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\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{97.101}\)
\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{97}-\frac{1}{101}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{4}.\frac{100}{101}\)
\(=\frac{25}{101}\)
\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+\frac{1}{13.17}+...+\frac{1}{41.45}\)
\(=\frac{1}{4}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\left(1-\frac{1}{45}\right)\)
\(=\frac{1}{4}.\frac{44}{45}\)
\(=\frac{11}{45}\)
Đặt \(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\) là A.
Ta có:
\(A=\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\)
\(4A=4\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}\right)\)
\(4A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}\)
\(4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}\)
\(4A=1-\frac{1}{45}\)
\(4A=\frac{44}{45}\)
\(A=\frac{44}{45}:4\)
\(A=\frac{11}{45}\)
Vậy \(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+\frac{1}{9\cdot13}+\frac{1}{13\cdot17}+...+\frac{1}{41\cdot45}=\frac{11}{45}\)
thời gian hồng giải bài toán 3 là
45x\(\frac{1}{3}\)=15(phút)
thời gian hồng giải cả 3 bài là
45+15=60(phút)
đs...
A= 1/5.9+1/9.13+1/13.17+1/17.21+1/21.25
4A= 4/5.9+4/9.13+4/13.17+4/17.21+4/21.25
4A= (1/5-1/9)+(1/9-1/13)+(1/13-1/17)+(1/17-1/21)+(1/21-1/25)
4A= 1/5- 1/25
4A= 4/25
A= 4/25 :4
A= 1/25
S=1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+...+1/(1+2+3+4+...+10)
S=1/(2*3/2)+1/(3*4/2)+1/(4*5/2)+...+1/(10*11/2)
S=2(1/(2*3)+1/(3*4)+1/(4*5)+1/(5*6)+...+1/(10*11)
S=2(1/2-1/3+1/3-1/4+1/4-1/5+...+1/10-1/11)
S=2(1/2-1/11)
S=2*9/22
S=9/11
nho k cho minh voi nha
1/1.5 + 1/5.9 + 1/9.13 + ... + 1/97.101
= 1/4.(4/1.5 + 4/5.9 + 4/9.13 + ... + 4/97.101)
= 1/4.(1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/97 - 1/101)
= 1/4.(1 - 1/101)
= 1/4.100/101
= 25/101
\(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+........+\frac{1}{97.101}\)
\(=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+........+\frac{1}{97}-\frac{1}{101}\right)\)
\(=\frac{1}{4}\left(1-\frac{1}{101}\right)\)
\(=\frac{1}{4}.\frac{100}{101}=\frac{25}{101}\)