1.

a. \(\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{99.100}\)

\(=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

\(=5.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=5.\left(1-\frac{1}{100}\right)\)

\(=5.\frac{99}{100}\)

\(=\frac{99}{20}\)

b. \(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{99.101}\)

\(=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)

\(=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{4}{2}.\left(1-\frac{1}{101}\right)\)

\(=2.\frac{100}{101}\)

\(=\frac{200}{101}\)

Đặt \(A=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{99.101}\)

\(\Rightarrow\frac{1}{2}A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(\Rightarrow\frac{1}{2}A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(\Rightarrow\frac{1}{2}A=1-\frac{1}{101}\)

\(\Rightarrow\frac{1}{2}A=\frac{100}{101}\)

\(\Rightarrow A=\frac{100}{101}.2\)

\(\Rightarrow A=\frac{200}{101}.\)

Câu hỏi của ✨♔♕ Saiko ♕♔✨ - Toán lớp 6 - Học toán với OnlineMath

a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

b) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=2.\left(1-\frac{1}{99}\right)\)

\(=2.\frac{98}{99}\)

\(=\frac{196}{99}=1\frac{97}{99}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(B=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)

\(=1-\frac{1}{99}\)

\(=\frac{98}{99}\)

A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)

=>\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

=>\(\frac{1}{1}-\frac{1}{100}\)

=>\(\frac{99}{100}\)

B=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{97.99}\)

=>\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\)

=>\(\frac{1}{1}-\frac{1}{99}\)

=>\(\frac{98}{99}\)

 nhung ma ko cothoi gian giai

\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)

\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)

\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)

\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)

làm tắt thế ai mà hỉu đc

em không biết 

Câu hỏi của ✨♔♕ Saiko ♕♔✨ - Toán lớp 6 - Học toán với OnlineMath

\(B=70\cdot\left(\frac{131313}{565656}+\frac{131313}{727272}+\frac{131313}{909090}\right)\)

\(B=70\cdot\left(\frac{13}{56}+\frac{13}{72}+\frac{13}{90}\right)\)

\(B=70\cdot\left[13\cdot\left(\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\right]\)

\(B=70\cdot\left[13\cdot\left(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\right)\right]\)

\(B=70\cdot\left[13\cdot\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\right]\)

\(B=70\cdot\left[13\cdot\left(\frac{1}{7}-\frac{1}{10}\right)\right]\)

\(B=70\cdot13\cdot\frac{3}{70}\)

\(B=70\cdot\frac{3}{70}\cdot13\)

\(B=3\cdot13\)

\(B=39\)

a) (-1)^a =1 với a chẵn, (-1)^a =-1 với a lẻ

\(A=\left(-1\right)^{1+2+3+4+..+2010+2011}=\left(-1\right)^{\frac{2011+1}{2}.2011}=\left(-1\right)^{1006.2011}=1\)

Vì 1006 là số chẵn => 1006.2011 là số chẵn

b) \(B=70.\left(\frac{13.10101}{56.10101}+\frac{13.10101}{72.10101}+\frac{13.10101}{90.10101}\right)=70.\left(\frac{13}{56}+\frac{13}{72}+\frac{13}{90}\right)=3.13=39\)

c) Áp dụng dãy tỉ số bằng nhau ta có:

\(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}=\frac{2a+3b+4c+5d}{3b+4c+5d+2a}=1\)

=> C=4

\(A=\left(-1\right)\cdot\left(-1\right)^2\cdot\left(-1\right)^3\cdot\left(-1\right)^4\cdot...\cdot\left(-1\right)^{2010}\cdot\left(-1\right)^{2011}\)

\(A=\left(-1\right)\cdot1\cdot\left(-1\right)\cdot1\cdot...\cdot1\cdot\left(-1\right)\)                \(\left(\text{Có 1006 số hạng }\left(-1\right)\right)\left(\text{Chẵn}\right)\) \(\Rightarrow\text{ }\left(-1\right)\cdot\left(-1\right)\cdot\left(-1\right)\cdot...\cdot\left(-1\right)=1\)

\(A=1\cdot1\cdot1\cdot...\cdot1\)

\(A=1\)

ai jup mih vs huhu

jup vs huhu mai phải nộp đề rồi

dễ mà cu

đợi đấy a lm cho

a. Đặt A \(=\left(-1\right).\left(-1^2\right).\left(-1^3\right)...\left(-1^{2010}\right).\left(-1^{2011}\right)\)

\(-1\)A\(=\).\(1.1^2.1^3.1^4....1^{2010}.1^{2011}\)

-1 A = 1

A = 1 : -1 = -1

Vậy A = -1

\(\frac{2a}{3b}.\frac{3b}{4c}.\frac{4c}{5d}.\frac{5d}{2a}=\frac{2a.3b.4c.5d}{2a.3b.4c.5d}=1\)

vì \(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}\Rightarrow\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}=-1;1\)

xét \(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}=-1\Rightarrow S=-1.4=-4\)

xét \(\frac{2a}{3b}=\frac{3b}{4c}=\frac{4c}{5d}=\frac{5d}{2a}=1\Rightarrow S=1.4=4\)

vậy S=-4;4

MÀ áp dụng tính chất tổng các số nghịch đảo

Phó GS; Tiến sĩ Vũ Hữu Bình ghi kết quả là 4 hoặc -4