Bài 2: Nhân đa thức với đa thức

\(AC=\sqrt{50^2-30^2}=40\left(m\right)\)

Xét ΔBAC có BD/BA=BE/BC

nên DE//AC và DE/AC=BD/BA=1/2

=>DE=1/2*40=20(m)

=(2^2-1)(2^2+1)(2^4+1)*...*(2^128+1)

=(2^4-1)*...*(2^128+1)

...

=2^256-1

`(2x+3)(2x-3)-(2x+1)^2`

`=4x^2 - 9 - 4x^2 - 4x -1`

`=-4x -10`

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

\(=4x^2-9-4x^2-4x-1\)

\(=4x^2-4x^2-4x-9-1\)

\(=-4x-10\)

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

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

\(=4x^2-9-4x^2-4x-1\)

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

\(=-10-4x=-4x-10\)

3:

a: =x^2+4x-2x-8=x^2+2x-8

b: =4x^2+8x-6x-12=4x^2+2x-12

c: =3x+15-x^2-5x=-x^2-2x+15

d: =x^2+6x+5x+30=x^2+11x+30

\(\left(x^2-2x-1\right)\left(x-3\right)\\ =x^3-2x^2-x-3x^2+6x+3\\ =x^3-5x^2+5x+3\)

\(\left(x^2-2x-1\right).\left(x-3\right)=x^3-3x^2-2x^2+6x-x+3\)

\(=x^3-5x^2+5x+3\)

\(=-x^3+5x^2+x^2-5x+x-5\\ =-x^2+6x^2-4x-5\)

\(=-x^3+x^2+x+5x^2-5x-5=-x^3+6x^2-4x-5\)

\(\left(-5x+2\right)\left(3x-4\right)=-15x^2+6x+20x-8=-15x^2+26x-8\)

\(=-15x^2+20x+6x-8=-15x^2+26x-8\)

\(\left(3x+5\right)\left(2x-7\right)=6x^2-21x+10x-35=6x^2-11x-35\)

= \(6x^2-11x-35\)

\(\left(3x+5\right)\left(2x-7\right)\\ =6x^2+10x-21x-35\\ =6x^2-11x-35\)

a: \(M=\left(2x-y\right)^3=\left(12+8\right)^3=20^3=8000\)

b: \(=2x^2+x-x^3-2x^2+x^3-x+2014=2014\)

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

`= x^2 + 2x + 1 - x^2 + 4 - 2x`

`= 5`