\(\frac{\frac{4}{17}}{\frac{5}{17}}-\frac{\frac{4}{177}}{\frac{5}{177}}-\frac{\frac{4}{1779}}{\frac{5}{1779}}+\frac{131313}{151515}\)
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a) Ta có: \(\frac{\frac{4}{17}-\frac{4}{177}-\frac{4}{1779}}{\frac{5}{17}-\frac{5}{177}-\frac{5}{1779}}=\frac{4.\left(\frac{1}{7}-\frac{1}{177}-\frac{1}{1779}\right)}{5.\left(\frac{1}{7}-\frac{1}{177}-\frac{1}{1779}\right)}=\frac{4}{5}\)
b) \(\frac{1330}{1331}-\frac{7}{1.8}-\frac{19}{8.27}-.....-\frac{331}{1000.1331}\)
\(=\frac{1330}{1331}-\left(\frac{8-7}{1.8}+\frac{27-8}{8.27}+.....+\frac{1331-1000}{1000.1331}\right)\)
\(=\frac{1330}{1331}-\left(1-\frac{1}{8}+\frac{1}{8}-\frac{1}{27}+....+\frac{1}{1000}-\frac{1}{1331}\right)\)
\(=\frac{1330}{1331}-\left(1-\frac{1}{1331}\right)\)
\(=\frac{1330}{1331}-\frac{1330}{1331}=0\)
Vậy \(\frac{1330}{1331}-\frac{7}{1.8}-\frac{19}{8.27}-....\frac{331}{1000.1331}=0\)
CHÚC BẠN HỌC TỐT
a) \(\frac{\frac{4}{17}-\frac{4}{177}-\frac{4}{1779}}{\frac{5}{17}-\frac{5}{177}-\frac{5}{1779}}\)
\(=\frac{4\left(\frac{1}{17}-\frac{1}{177}-\frac{1}{1779}\right)}{5\left(\frac{1}{17}-\frac{1}{177}-\frac{1}{1779}\right)}\)
\(=\frac{4}{5}\)
\(\frac{3}{5}x-\frac{13}{9}:\left(\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}\right)=-10\)
<=> \(\frac{3}{5}x-\frac{13}{9}:\left(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}\right)=-10\)
<=> \(\frac{3}{5}x-\frac{13}{9}:13:\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}\right)=-10\)
<=> \(\frac{3}{5}x-\frac{1}{9}:\left(\frac{21}{315}+\frac{9}{315}+\frac{5}{315}\right)=-10\)
<=> \(\frac{3}{5}x-\frac{1}{9}:\frac{35}{315}=-10\)
<=> \(\frac{3}{5}x-\frac{1}{9}:\frac{1}{9}=-10\)
<=> \(\frac{3}{5}x-1=-10\)
<=> \(\frac{3}{5}x=-9\)
<=> \(x=-15\)
Vậy x = -15.
\(\frac{3}{5}x-1\frac{4}{9}:\left(\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}\right)=-10\)
\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}=-10\right)\)
\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left[\frac{13}{2}\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\right)\right]=-10\)
\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left[\frac{13}{2}\left(\frac{2}{3.5}+\frac{2}{.57}+\frac{2}{7.9}\right)\right]=-10\)
\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left[\frac{13}{2}\left(\frac{1}{3}-\frac{1}{9}\right)\right]=-10\)
\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\left(\frac{13}{2}.\frac{2}{9}\right)=-10\)
\(\Leftrightarrow\frac{3}{5}x-\frac{13}{9}:\frac{26}{18}=-10\)
\(\Leftrightarrow\frac{3}{5}x-1=-10\)
\(\Leftrightarrow\frac{3}{5}x=-10+1\)
\(\Leftrightarrow\frac{3}{5}x=-9\)
\(\Rightarrow x=-9:\frac{3}{5}\)
\(\Rightarrow x=-15\)
Vậy \(x=-15\)
1/3xD=1/(2x4)+1/(4x6)+...+1/(98x100)
2/3xD=2/(2x4)+2/(4x6)+...+1/(98x100)
2/3xD= 1/2-1/4+1/4-1/6+...+1/98-1/100
2/3xD=1/2-1/100
2/3xD=49/100
D=147/200
\(\frac{3}{2}.x-70\frac{10}{11}:\left(\frac{131313}{151515}+\frac{131313}{353535}+\frac{131313}{636363}+\frac{131313}{999999}=-5\right)\)
\(\frac{3}{2}.x-70\frac{10}{11}:\left(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\right)=-5\)
\(\frac{3}{2}.x-70\frac{10}{11}:\left[13\times\left(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}\right)\right]=-5\)
\(\frac{3}{2}.x-70\frac{10}{11}:\left[13\times\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)\right]=-5\)
\(\frac{3}{2}.x-70\frac{10}{11}:\left[13\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\right]=-5\)
\(\frac{3}{2}.x-70\frac{10}{11}:\left[13\times\left(\frac{1}{3}-\frac{1}{11}\right)\right]=-5\)
\(\frac{3}{2}.x-\frac{45}{2}=-5\)
\(\frac{3}{2}.x=\frac{35}{2}\)
\(x=\frac{35}{3}\)
Rút gọn biểu thức S, ta có:
\(S=\frac{13}{30}+\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)
\(\Leftrightarrow S=\frac{13}{30}+\left(\frac{13}{3\cdot5}+\frac{13}{5\cdot7}+\frac{13}{7\cdot9}+\frac{13}{9\cdot11}\right)\)
Đặt \(P=\frac{13}{3\cdot5}+\frac{13}{5\cdot7}+\frac{13}{7\cdot9}+\frac{13}{9\cdot11}\)
\(\Rightarrow P\cdot\frac{2}{13}=\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}\)
\(\Rightarrow P\cdot\frac{2}{13}=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+\frac{2}{9}-\frac{2}{11}\)
\(\Rightarrow P\cdot\frac{2}{13}=\frac{2}{3}-\frac{2}{11}\)
\(\Rightarrow P\cdot\frac{2}{13}=\frac{16}{33}\)
\(\Rightarrow P=\frac{104}{33}=3\frac{5}{33}\)
Ta có: \(P+1>P+\frac{13}{30}\)
Mà \(P+\frac{13}{30}=S\)
Còn \(P+1=3\frac{5}{33}+1=4\frac{5}{33}<5\)
\(\Rightarrow S<4\frac{5}{33}<5\)
Vậy đề bài sai.
\(\frac{\frac{4}{17}+\frac{4}{19}-\frac{4}{2111}}{\frac{5}{17}+\frac{5}{19}-\frac{5}{2111}}-\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{-\frac{5}{123}+\frac{5}{19}-\frac{5}{371}+1}\)
\(=\frac{4.\left(\frac{1}{17}+\frac{1}{19}-\frac{1}{2111}\right)}{5.\left(\frac{1}{17}+\frac{1}{19}-\frac{1}{2111}\right)}+\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{5.\left(\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}\right)}=\frac{4}{5}+\frac{1}{5}=1\)
Cho tam giác ABC có đường cao AD .Gọi E là trung điểm của AB .F đối xứng vs D qua E c/m AB = DF
\(\frac{\frac{4}{17}}{\frac{5}{17}}-\frac{\frac{4}{177}}{\frac{5}{177}}-\frac{\frac{4}{1779}}{\frac{5}{1779}}+\frac{131313}{151515}\)
\(=\)\(\frac{4}{1445}-\frac{4}{156645}-\frac{4}{15824205}\)\(+\frac{131313}{151515}\)
\(=\)\(0,8694090445\)