Đáp án:
a)2x(2x-1)²-3x(x+3)(x-3)-4x(x+1)²
=2x(4x²-4x+1)-3x(x²-9)-4x(x²+2x+1)
=8x³-8x²+2x-3x³+27x-4x³-8x²-4x
=x³-16x²+25x
b) (a-b+c) ²-(b-c)²+2ab-2ac
=a²+b²+c²-2ab-2bc+2ac-b²+2bc-c²+2ab-2ac
=a²
c)(3x+1)²-2(3x+1)(3x+5)+(3x+5)²
=9x²+6x+1-2(3x+3-2)(3x+3+2)+9x²+30x+25
=18x²+36x+26-2((3x+3)²-4)
=18x²+36x+26-2(9x²+18x+9-4)
=18x²+36x+26-2(9x²+18x+5)
=18x²+36x+26-18x²-36x-10
=16
d) (2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)
=1(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)
=(2-1)(2+1)(2²+1)(2^4+1)(2^8+1)(2^16+1)
=(2²-1)(2²+1)(2^4+1)(2^8+1)(2^16+1)
=(2^4-1)(2^4+1)(2^8+1)(2^16+1)
=(2^8-1)(2^8+1)(2^16+1)
=(2^16-1)(2^16+1)
=2^32-1
e) (3+1)(3²+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)
=$\frac{1}{2}$ (3-1)(3+1)(3²+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)
=$\frac{1}{2}$(3²-1) (3²+1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)
=$\frac{1}{2}$(3^4-1)(3^4+1)(3^8+1)(3^16+1)(3^32+1)
=$\frac{1}{2}$(3^8-1)(3^8+1)(3^16+1)(3^32+1)
=$\frac{1}{2}$(3^16-1)(3^16+1)(3^32+1)
=$\frac{1}{2}$(3^32-1)(3^32+1)
=$\frac{1}{2}$(3^64-1)
g)(a+b-c)²+(a-b+c)²-2(b-c)²
=(a+b)²-2(a+b).c+c²+(a-b)²-2(a-b).c-2(b²-2bc+c²)
=a²+2ab+b²-2ac-2bc+a²-2ab+b²+2ac-2bc+c²-2b²+4bc-2c²
=2a²-c²
h) (x-2)(x²-2x+4)(x+2)(x²+2x+4)
=((x-2)(x²+2x+4))((x+2)(x²-2x+4))
=(x³-8)(x³+8)
=x^6-64
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