Bài 1:
Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=19\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)-19=0\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6-19=0\)
\(\Leftrightarrow12x-23=0\)
\(\Leftrightarrow12x=23\)
hay \(x=\frac{23}{12}\)
Vậy: \(x=\frac{23}{12}\)
Bài 2: Rút gọn biểu thức
Ta có: \(3\cdot\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
