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\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)
\(A=\frac{1}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{16}-\frac{1}{19}\right)\)
\(A=\frac{1}{3}\cdot\left(1-\frac{1}{19}\right)\)
\(A=\frac{1}{3}\cdot\frac{18}{19}=\frac{6}{19}\)
\(B=\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)
\(B=\frac{1}{4\cdot8}+\frac{1}{8\cdot12}+\frac{1}{12\cdot16}+\frac{1}{16\cdot20}+\frac{1}{20\cdot24}\)
\(B=\frac{1}{4}\cdot\left(\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+\frac{1}{12}-\frac{1}{16}+\frac{1}{16}-\frac{1}{20}+\frac{1}{20}-\frac{1}{24}\right)\)
\(B=\frac{1}{4}\cdot\left(\frac{1}{4}-\frac{1}{24}\right)\)
\(B=\frac{1}{4}\cdot\frac{5}{24}=\frac{5}{96}\)
\(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}\)
\(A=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{16}-\frac{1}{19}\right)\)
\(A=\frac{1}{3}\left(1-\frac{1}{19}\right)\)
\(A=\frac{1}{3}.\frac{18}{19}\)
\(A=\frac{6}{19}\)
\(B=\frac{1}{32}+\frac{1}{96}+\frac{1}{192}+\frac{1}{320}+\frac{1}{480}\)
\(B=\frac{1}{4.8}+\frac{1}{8.12}+\frac{1}{12.16}+\frac{1}{16.20}+\frac{1}{20.24}\)
\(B=\frac{1}{4}\left(\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{12}+...+\frac{1}{20}-\frac{1}{24}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{4}-\frac{1}{24}\right)\)
\(B=\frac{1}{2}.\frac{5}{24}\)
\(B=\frac{5}{48}\)
đây là toán lớp 5 cơ mà
a)A=\(\frac{1}{1x4}\)+\(\frac{1}{4x7}\)+...+\(\frac{1}{16x19}\)
A=\(\frac{1}{3}\)x3x(\(\frac{1}{1.4}\)+\(\frac{1}{4.7}\)+.......+\(\frac{1}{16.19}\)
A=\(\frac{1}{3}\)x(\(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+............+\(\frac{3}{16.19}\))
A=\(\frac{1}{3}\)x(1-1/4+1/4-1/7+......+1/13-1/16+1/16-1/19)
A=\(\frac{1}{3}\)x(1-\(\frac{1}{19}\))
A=\(\frac{1}{3}\)x\(\frac{18}{19}\)
A=\(\frac{6}{19}\)
\(B=2016:\left(\frac{0.4-\frac{2}{9}+\frac{2}{11}}{1.4-\frac{7}{9}+\frac{7}{11}}.\frac{-1\frac{1}{6}+0.875-0.7}{\frac{1}{3}-0.25+\frac{1}{5}}\right)\)
<=>\(B=2016:\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}.\frac{\frac{-7}{6}+\frac{7}{8}-\frac{7}{10}}{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}\right)\)
<=>\(B=2016:\left(\frac{2.\left(\frac{1}{5}.\frac{1}{9}.\frac{1}{11}\right)}{5.\left(\frac{1}{5}.\frac{1}{9}.\frac{1}{11}\right)}.\frac{\frac{7}{6}-\frac{7}{8}-\frac{7}{10}}{\frac{2}{6}-\frac{2}{8}-\frac{2}{10}}\right)\)
<=>\(B=2016:\left(\frac{2}{5}.\frac{7.\left(\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\right)}{2.\left(\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\right)}\right)\)
<=>\(B=2016:\left(\frac{2}{5}.\frac{7}{2}\right)\)
<=>\(B=2016:\frac{7}{5}\)
<=>\(B=2016.\frac{5}{7}\)
<=>\(B=1440\)
Vậy B=1440
k cho mink nha
a)
\(A=\left(\frac{1}{9}-\frac{1}{10}\right)-\left(\frac{1}{8}-\frac{1}{9}\right)-....-\left(1-\frac{1}{2}\right)=\frac{1}{9}-\frac{1}{10}-\frac{1}{8}+\frac{1}{9}-....-1+\frac{1}{2}\)
\(A=-\left(\frac{1}{10}+1\right)=-\frac{11}{10}\)
a)\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\\ \Rightarrow A=-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}-\frac{1}{90}\\ \Rightarrow A=-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)Đặt \(B=\frac{1}{2}+\frac{1}{6}+...+\frac{1}{72}+\frac{1}{90}\)
\(\Rightarrow B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)
\(\Rightarrow B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow B=1-\frac{1}{10}=\frac{9}{10}\)
Ta có : \(A=-B\)
\(\Rightarrow A=-\frac{9}{10}\)
a, \(\frac{1}{1.4}\)+\(\frac{1}{4.7}\)+......+\(\frac{1}{97.100}\)= |\(\frac{x}{3}\)|
\(\Rightarrow\)\(\frac{1}{3}\) ( \(\frac{3}{1.4}\)+\(\frac{3}{4.7}\)+.......+\(\frac{3}{97.100}\))= |\(\frac{x}{3}\)|
\(\Rightarrow\)\(\frac{1}{3}\) ( 1 - \(\frac{1}{4}\)+ \(\frac{1}{4}\)-\(\frac{1}{7}\)+......+\(\frac{1}{97}\)-\(\frac{1}{100}\)) = |\(\frac{x}{3}\)|
\(\Rightarrow\)\(\frac{1}{3}\) ( 1-\(\frac{1}{100}\)) = |\(\frac{x}{3}\)|
\(\Rightarrow\)\(\frac{1}{3}\) . \(\frac{99}{100}\) = |\(\frac{x}{3}\)|
\(\Rightarrow\)\(\frac{33}{100}\) = |\(\frac{x}{3}\)|
\(\Rightarrow\)\(\frac{x}{3}\)= \(\orbr{\begin{cases}\frac{33}{100}\\\frac{-33}{100}\end{cases}}\)
Với \(\frac{x}{3}\) = \(\frac{33}{100}\)
\(\Rightarrow\)100x= 33.3
\(\Rightarrow\)100x=99
\(\Rightarrow\)x=\(\frac{99}{100}\)
Với \(\frac{x}{3}\)=\(\frac{-33}{100}\)
\(\Rightarrow\)100x=-33.3
\(\Rightarrow\)100x=-99
\(\Rightarrow\)x=\(\frac{-99}{100}\)
Vậy x=\(\orbr{\begin{cases}\frac{99}{100}\\\frac{-99}{100}\end{cases}}\)
b, \(\frac{4}{1.5}\)+ \(\frac{4}{5.9}\)+......+ \(\frac{4}{97.101}\)= |\(\frac{5x-4}{101}\)|
\(\Rightarrow\)1-\(\frac{1}{5}\)+\(\frac{1}{5}\)-\(\frac{1}{9}\)+......+\(\frac{1}{97}\)-\(\frac{1}{101}\)= |\(\frac{5x-4}{101}\)|
\(\Rightarrow\)1-\(\frac{1}{101}\)= |\(\frac{5x-4}{101}\)
\(\Rightarrow\) \(\frac{100}{101}\)= |\(\frac{5x-4}{101}\)|
\(\Rightarrow\)\(\frac{5x-4}{101}\) =\(\orbr{\begin{cases}\frac{100}{101}\\\frac{-100}{101}\end{cases}}\)
Với \(\frac{5x-4}{101}\) =\(\frac{100}{101}\)
\(\Rightarrow\)(5x-4).101=100.101
\(\Rightarrow\)505x-404=10100
\(\Rightarrow\)505x=10504
\(\Rightarrow\)x=\(\frac{104}{5}\)
Với \(\frac{5x-4}{101}\)=\(\frac{-100}{101}\)
\(\Rightarrow\)(5x-4). 101=-100.101
\(\Rightarrow\)505x-404=-10100
\(\Rightarrow\)505x=-9696
\(\Rightarrow\)x=\(\frac{-96}{5}\)
Vậy x=\(\orbr{\begin{cases}\frac{104}{5}\\\frac{-96}{5}\end{cases}}\)
bạn ơi như là cô giáo cho đề sai rồi kết quả phải là \(\frac{375}{376}\)thì mới giải được
Ta có:
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{125}{376}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{125}{376}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{x+3}=\frac{125}{376}:\frac{1}{3}=\frac{375}{376}\)
\(\Rightarrow\frac{1}{x+3}=1-\frac{375}{376}=\frac{1}{376}\Leftrightarrow x+3=376\Leftrightarrow x=373\)
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{125}{376}\)
\(3.\left(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}\right)=3.\frac{125}{376}\)
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{375}{376}\)
\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{375}{376}\)
\(1-\frac{1}{x+3}=\frac{375}{376}\)
\(\frac{x+2}{x+3}=\frac{375}{376}\)
=> x + 2 = 375
=> x = 375 - 2
=> x = 373
B=1/1x4+1/4x3+1/3x8+...+1/7x16+1/16x9+1/9x20
2B=2x(1/4+1/12+1/24+...+1/112+1/144+1/180
2B=2/8+2/24+2/48+...+2/224+2/288+2/360
2B=2/2x4+2/4x6+2/6x8+...+2/14x16+2/16x18+2/18x20
2B=1/2-1/4+1/4-1/6+1/6-1/8+...+1/14-1/16+1/16-1/18+1/18-1/20
2B=1/2-1/20
2B=9/20
B=9/20:2
B=9/40