m.n giúp mk mấy câu vio 7 với:
1)Cho \(D=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+......+\frac{10}{1400}\)
\(E=\frac{1}{1+2}+\frac{1}{1+2+3}+......+\frac{1}{1+2+3+....+24}\)
Tính tỉ số D/E?
2)Giả sử \(\frac{a+b}{a+c}=\frac{a-b}{a-c}\) và a khác c;a khác -c;ac khác 0
Tính giá trị biểu thức \(A=\frac{10b^2+9bc+c^2}{2b^2+bc+2c^2}\)
1) \(D=\frac{10}{56}+\frac{10}{140}+\frac{10}{260}+....+\frac{10}{1400}\)
\(D=\frac{5}{28}+\frac{5}{70}+\frac{5}{130}+.....+\frac{5}{700}\)
\(D=\frac{5}{4.7}+\frac{5}{7.10}+\frac{5}{10.13}+......+\frac{5}{25.28}\)
\(D=\frac{5}{3}.\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+.....+\frac{3}{25.28}\right)\)
\(D=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{25}-\frac{1}{28}\right)\)
\(D=\frac{5}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{5}{3}.\frac{6}{28}=\frac{5}{14}\)
\(E=\frac{1}{1+2}+\frac{1}{1+2+3}+.......+\frac{1}{1+2+3+....+24}\)
Ta có: \(1+2=\)\(\frac{2.\left(2+1\right)}{2}=3\);\(1+2+3=\frac{3.\left(3+1\right)}{2}=6\);\(1+2+3+...+24=\frac{24.\left(24+1\right)}{2}=300\)
\(E=\frac{1}{3}+\frac{1}{6}+....+\frac{1}{300}\)
=>\(\frac{1}{2}E=\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{600}=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{24.25}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{24}-\frac{1}{25}=\frac{1}{2}-\frac{1}{25}=\frac{23}{50}\)
=>\(E=\frac{46}{50}\)
Vậy \(\frac{D}{E}=\frac{5}{14}:\frac{46}{50}=\frac{250}{644}=\frac{125}{322}\)
2) Theo t/c dãy tỉ số=nhau:
\(\frac{a+b}{a+c}=\frac{a-b}{a-c}=\frac{a+b-\left(a-b\right)}{a+c-\left(a-c\right)}=\frac{a+b-a+b}{a+c-a+c}=\frac{2b}{2c}=1\)
=>b=c
do đó \(A=\frac{10b^2+9bc+c^2}{2b^2+bc+2c^2}=\frac{10b^2+9b^2+b^2}{2b^2+b^2+2b^2}=\frac{\left(10+9+1\right).b^2}{\left(2+1+2\right).b^2}=4\)
