Bài 1:
a, Chứng tỏ rằng với n thuộc N, n khác 0 thì:
\(\dfrac{1}{n\left(n+1\right)}\)=\(\dfrac{1}{n}\) - \(\dfrac{1}{n+1}\)
b, Áp dụng kết quả ở câu a để tính nhanh:
A=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+.....+\(\dfrac{1}{9.10}\)
Bài 2: Tính nhanh:
C=\(\dfrac{1}{2}\)+\(\dfrac{1}{14}\)+\(\dfrac{1}{35}\)+\(\dfrac{1}{65}\)+\(\dfrac{1}{104}\)+\(\dfrac{1}{152}\)
Bài 3:
a, Cho 2 phân số \(\dfrac{1}{n}\) và \(\dfrac{1}{n+1}\) (n thuộc Z, n > 0). Chứng tỏ rằng tích của 2 phân số này bằng hiệu của chúng.
b, Áp dụng kết quả trên để tính giá trị các biểu thức sau:
A=\(\dfrac{1}{2}\) . \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) . \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) . \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) . \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) . \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) . \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) . \(\dfrac{1}{9}\)
B=\(\dfrac{1}{30}\)+\(\dfrac{1}{42}\)+\(\dfrac{1}{56}\)+\(\dfrac{1}{72}\)+\(\dfrac{1}{90}\)+\(\dfrac{1}{110}\)+\(\dfrac{1}{132}\)
Các bạn giúp mk với nha!
Bài 1:
a) \(\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\)
Quy đồng \(VP\) ta được:
\(VP=\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(\Rightarrow VP=\dfrac{n+1}{n\left(n+1\right)}-\dfrac{n}{n\left(n+1\right)}\)
\(\Rightarrow VP=\dfrac{n+1-n}{n\left(n+1\right)}=\dfrac{1}{n\left(n+1\right)}\)
\(\Rightarrow VP=VT\)
Vậy \(\forall n\in Z,n>0\Rightarrow\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\) (Đpcm)
b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
Bài 3:
a) \(\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{1+1}{n\left(n+1\right)}-\dfrac{n}{n\left(n+1\right)}=\dfrac{1}{n\left(n+1\right)}\)
b) A=\(\dfrac{1}{2}.\dfrac{1}{3}+\dfrac{1}{3}.\dfrac{1}{4}+\dfrac{1}{4}.\dfrac{1}{5}+\dfrac{1}{5}.\dfrac{1}{6}+\dfrac{1}{6}.\dfrac{1}{7}+\dfrac{1}{7}.\dfrac{1}{8}+\dfrac{1}{8}.\dfrac{1}{9}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\)
\(=\dfrac{1}{2}-\dfrac{1}{9}\)
\(=\dfrac{7}{18}\)
B=\(\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)
\(=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}\)
\(=\dfrac{1}{5}-\dfrac{1}{12}\)
\(=\dfrac{7}{60}\)
Bài 1:
\(\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{n+1}{n\left(n+1\right)}-\dfrac{n}{\left(n+1\right)n}\)
\(=\dfrac{n+1-n}{n\left(n+1\right)}=\dfrac{1}{n\left(n+1\right)}\)
b)A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(A=1\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(\Rightarrow A=1\left(1-\dfrac{1}{10}\right)\)
A=\(1\left(\dfrac{10}{10}-\dfrac{1}{10}\right)\)
\(\Rightarrow A=\dfrac{9}{10}\)
Vậy A=\(\dfrac{9}{10}\)
~ chúc bạn học tốt~