thu gọn biẻu thức S = 1 + 2 + 2^2 + 2^3 + .. + 2^50
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a )
\(x\left(2x^2-3\right)-x^2\left(5x+1\right)+x^2\)
\(=2x^3-3x-5x^2-x^2+x^2\)
\(=-3x^3-3x\)
b )
\(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=-11x+24\)

a, \(2x\left(x+2\right)-\left(x+2\right)\left(x-2\right)=\left(x+2\right)^2=x^2+4x+4\)
b, \(\left(x-3\right)\left(x^2+3x+9\right)-\left(x^2-27x\right)=x^3-27-x^2+27x\)
c, \(\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x-y\right)\left(x^2+xy+y^2\right)=x^3+y^3-x^3+y^3=2y^3\)

\(a,\left(x-3\right)\left(x^2+3x+9\right)-\left(x^2-1\right)\left(x+27\right)\)
\(=\left(x^3-27\right)-x^3-27x^2+x+27=x-27x^2\)
\(b,\left(3-x\right)^3-\left(x+3\right)\left(x^2-3x+9\right)\)
\(=27-9x+3x^2-x^3-\left(x^3+27\right)=3x^2-9x-2x^3\)
\(c,\left(x-2\right)\left(x^2+2x+4\right)-x\left(x-3\right)\left(x+3\right)\)
\(=\left(x^3-8\right)-x\left(x^2-9\right)=x^3-8-x^3+9x=9x-8\)
a) (x-3)(x2+3x+9)-(x2-1)(x+27)
=(x3-27)-(x3+27x2-x-27)
=x3-27-x3-27x2+x+27
=-27x2+x
=x(-27x+1)
b) (3-x)3-(x+3)(x2-3x+9)
=27-27x+9x2-x3-x3-27
=-2x3+9x2-27x
=x(-2x+9x-27)
c) (x-2)(x2+2x+4)-x(x-3)(x+3)
=x3-8-x(x2-9)
=x3-8-x3+9x
=9x-8
#H

Ta có: \(S=1+2^2+2^3+....+2^{50}\)
\(\Rightarrow2S=2+2^3+2^4+2^5+....+2^{51}\)
\(\Rightarrow2S-S=\left(2+2^3+2^4+...+2^{51}\right)-\left(1+2^2+...+2^{50}\right)\)
\(\Rightarrow S=2^{51}-1\)
Vậy \(S=2^{51}-1\)

\(\left(x^2-2\right)\left(x^2-2x+2\right)\left(x^2+2x+2\right)\left(x^2+2\right)\)
\(=\left[\left(x^2-2\right)\left(x^2+2\right)\right].\left\{\left[\left(x^2+2\right)-2x\right].\left[\left(x^2+2\right)+2x\right]\right\}\)
\(=\left(x^4-4\right).\left[\left(x^2+2\right)^2-4x^2\right]\)
\(=\left(x^4-4\right)\left(x^4+4x^2+4-4x^2\right)=\left(x^4-4\right)\left(x^4+4\right)\)
\(=x^8-16\)

sai đề rồi làm j có 1! hay 2! hay ...
Sửa đề đi rồi tui làm cho
\(S=1+2+2^2+2^3+...+2^{50}\)
\(\Rightarrow2S=2+2^2+2^3+2^4+...+2^{51}\)
\(\Rightarrow2S-S=2^{51}-1\)
\(\Rightarrow S=2^{51}-1\)