a,    =(x+4)³

b)  = (x-2)³

c)  =\(-\left(x-1\right)^3\)

\(S=1^3+2^3+3^3+...+n^3=\left(1+2+3+...+n\right)^2\)

\(=\left[\dfrac{n\left(n+1\right)}{2}\right]^2=\dfrac{n^2\cdot\left(n+1\right)^2}{4}\)

\(27x^3-54x^2y+36xy^2-8y^3\)

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

Tại  \(x=4;\)\(y=6\) thì gtbt là:

    \(\left(3.4-2.6\right)^3=0\)

91³+3×91×9+3×91×9²+9³  tínnh nhanh nhé

\(a,\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)\)

\(\sqrt{x}^2-6^2\)

\(x-36\)

\(b,\left(2\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\)

\(\left(2\sqrt{x}\right)^2-1\)

\(4x-1\)

\(\left(\sqrt{x}-6\right)\left(6+\sqrt{x}\right)\)

\(=\left(\sqrt{x}-6\right)\left(\sqrt{x}+6\right)\)

\(=\left(\sqrt{x}\right)^2-6^2\)

\(=x-36\)

b.\(\left(2\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)\)

\(=\left(2\sqrt{x}\right)^2-1^2\)

\(=4x-1\)

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

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

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

\(=x-1-x^2+2^2\) 

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

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

 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

1) \(\left(3x^2-1\right)\left(9x^4+3x^2+1\right)\)

\(=27x^6+9x^4+3x^2-9x^4-3x^2-1\)

\(=27x^6-1\) (hằng đẳng thức dạng a³ - b³)

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

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

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

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

\(=x^6-64\)

a) \(\left(3x^2-1\right)\left(9x^4+3x^2+1\right)=\left(3x^2-1\right)\left[\left(3x^2\right)^2+3x^2.1+1^2\right]=\left(3x^2\right)^3-1^3=3x^6-1\)

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

\(\left(x^2+\frac{1}{x}+\frac{1}{9}\right)\left(x-\frac{1}{3}\right)-\left(x-\frac{1}{3}\right)^3\)

\(=\left[x^3-\left(\frac{1}{3}\right)^3\right]-\left(x-\frac{1}{3}\right)^3\)

\(=\left(x-\frac{1}{3}\right)^3-\left(x-\frac{1}{3}\right)^3\)

\(=\left(x-\frac{1}{3}\right)\left[x^2+\frac{1}{x}+\frac{1}{9}-\left(x-\frac{1}{3}\right)^2\right]\)

\(=\left(x-\frac{1}{3}\right)\left(\frac{1}{x}+\frac{2x}{3}\right)\)

\(=\frac{3x-1}{3}\times\frac{3+2x^2}{3x}\)

\(=\frac{9x+6x^2-3-2x^2}{9x}\)

\(=\frac{4x^2+9x-3}{9x}\)

Trả lời:

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

b, \(\left(\sqrt{x}-2\sqrt{y}\right)\left(2\sqrt{y}+\sqrt{x}\right)=\left(\sqrt{x}-2\sqrt{y}\right)\left(\sqrt{x}+2\sqrt{y}\right)=\left(\sqrt{x}\right)^2-\left(2\sqrt{y}\right)^2\)

\(=x-4y\)