Question

[Alice Sophia]

Rút gọn biểu thức\(B=\frac{x^{39} x^{36} x^{33} ... x^3 1}{x^{40} x^{38} x^{36} ... x^2 1}\)\(A=\frac{x^{95} x^{94} x^{93} ... x 1}{x^{31} x^{30} x^{29} ... x 1}\)

Answer 1

[Nguyễn Như Nam]

Rút gọn.

\(B=\dfrac{x^{39}x^{36}x^{33}...x^31}{x^{40}x^{38}x^{36}...x^21}=\dfrac{x^{\left(39+36+33+...+3\right)}}{x^{\left(40+38+36+...+2\right)}}\)

ta có: \(39+36+33+...+3=\dfrac{\left(39+3\right)\left(\dfrac{39-3}{3}+1\right)}{2}=273\)

\(40+38+36+....+2=\dfrac{\left(40+2\right)\left(\dfrac{40-2}{2}+1\right)}{2}=420\)

=> \(B=\dfrac{x^{273}}{x^{420}}=\dfrac{1}{x^{147}}\)

Tương tự như B => \(A=\dfrac{x^{4560}}{x^{496}}=x^{4064}\)

Answer 2

[Nguyễn Như Nam]

Ta có:

\(B=\dfrac{x^{\left(39+36+33+....+3\right)}}{x^{\left(40+38+36+....+2\right)}}\)

\(39+36+33+....+3=\dfrac{\left(39+3\right)\left(\dfrac{39-3}{3}+1\right)}{2}=273\)

\(40+38+36+....+2=\dfrac{\left(40+2\right)\left(\dfrac{40-2}{2}+1\right)}{2}=420\)

\(\Rightarrow B=\dfrac{x^{273}}{x^{420}}=\dfrac{1}{x^{147}}\)

tương tự => \(A=\dfrac{x^{4560}}{x^{496}}=x^{4064}\)