Tính nhanh:
\(\frac{3-3^2+3^3-3^4+...+3^{99}}{\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}}.\left(11-\sqrt{91}\right)\left(11-\sqrt{95}\right)\left(11+\sqrt{99}\right)\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)\left(11-\sqrt{113}\right)...\left(11-\sqrt{113}\right)\left(11-\sqrt{104}\right)\)
Đặt \(A=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)\left(11-\sqrt{113}\right)....\left(11-\sqrt{104}\right)\)
\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-\sqrt{121}\right)....\left(11-\sqrt{104}\right)\)
\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-11\right)....\left(11-\sqrt{104}\right)\)
\(=0\)
Do đó biểu thức trên đầu bài bằng 0
bạn ơi, trong dãy này không có số \(\sqrt{121}\)đâu
