot b^2-b^3\)

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

Bài 2: 

a: \(A=\left(x+1\right)^3+5=20^3+5=8005\)

b: \(B=\left(x-1\right)^3+1=10^3+1=1001\)