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A=\(\frac{4}{11}-\frac{4}{13}+\frac{4}{13}-\frac{4}{15}+...+\frac{4}{99}-\frac{4}{101}\)
\(A=4\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=4.\left(\frac{1}{11}-\frac{1}{101}\right)\)
A=4. 90/1111=360/1111
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10 + (32 + 98) + (54 + 76) = 10 + 130 + 130 = 270
(12 + 78) + (36 + 54)+ 90 = 90 + 90 + 90 = 3 x 90 = 270
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35 × 11 = 385 1298 × 0 < 150
17 × 100 < 1800 54 × 72 = 72 × 54
1600 : 10 > 106 24 = 2400 : 100
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Ta có :
\(G=\frac{2}{11.13}+\frac{2}{13.15}+\frac{2}{15.17}+...+\frac{2}{97.99}+\frac{2}{99.101}\)
\(G=\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)
\(G=\left(\frac{1}{13}-\frac{1}{13}\right)+\left(\frac{1}{15}-\frac{1}{15}\right)+\left(\frac{1}{17}-\frac{1}{17}\right)+...+\left(\frac{1}{99}-\frac{1}{99}\right)+\left(\frac{1}{11}-\frac{1}{101}\right)\)
\(G=\frac{1}{11}-\frac{1}{101}\)
\(G=\frac{101}{1111}-\frac{11}{1111}\)
\(G=\frac{101-11}{1111}\)
\(G=\frac{90}{1111}\)
Vậy \(G=\frac{90}{1111}\)
Chúc bạn học tốt ~
\(G=\frac{2}{11\times13}+\frac{2}{13\times15}+\frac{2}{15\times17}+...+\frac{2}{97\times99}+\frac{2}{99\times101}\)
\(G=2\times\left(\frac{1}{11\times13}+\frac{1}{13\times15}+\frac{1}{15\times17}+...+\frac{1}{97\times99}+\frac{1}{99\times101}\right)\)
\(G=2\times\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\right)\)
( GẠCH BỎ CÁC PHÂN SỐ GIỐNG NHAU TRONG NGOẶC)
\(G=2\times\left(\frac{1}{11}-\frac{1}{101}\right)\)
\(G=2\times\frac{90}{1111}\)
\(G=\frac{180}{1111}\)
MK VIẾT ĐỀ BÀI NHƯ THẾ CÓ ĐÚNG KO BN!
MK CHỈ NGHĨ RA VẬY THÔI
CHÚC BN HỌC TỐT!!!!
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10+32+54+76+98=10+(32+98)+(54+76)=10+130+130=270
54+90+36+12+78=(54+36)+(12+78)+90=90+90+90=270
74+18+92+30+56=(74+56)+(18+92)+30=130+110+30=270
Vậy 10+32+54+76+98=54+90+36+12+78=74+18+92+30+56
\(\dfrac{15\times27-15\times17}{10\times64-10\times54}=\dfrac{15\times\left(27-17\right)}{10\times\left(64-54\right)}=\dfrac{15}{10}=\dfrac{3}{2}\)