\(\frac{1}{n}-\frac{1}{n +a}=\frac{n+a}{n\left(n+a\right)}-\frac{n}{n\left(n+a\right)}=\frac{n+a-n}{n\left(n+a\right)}=\frac{a}{n\left(n+a\right)}\)
\(\frac{1}{n}-\frac{1}{n +a}=\frac{n+a}{n\left(n+a\right)}-\frac{n}{n\left(n+a\right)}=\frac{n+a-n}{n\left(n+a\right)}=\frac{a}{n\left(n+a\right)}\)
chứng minh rằng :
\(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\)(a,n thuộc N*)
Chứng minh rằng:
a)\(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2^{20}}\)
b)\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)...2n}=\frac{1}{2^n}\)với n thuộc N*
CHỨNG MINH RẰNG :
\(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\)(n,a thuộc N*)
NHỚ CÓ LỜI GIẢI HỘ NHÉ!!!
Chứng minh rằng:
a,\(\frac{1.3.5...39}{21.22.23...40}=\frac{1}{2^{20}}\)
b,\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3\right)...2n}=\frac{1}{2^n}\)
Biết rằng n thuộc N*
Chứng Minh Rằng :
\(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\) \(\left(n,a\inℕ^∗\right)\)
Chứng minh rằng :
a, \(\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n\left(n+1\right)}\)
b, \(\frac{1}{n\left(n+q\right)}=\frac{1}{q}\left(\frac{1}{n}-\frac{1}{n+q}\right)\)
Chứng minh rằng:
\(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\)
\(3.Chứng\) Minh rằng
\(A.\) \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
\(B.\)\(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\)
chứng minh : \(\frac{a}{n\times\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\left(n;a\in Nsao\right)\)