Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A em tự tính nhé
b) B = 1+ 3 + 32+...+399
3B = 3+ 32+33+...+3100
do đó 3B-B= (3+32+33+...+3100) - ( 1+3+32+...+399)
2B= 3100-1
B= (3100-1) : 2
c) \(C=1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x.\left(x+1\right)}\)
\(C=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}\)
\(C=1+\frac{1}{2}.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)\)
\(C=1+\frac{1}{2}.\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x.\left(x+1\right)}\right)\)
\(C=1+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)\)
\(C=1+\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{x+1}\right)\)
Phần c thế này thôi vì ko có giá trị x cụ thể .
d) \(D=\frac{9}{8}.\frac{16}{15}.\frac{25}{24}.....\frac{8100}{8099}\)
\(D=\frac{9.16.25....8100}{8.15.24....8099}\)
\(D=\frac{3.3.4.4.5.5....90.90}{2.4.3.5.4.6.....89.91}\)
\(D=\frac{\left(3.4.5...90\right).\left(3.4.5...90\right)}{\left(2.3.5...89\right).\left(4.5.6...91\right)}\)
\(D=\frac{3.4.5...90}{2.3.4...89}.\frac{3.4.5...90}{4.5.6...91}\)
\(D=\frac{90}{2}.\frac{3}{91}\)
\(D=45.\frac{3}{91}=\frac{135}{91}\)
a)A=1/10+1/15+...+1/120
=2(1/20+1/30+...+1/240)
=2(1/4*5+1/5*6+...+1/15*16)
=2*(1/4-1/5+1/5-1/6+...+1/15-1/16)
=2*[(1/4-1/16)+(1/5-1/5)+...+(1/15-1/15)]
=2*[(4/16-1/16)+0+...+0]
=2*3/16=3/8
b) B=1+1/3+1/6+...+1/1225
=2(1/2+1/6+1/12+...+1/2450)
=2(1/1*2+1/2*3+...+1/49*50)
=2*[1-1/2+1/2-1/3+...+1/49-1/50]
=2*[(1-1/50)+(1/2-1/2)+...+(1/49-1/49)]
=2*[(50/50-1/50)+0+...+0]
=2*49/50=49/25
a,\(\frac{1}{2}A=\frac{1}{2}\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\right)\)
\(\frac{1}{2}A=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\)
\(\frac{1}{2}A=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\)
\(\frac{1}{2}A=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\)
\(\frac{1}{2}A=\frac{1}{4}-\frac{1}{16}\)\(\frac{1}{2}A=\frac{3}{16}\)suy ra \(A=\frac{3}{16}:\frac{1}{2}=\frac{3}{8}\)
B thì cậu có thể làm nhiều cách
1) a) A=\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{3}-\frac{1}{8}=\frac{5}{24}\)
c) C=\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(C=1-\frac{1}{101}\)
\(C=\frac{100}{101}\)
d) Sửa đề: thay \(\frac{3}{92.98}\)=\(\frac{3}{92.95}\)
\(D=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{92}-\frac{1}{95}\)
\(D=\frac{1}{2}-\frac{1}{95}\)
\(D=\frac{95-2}{190}=\frac{93}{190}\)
Các bài trên áp dụng theo tính chất: \(\frac{a}{b\left(b+a\right)}\frac{1}{b}-\frac{1}{b+a}\)
\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
\(A=2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)\)
\(A=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)
\(A=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(A=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(A=2.\frac{3}{16}=\frac{3}{8}\)
\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
\(=2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)\)
\(=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(=2.\frac{3}{16}=\frac{3}{8}\)
c) \(A=\frac{6}{4}+\frac{6}{28}+\frac{6}{70}+\frac{6}{130}+\frac{6}{208}\)
\(=\frac{6}{1.4}+\frac{6}{4.7}+\frac{6}{7.10}+\frac{6}{10.13}+\frac{6}{13.16}\)
\(=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\right)\)
\(=2\left(1-\frac{1}{16}\right)\)
\(=2.\frac{15}{16}\)
\(=\frac{15}{8}\)
Vậy A=\(\frac{15}{8}\)
a) \(\frac{3^2}{1.4}+\frac{3^2}{4.7}+\frac{3^2}{7.10}+...+\frac{3^2}{97.100}\)
\(=3\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}\right)\)
\(=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(=3\left(1-\frac{1}{100}\right)\)
\(=3.\frac{99}{100}=\frac{297}{100}\)
\(A=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{120}\)
\(A=\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+...+\frac{2}{240}\)
\(A=2.\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{240}\right)\)
\(A=2.\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\right)\)
\(A=2.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(A=2.\left(\frac{1}{4}-\frac{1}{16}\right)\)
\(A=2.\frac{3}{16}\)
\(A=\frac{3}{8}\)
\(B=\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+...+\frac{2}{399}\)
\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{19.21}\)
\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{19}-\frac{1}{21}\)
\(B=\frac{1}{3}-\frac{1}{21}\)
\(B=\frac{2}{7}\)
#)Trả lời :
\(A=\frac{\left(140+70+42+28+20+15\right)}{420}\)
\(A=\frac{315}{420}=\frac{\left(315:105\right)}{\left(420:105\right)}=\frac{3}{4}\)
Vậy : \(A=\frac{3}{4}\)
#~Will~be~Pens~#