a) Vế trái  \(=\dfrac{1.3.5...39}{21.22.23...40}=\dfrac{1.3.5.7...21.23...39}{21.22.23....40}=\dfrac{1.3.5.7...19}{22.24.26...40}\)

               \(=\dfrac{1.3.5.7....19}{2.11.2.12.2.13.2.14.2.15.2.16.2.17.2.18.2.19.2.20}\\ =\dfrac{1.3.5.7.9.....19}{\left(1.3.5.7.9...19\right).2^{20}}=\dfrac{1}{2^{20}}\left(đpcm\right)\)

b) Vế trái

 \(=\dfrac{1.3.5...\left(2n-1\right)}{\left(n+1\right).\left(n+2\right).\left(n+3\right)...2n}\\ =\dfrac{1.2.3.4.5.6...\left(2n-1\right).2n}{2.4.6...2n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1.2.3.4...\left(2n-1\right).2n}{2^n.1.2.3.4...n.\left(n+1\right)\left(n+2\right)...2n}\\ =\dfrac{1}{2^n}.\\ \left(đpcm\right)\)

@Akai Haruma

Bài đã đăng bạn hạn chế không đăng lại nữa nhé.

Lời giải:

a) \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{n-1}-1\right)\left(\frac{1}{n}-1\right)\)

\(=\frac{1-2}{2}.\frac{1-3}{3}.\frac{1-4}{4}...\frac{-(n-2)}{n-1}.\frac{-(n-1)}{n}\)

\(=\frac{(-1)(-2)(-3)...[-(n-2)][-(n-1)]}{2.3.4...(n-1)n}\)

\(=\frac{(-1)^{n-1}(1.2.3....(n-2)(n-1))}{2.3.4...(n-1)n}=(-1)^{n-1}.\frac{1}{n}\)

b) \(B=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)...\left(\frac{1}{n^2}-1\right)\)

\(=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}.....\frac{1-n^2}{n^2}\)

\(=\frac{(-1)(2^2-1)}{2^2}.\frac{(-1)(3^2-1)}{3^2}....\frac{(-1)(n^2-1)}{n^2}\)

\(=(-1)^{n-1}.\frac{(2^2-1)(3^2-1)...(n^2-1)}{2^2.3^2....n^2}\)

\(=(-1)^{n-1}.\frac{(2-1)(2+1)(3-1)(3+1)...(n-1)(n+1)}{2^2.3^2....n^2}\)

\(=(-1)^{n-1}.\frac{(2-1)(3-1)...(n-1)}{2.3...n}.\frac{(2+1)(3+1)...(n+1)}{2.3...n}\)

\(=(-1)^{n-1}.\frac{1.2.3...(n-1)}{2.3...n}.\frac{3.4...(n+1)}{2.3.4...n}\)

\(=(-1)^{n-1}.\frac{1}{n}.\frac{n+1}{2}=(-1)^{n-1}.\frac{n+1}{2n}\)

Không.

Có, khi a = 0

Có thể có phân số a/b (a, b ∈ Z, b ≠ 0) sao cho:

(m, n ∈ Z, m , n ≠ 0 , m ≠ n) khi và chỉ khi a = 0

Vì (m, n ∈ Z, m , n ≠ 0 , m ≠ n)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\left(\dfrac{1}{c}-\dfrac{1}{a+b+c}\right)=0\)

\(\Leftrightarrow\dfrac{a+b}{ab}+\dfrac{a+b+c-c}{c\left(a+b+c\right)}=0\)

\(\Leftrightarrow\dfrac{a+b}{ab}+\dfrac{a+b}{c\left(a+b+c\right)}=0\)

\(\Leftrightarrow\left(a+b\right)\times\dfrac{ac+bc+c^2+ab}{abc\left(a+b+c\right)}=0\)

\(\Leftrightarrow\dfrac{\left(a+b\right)\left(a+c\right)\left(b+c\right)}{abc\left(a+b+c\right)}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)

\(\Rightarrow N=0\)

Đặt \(f\left(1\right)=d\)

\(f\left(n+1\right)=af^2\left(n\right)+bf\left(n\right)+\dfrac{b^2}{4a}-\dfrac{b}{2a}\)

\(\Leftrightarrow f\left(n+1\right)+\dfrac{b}{2a}=a\left[f\left(n\right)+\dfrac{b}{2a}\right]^2\)

Đặt \(f\left(n\right)+\dfrac{b}{2a}=g\left(n\right)\Rightarrow\left\{{}\begin{matrix}g\left(1\right)=d+\dfrac{b}{2a}\\g\left(n+1\right)=a.g^2\left(n\right)\end{matrix}\right.\)

\(\Rightarrow g\left(n\right)=a.g^2\left(n-1\right)=a\left[a.g^2\left(n-2\right)\right]^2=a^{2^2-1}.g^{2^2}\left(n-2\right)=...=a^{2^{n-1}-1}.\left[g\left(1\right)\right]^{2^{n-1}}\)

\(\Rightarrow g\left(n\right)=a^{2^{n-1}-1}.\left(d+\dfrac{b}{2a}\right)^{2^{n-1}}\)

\(\Rightarrow f\left(n\right)=a^{2^{n-1}-1}.\left(d+\dfrac{b}{2a}\right)^{2^{n-1}}-\dfrac{b}{2a}\) (1)

Sau đó kiểm tra lại công thức (1) bằng quy nạp là được