\(\frac{5}{1x3}+\frac{5}{3x5}+\frac{5}{5x7}\)
\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}=5\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\right)\)
\(=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}\right)\)
\(=\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\right)\)
\(=\frac{5}{2}\left(1-\frac{1}{7}\right)=\frac{5}{2}\left(\frac{7}{7}-\frac{1}{7}\right)=\frac{5}{2}.\frac{6}{7}=\frac{15}{7}\)
1)\(\frac{4}{7}:\frac{-15}{28}x\left(-3\right)^2\) 2)\(\frac{15}{49}x1,4-\left(\frac{2}{3}+\frac{4}{5}\right);\frac{22}{10}\) 3)\(\frac{1}{4}-\frac{7}{4}:\left(-7\right)-3:\frac{3}{4}x\left(-2\right)^{^2}\)
4)\(\frac{5}{1x3}+\frac{5}{3x5}+\frac{5}{5x7}+....\frac{5}{197x199}\)
ok ai giải được giúp mik nha chiều mai mik phải nộp rồi
Tính nhanh hợp lí:
a)\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{600}\)
b)\(B=\frac{-5}{3}+\frac{-5}{15}+\frac{-5}{35}+...+\frac{-5}{2499}\)
{\(\frac{2}{15}+\frac{5}{3}-\frac{3}{5}\)} : {\(4\frac{2}{3}-2\frac{1}{5}\)}
{\(\frac{2}{1x3}+\frac{2}{3x5}.\).......\(+\frac{2}{19x21}\)} x {\(\frac{2}{21}x\frac{3}{8}-\frac{5}{2}:70\)}
{9/5-3/5} : {15/3-11/5}
6/5:14/5
6/5x5/14=3/7
Câu 2:
{2/1x3+2/3x5+...+2/19x21} x{1/28-5/2x1/70}
{2/1x3+2/3x5+...+2/19x21} x{1/28-1/28}
{2/1x3+2/3x5+...+2/19x21} x0=0
Tính:
a) \(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
b) \(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
a)\(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}=\frac{5}{3}\cdot\left(\frac{3}{1.4}+\frac{4}{4.7}+...+\frac{3}{100.103}\right)\)
\(=\frac{5}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}\cdot\left(1-\frac{1}{103}\right)=\frac{5}{3}\cdot\frac{102}{103}=\frac{170}{103}\)b)\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{1}{2}\cdot\frac{16}{51}=\frac{8}{51}\)
Câu a) bạn Ác Mộng làm rồi nên mình làm b) nha
b)Gọi A = \(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(2A=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\right)\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(2A=\frac{1}{3}-\frac{1}{51}\)
\(2A=\frac{16}{51}\)
\(A=\frac{16}{51}:2\)
\(A=\frac{8}{51}\)
5*(5-5/4+5/4-5/7+.......+5/100-5/103)
5*(5-5/103)
5*......... bạn tự tính nhé
câu b 1/3*5+1/5*7+............+1/49*51
1*(1/1-1/3+1/3-1/5+............+1/49-1/51)
1/1-1/51 tính ra rồi ra kết quả
tk nha
Tính nhanh hợp lí:
\(B=\frac{-5}{3}+\frac{-5}{15}+\frac{-5}{35}+....+\frac{-5}{2499}\)
Ta có:
B = \(\frac{-5}{3}+\frac{-5}{15}+\frac{-5}{35}+...+\frac{-5}{2499}.\)
= \(-5\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\right)\)
= \(-5\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\right)\)
= \(-5\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right):2\)
= \(\frac{-5}{2}\left(1-\frac{1}{51}\right)\)
= \(\frac{-5}{2}.\frac{50}{51}\) = -6375
Cho: S=\(\frac{3}{1x4}+\frac{3}{4x7}+\frac{3}{7x10}+...+\frac{3}{100x103}\). Chứng minh S<1
S=1/1-1/4+1/4+1/7-1/7+1/10+...+1/100-1/103
S=1/1-1/103
S=102/103
Vì 102/103<1 nên S<1
\(S=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{100\cdot103}\)
\(S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\)
\(S=1-\frac{1}{103}\)
\(S=\frac{102}{103}< 1\)
\(\frac{3}{1x4}+\frac{3}{4x7}+\frac{3}{7x10}+.......+\frac{3}{100x103}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}....+\frac{1}{100}-\frac{1}{103}\)
\(=\frac{1}{1}-\frac{1}{103}\)
=\(\frac{102}{103}\)
chứng minh rằng :
a) \(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\) ( n , a ϵ N* )
b) áp dụng câu a tính ;
\(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(B=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\)
\(C=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
a) \(\frac{1}{n}-\frac{1}{n+a}=\frac{\left(n+a\right)-n}{n\left(n+a\right)}=\frac{a}{a\left(n+a\right)}\) (đpcm)
b) \(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(B=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}.\left(1-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}=\frac{1}{3}-\frac{1}{51}=\frac{16}{51}\)
Tính nhanh hợp lí:
\(B=\frac{-5}{3}+\frac{-5}{15}+\frac{-5}{35}+...+\frac{-5}{2499}\)
\(B=\frac{-5}{3}+\frac{-5}{15}+\frac{-5}{35}+...+\frac{-5}{2499}\)
\(B=\frac{-5}{1.3}+\frac{-5}{3.5}+\frac{-5}{5.7}+...+\frac{-5}{49.51}\)
\(B=\frac{-5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(B=\frac{-5}{2}.\left[\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+...+\left(\frac{1}{49}-\frac{1}{51}\right)\right]\)
\(B=\frac{-5}{2}.\left[1-\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+...+\left(\frac{1}{49}-\frac{1}{49}\right)-\frac{1}{51}\right]\)
\(B=\frac{-5}{2}.\left(1-\frac{1}{51}\right)=\frac{-5}{2}.\frac{50}{51}=\frac{-125}{51}\)
=-5/1.3+-5/3.5+-5/5.7+.............+-5/49.51
-5/2(1-1/51)
=-125/51
ai k mk mk k lai
ok
B = -5/1.3 + -5/3.5 +.....+ -5/49.51
B = -5/2(1 - 1/51)
B = -5/2 . 50/51
B = -125/51 nha!
