tính \(\frac{\left(\frac{-5}{7}^{ }\right)n+1^{ }}{\left(\frac{-5}{7}\right)^{ }n}\)(n>=1)
Tính :
a, \(\frac{\frac{\left(-5\right)^n}{\left(7\right)}}{\frac{\left(-5\right)^{n-1}}{7}}\left(n>=1\right)\) b,\(\frac{\frac{\left(-1\right)^{2n}}{2}}{\frac{\left(-1\right)^n}{2}}\left(n\in N\right)\)
Phân số \(\frac{-5}{7}\)và \(\frac{-1}{2}\)nằm trong ngoặc nhưng mình chỉ đóng ngoặc đc tử nên đừng hiểu sai nha
Ai nhanh mình tick
Tính:
\(\frac{\left(\frac{-5}{7}\right)^{n+1}}{\left(\frac{-5}{7}\right)^n}\)
\(\frac{\left(\frac{-5}{7}\right)^{n+1}}{\left(\frac{-5}{7}\right)^n}\)
\(=\frac{-5}{7}\)
#)Giải :
\(\frac{\left(\frac{-5}{7}\right)^{n+1}}{\left(\frac{-5}{7}\right)^n}=\frac{\left(\frac{-5}{7}\right)^n\times\left(\frac{-5}{7}\right)}{\left(\frac{-5}{7}\right)^n}=\frac{-5}{7}\)
\(\frac{\left(\frac{-5}{7}\right)^{n+1}}{\left(\frac{-5}{7}\right)^n}=\left(\frac{-5}{7}\right)^{n+1-n}=\left(\frac{-5}{7}\right)^1=\frac{-5}{7}\)
Tính :\(\frac{\left(\frac{-5}{7}\right)^{n+1}}{\left(\frac{-5}{7}\right)^n}\)\(\left(n\ge1\right)\)
\(\frac{\left(-\frac{5}{7}\right)^{n+1}}{\left(-\frac{5}{7}\right)^n}=\frac{\left(-\frac{5}{7}\right)^n.\left(-\frac{5}{7}\right)}{\left(-\frac{5}{7}\right)^n}=\frac{-\frac{5}{7}}{1}=-\frac{5}{7}\)
Tính :
a) \(\frac{\left(\frac{-5}{7}\right)^n}{\left(\frac{-5}{7}\right)^{n-1}}\)( n\(\ge\)1 )
b) \(\frac{\frac{-1}{2}^{2n}}{\left(\frac{-1}{2}\right)^n}\) ( n \(\in\)N )
a: \(=\dfrac{\left(-\dfrac{5}{7}\right)^n}{\left(-\dfrac{5}{7}\right)^n\cdot\dfrac{-7}{5}}=1:\dfrac{-7}{5}=-\dfrac{5}{7}\)
b: \(=\dfrac{\dfrac{1}{4}^n}{\left(-\dfrac{1}{2}\right)^n}=\left(-\dfrac{1}{2}\right)^n\)
1) Cho tổng:
A = 4n + 4 \(\left(n\in Z\right)\) . Tìm n để A chia hết cho n
B = 5n + 6 \(\left(n\in Z\right)\) . Tìm n để B chia hết cho n
2) Tính nhanh
a) \(\left(\frac{3}{29}-\frac{1}{5}\right).\frac{29}{3}\)
b) \(\frac{1}{7}.\frac{5}{9}+\frac{5}{9}.\frac{1}{7}+\frac{5}{9}.\frac{3}{7}\)
\(\frac{A}{n}=\frac{4n+4}{n}=4+\frac{4}{n}\)
\(\Rightarrow n\in U\left(4\right)\)
Lập bảng tiếp nhé!
\(\frac{B}{n}=\frac{5n+6}{n}=5+\frac{6}{n}\)
Lập bảng
\(2.\)
a)\(\left(\frac{3}{29}-\frac{1}{5}\right)\cdot\frac{29}{3}=\frac{3}{29}\cdot\frac{29}{3}-\frac{1}{5}\cdot\frac{29}{3}=1-\left(1+\frac{14}{15}\right)=1-1-\frac{14}{15}=\frac{14}{15}\)
b)\(\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}=\frac{5}{9}\cdot\left(\frac{1}{7}+\frac{1}{7}+\frac{3}{7}\right)=\frac{5}{9}\cdot\frac{5}{7}=\frac{25}{63}\)
Rút gọn \(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
Ta có:
\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{2n+1}{n^2\left(n+1\right)^2}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{2n+1}{n^2}-\frac{2n+1}{\left(n+1\right)^2}\)
\(=1-\frac{2n+1}{\left(n+1\right)^2}\)
Vậy \(A=\frac{2n+1}{\left(n+1\right)^2}\)
Bài 1 Thưc hiện phép tính ( tính nhanh nếu có thể)
a)\(\frac{-1}{24}-\left[\frac{1}{4}-\left(\frac{1}{2}-\frac{7}{8}\right)\right]\)
b)\(\left(\frac{5}{7}-\frac{7}{5}\right)-\left[\frac{1}{2}-\left(\frac{-2}{7}-\frac{1}{10}\right)\right]\)
C)\(\left(\frac{-1}{2}\right)-\left(\frac{-3}{5}\right)+\left(\frac{-1}{9}\right)+\frac{1}{17}-\left(\frac{-2}{7}\right)+\frac{4}{35}-\frac{7}{18}\)
d)\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
Tính tổng:
\(S=\frac{3}{1^2.3}+\frac{5}{\left(1^2+2^2\right).4}+\frac{7}{\left(1^2+2^2+3^2\right).5}+...+\)\(\frac{2n+1}{\left(1^2+2^2+3^2+...+n^2\right).\left(n+2\right)}\)
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