Mình làm ngắn gọn nhé.

\(A=1+2+2^2+...+2^{50}\)

\(\Rightarrow2A=2+2^2+...+2^{51}\)

\(\Rightarrow2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}\)

\(\Rightarrow A=2^{51}-1\)

\(B=1+3+...+3^{66}\)

\(3B=3+3^2+...+3^{67}\)

\(2B=3+3^2+...+3^{67}-1-3-...-3^{66}\)

\(2B=3^{67}-1\)

\(B=\frac{3^{67}-1}{2}\)

Bài 2: 

a: Ta có: \(M=\left(x+y\right)^3+2x^2+4xy+2y^2\)

\(=\left(x+y\right)^3+2\cdot\left(x+y\right)^2\)

\(=7^3+2\cdot7^2=441\)

a: \(=25x^4-10x^3+5x^2\)

c: \(=2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)

   13⁵.17³.a⁹.(13.17².a³)

= 13⁵.17³.13.17².a⁹.a³

= (13⁵.13).(17³.17²).(a⁹.a³)

= 13⁶.17⁵.a¹²

1a) \(\left(x+1\right)^2\left(x-2\right)^2=0\)

=> \(\orbr{\begin{cases}\left(x+1\right)^2=0\\\left(x-2\right)^2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)

b) \(\left(x-9\right)^5\left(x-5\right)^8=0\)

=> \(\orbr{\begin{cases}\left(x-9\right)^5=0\\\left(x-5\right)^8=0\end{cases}}\)

=> \(\orbr{\begin{cases}x-9=0\\x-5=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=9\\x=5\end{cases}}\)

\(a,\left(x-3\right)\left(x^2+3x+9\right)-\left(x^2-1\right)\left(x+27\right)\)

\(=\left(x^3-27\right)-x^3-27x^2+x+27=x-27x^2\)

\(b,\left(3-x\right)^3-\left(x+3\right)\left(x^2-3x+9\right)\)

\(=27-9x+3x^2-x^3-\left(x^3+27\right)=3x^2-9x-2x^3\)

\(c,\left(x-2\right)\left(x^2+2x+4\right)-x\left(x-3\right)\left(x+3\right)\)

\(=\left(x^3-8\right)-x\left(x^2-9\right)=x^3-8-x^3+9x=9x-8\)

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