1) Rút gọn \(A=\frac{7.9+14.27+21.36}{21.27+42.81+63.108}\)
2) Cho \(S=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{n\left(n+3\right)}\)
Chứng minh S<1
3) So sánh: \(\frac{2003.2004-1}{2003.2004}\) và \(\frac{2004.2005-1}{2004.2005}\)
1 rút gọn :
A= \(\frac{7.9+14.27+21.36}{21.27+42.81+63.108}\)
2 cho s= \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+.......+\frac{3}{n\left(n+3\right)}\)n thuộc N chứng minh s<1
Câu 1:
1)Rút gọn A=7.9+14.27+21.36/21.27+42.81+63.108
2)Cho S=3/1.4+3/4.7+3/7.10+...+3/n.(n+3). (n thuộc N*)
3)So sánh :2003.2004-1/2003.2004 và 2004.2005-1/2004.2005
1, mình không ghi đề nha
A= \(\frac{1.1+1.1+1.1}{3+3.3+3.3+3}\)
A=\(\frac{1.3}{9.3}\)
A=\(\frac{1}{9}\)
Câu 1:
1, Rút gọn: \(A=\dfrac{7.9+14.27+21.36}{21.27+42.81+63.108}\)
2,Cho \(S=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{n.(n+3)}\)\
Chứng minh S < 1
3, So sánh: \(\dfrac{2003.2004-1}{2003.2004} và \dfrac{2004.2005-1}{2004.2005}\)
Câu 2:
Cho \(n\in Z\) chứng minh rằng: \(5^{n}-1\) chia hết cho 4
Rút gọn phân số :
A = \(\frac{7.9+14.27+21.36}{21.27+42.81+63.108}\)
\(A=\frac{7.9+14.27+21.36}{21.27+42.81+63.108}\)
\(A=\frac{7.9+7.2.9.3+7.3.9.4}{21.27+21.2.27.3+21.3.27.4}\)
\(A=\frac{7.9.\left(1+2.3+3.4\right)}{21.27+\left(1+2.3+3.4\right)}\)
\(A=\frac{7.9}{3.7.3.9}\)
\(A=\frac{1}{3.3}=\frac{1}{9}\)
rút gọn:
\(A=\frac{7.9+14.27+21.36}{21.27+42.81+63.108}\)
Ta có:A=\(\frac{7.9.\left(1+2.3+3.4\right)}{21.27.\left(1+2.3+3.4\right)}=\frac{7.9}{21.27}=\frac{1}{9}\)
Vậy A=\(\frac{1}{9}\)
rút gọn phân số \(\frac{7.9+14.27+21.36}{21.27+42.81+63.108}\)
Ta có:
\(\frac{7.9+14.27+21.36}{21.27+42.81+63.108}\)
\(=\frac{7.9+14.27+21.36}{7.3.9.3+14.3.27.3+21.3.36.3}\)
\(=\frac{7.9+14.27+21.36}{\left(.9+14.27+21.36\right).3.3}\)
\(=\frac{7.9+14.27+21.36}{\left(7.9+14.27+21.36\right).9}\)
\(=\frac{1}{9}\)
Rút gọn phân số: \(\frac{7.9+14.27+21.36}{21.27+42.81+63.108}\)
ta có:
7.9+14.27+21.36/21.27+42.81+63.108=1/9
Đúng thì like cho mình nhé
rút gon \(A=\frac{7.9+14.27+21.36}{21.27+42.81+63.108}\)
CẢM ƠN !
\(A=\frac{7.9+14.27+21.36}{21.27+42.81+63.108}\)
\(A=\frac{7.9+7.2.27+7.3.36}{7.3.27+7.6.81+7.9.108}\)
\(A=\frac{7.9+7.54+7.108}{7.81+7.486+7.972}\)
\(A=\frac{7.\left(9+54+108\right)}{7.\left(81+486+972\right)}\)
\(A=\frac{7.171}{7.1539}=\frac{171}{1539}=\frac{1}{9}\)
So sánh:\(M=\frac{7.9+14.27+21.36}{21.27+42.81+63.108}và M=\frac{37}{333}\)
ta có: \(M=\frac{7.9+14.27+21.36}{21.27+42.81+63.108}\)
\(M=\frac{7.9+7.2.9.3+7.3.9.4}{21.27+21.227.3+21.3.27.4}\)
\(M=\frac{7.9.\left(1+2.3+3.4\right)}{21.27.\left(1+2.3+3.4\right)}\)
\(M=\frac{7.9}{3.7.3.9}\)
\(M=\frac{1}{3^2}\)
\(M=\frac{1}{9}\)
và \(N=\frac{37}{333}=\frac{37}{37.9}=\frac{1}{9}\)
\(N=\frac{1}{9}\)
\(\Rightarrow M=N\left(=\frac{1}{9}\right)\)
M= \(\frac{7.9+14.27+21.36}{21.27+42.81+63.108}\).
M= \(\frac{7.9+14.27+21.36}{7.3.9.3+14.3.27.3+21.3+36.3}\).
M= \(\frac{7.9+14.27+21.36}{7.9.9+14.27.9+21.36.9}\).
M= \(\frac{7.9+14.27+21.36}{9\left(7.9+14.27+21.36\right)}\).
M= \(\frac{1}{9}\).
Ta có: N= \(\frac{37}{333}\)= \(\frac{1}{9}\).
Vì \(\frac{1}{9}\)= \(\frac{1}{9}\) nên M= N.
Vậy M= N.