\(\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)
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Ta có :
\(\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)
\(=\frac{1}{1.2}+\frac{1}{2.7}+\frac{1}{7.5}+\frac{1}{5.13}+\frac{1}{13.8}+\frac{1}{8.19}\)
Giá trị không đổi khi cả tử và mẫu cùng nhân với 2, ta được :
\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+\frac{2}{13.16}+\frac{2}{16.19}\)
\(=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{16}-\frac{1}{19}\right)\)
\(=\frac{2}{3}.\left(1-\frac{1}{19}\right)=\frac{2}{3}.\frac{18}{19}=\frac{12}{19}\)
\(A=\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}=\frac{1}{2}.\left(\frac{1}{4}+\frac{1}{28}+\frac{1}{70}+\frac{1}{130}+\frac{1}{208}+\frac{1}{304}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{19}\right)=\frac{9}{19}\)
N = 2/4+2/28+2/70+2/130+2/208+2/304
N = 2/1.4+2/4.7+2/7.10+2.......
C=\(\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)
trình bày mới tk
\(C=\frac{1}{2}+\frac{1}{14}+\frac{1}{35}+\frac{1}{65}+\frac{1}{104}+\frac{1}{152}\)
\(C=\frac{1}{1.2}+\frac{1}{2.7}+\frac{1}{7.5}+\frac{1}{5.13}+\frac{1}{13.8}+\frac{1}{8.19}\)
\(C=\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+\frac{2}{13.16}+\frac{2}{16.19}\)
\(C=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}\right)\)
\(C=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}\right)\)
\(C=\frac{2}{3}.\left(1-\frac{1}{19}\right)\)
\(C=\frac{2}{3}.\frac{18}{19}=\frac{12}{19}\)
M=1/2+1/2.7+1/7.5+1/5.13+1/13.8+1/8.19
M=1/2-1/2+1/7-1/7+1/5-1/5+1/13-1/13+1/8-1/8+1/19
M=1/2-1/19
M=17/38
C= 1/2 + 1/14 + 1/35+1/65+1/104+1/152 = 12/19 nhé bn
Nhân cả tử cả mẫu của các phân số trong A lên 2 ta có:
A=\(\frac{2}{4}+\frac{2}{28}+\frac{2}{70}+\frac{2}{130}+\frac{2}{208}+\frac{2}{304}\)
=\(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+\frac{2}{10.13}+\frac{2}{13.16}+\frac{2}{16.19}\)
=\(\frac{2}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+\frac{3}{16.19}\right)\)
=\(\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}\right)\)
=\(\frac{2}{3}.\left(1-\frac{1}{19}\right)\)
=\(\frac{2}{3}.\frac{18}{19}\)
=\(\frac{12}{19}\)
đề kiu tính hợp lý hay tính bình thường
12/19