1,

Đặt \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(\left(2-1\right)A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(1A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(A=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(A=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(A=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(A=2^{32}-1\)

Vậy \(A=2^{32}-1\)

2, \(x^2-6x=-9\)

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

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

\(x-3=0\)

\(x=3\)

Vậy \(x=3\)

Bài 2 

a. (x-2y)2 =2x-4y

b. (2x^2 +3)2 =4x^2+6

c. (x-2) (x^2+2x+4) = x^3-8 (hằng đẳng thức)

d. (2x-1)3 = 6x-3

 Xin lỗi mik chỉ lm ổn bài 2 thôi!

\(a\\ -5x^2+3x.\left(x+2\right)=-5x^2+3x^2+6x=-2x^2+6x\\ b\\ -2x.\left(1-x^2\right)-2x^3=-2x+2x^3-2x^3=-2x\\ c\\ 4x.\left(x-1\right)-4.\left(x^2+2x-1\right)\\ =4x^2-4x-4x^2-8x+4=-12x+4\)

\(d\\ 6x^3-2x^2.\left(-x^2-3x\right)=6x^3+2x^4+6x^3=2x^4+12x^3\\ e\\ 3x.\left(x-1\right)-\left(1+2x\right).5x\\ =3x^2-3x-5x-10x^2=-7x^2-8x\\ f\\ -5x^2-\left(x-6\right).\left(-2x^2\right)=-5x^2+2x^3-12x^2=2x^3-17x^2\)

\(a,\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=4xy\\ b,\left(x+y\right)^2+\left(x-y\right)^2-2\left(x+y\right)\left(x-y\right)=\left(x+y-x+y\right)^2=4y^2\\ c,\left(x^2-1\right)\left(x^2-x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\\ =\left(x-1\right)\left(x^3+1\right)\\ =x^4-x^3+x-1\)

a. (x + y)² - (x - y)²

= (x + y - x + y)(x + y + x - y)

= 2y . 2x

= 4xy

b. (x + y)² + (x - y)² - 2(x + y)(x - y)

= (x² + 2xy + y²) + (x² - 2xy + y²) - 2(x² - y²)

= x² + 2xy + y² + x² - 2xy + y² - 2x² + 2y²

= x² + x² - 2x² + 2xy - 2xy + y² + y² + 2y²

= 4y²

c. (x² - 1)(x² - x + 1)

= x⁴ - x³ + x² - x² + x - 1

= x⁴ - x³ + x - 1

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

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

c: \(\left(x^2-1\right)\cdot\left(x^2-x+1\right)\)

\(=\left(x^3+1\right)\left(x-1\right)\)

\(=x^4-x^3+x-1\)

a: \(=\dfrac{\left(x+1\right)\left[\left(3x-2\right)-\left(2x+5\right)\left(x-1\right)\right]}{x+1}\)

=3x-2-2x^2+2x-5x+5

=-2x^2+3

b: \(=\left(2x+1-3+x\right)^2=\left(3x-2\right)^2=9x^2-12x+4\)

c: =x^3-3x^2+3x-1-x^3-1+9x^2-1

=6x^2+3x-3

\(a,\left[\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x^2-1\right)\right]:\left(x+1\right)\)

\(=\left[\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x-1\right)\left(x+1\right)\right]:\left(x+1\right)\)

\(=\left[\left(x+1\right)\left(3x-2-\left(2x+5\left(x-1\right)\right)\right)\right]:\left(x+1\right)\)

\(=\left[\left(x+1\right)\left(3x-2-2x^2+2x-5x+5\right)\right]:\left(x+1\right)\)
\(=\left[\left(x+1\right)\left(-2x^2+3\right)\right].\dfrac{1}{x+1}\)

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

\(b,\left(2x+1\right)^2-2\left(2x+1\right)\left(3-x\right)\)

\(=\left(2x+1\right)\left[\left(2x+1\right)-2\left(3-x\right)\right]\)

\(=\left(2x+1\right)\left(2x+1-6+2x\right)\)

\(=\left(2x+1\right)\left(4x-5\right)\)

\(c,\left(x-1\right)^3-\left(x+1\right)\left(x^2-x+1\right)-\left(3x+1\right)\left(1-3x\right)\)

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

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

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

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

\(=\dfrac{3x}{\left(x+1\right)\left(x^2+x+1\right)}\)

\(A=x^2+4x-21-x^2-4x+5=-16\\ B=-2\left(4x^2+20x+25\right)-\left(1-16x^2\right)\\ B=-8x^2-40x-50-1+16x^2=8x^2-40x-51\\ C=x^2\left(x^2-16\right)-\left(x^4-1\right)=x^4-16x^2-x^4+1=1-16x^2\\ D=x^3+1-\left(x^3-1\right)=2\\ E=x^3-3x^2+3x-1-x^3+1-9x^2+1=-12x^2+3x+1\)

\(a,x^2+4x-21-x^2-4x+5=-16\\ b,=\left(x+8-x+2\right)^2=10^2=100\\ c,=x^2\left(x^2-16\right)-\left(x^4-1\right)\\ =x^4-16x^2-x^4+1=1-16x^2\\ d,=x^3+1-x^3+1=2\)

\(a,=x^2+4x-21-x^2-4x+5=-16\\ b,=\left(x+8-x+2\right)^2=10^2=100\\ c,=x^2\left(x^2-16\right)-\left(x^4-1\right)\\ =x^4-16x^2-x^4+1=1-16x^2\\ d,=x^3+1-x^3+1=2\)