Tính:
a) \(\frac{\frac{4}{17}-\frac{4}{177}-\frac{4}{1779}}{\frac{5}{17}-\frac{5}{177}-\frac{5}{1779}}\)
b) \(\frac{1330}{1331}-\frac{7}{1.8}-\frac{19}{8.27}-.....-\frac{331}{1000.1331}\)
K
Khách
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Những câu hỏi liên quan
27 tháng 9 2016
\(\frac{\frac{4}{17}}{\frac{5}{17}}-\frac{\frac{4}{177}}{\frac{5}{177}}-\frac{\frac{4}{1779}}{\frac{5}{1779}}+\frac{131313}{151515}\)
\(=\)\(\frac{4}{1445}-\frac{4}{156645}-\frac{4}{15824205}\)\(+\frac{131313}{151515}\)
\(=\)\(0,8694090445\)
U
10 tháng 8 2018
\(B=-1\frac{1}{5}\cdot\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}}\div\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
\(B=\frac{-6}{5}\cdot4\div\frac{4\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
\(B=\frac{-24}{5}\div\frac{4}{5}\)
\(B=-6\)
10 tháng 8 2018
\(B=-1\frac{1}{5}.\frac{4.\frac{3}{7}}{\frac{3}{37}}:\frac{4+3.\frac{4}{1}}{5+3.\frac{5}{1}}\)
\(B=-\frac{6}{5}.\frac{148}{7}:\frac{4}{5}\)
\(B=-\frac{222}{7}\)
AK
31 tháng 5 2018
P/s : Đề của bạn sai nên mik đã sửa lại rồi
Ta có :
\(B=-1\frac{1}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}}:\frac{4+\frac{4}{17}+\frac{4}{19}+\frac{4}{2003}}{5+\frac{5}{17}+\frac{5}{19}+\frac{5}{2003}}\)
\(\Rightarrow B=-\frac{6}{5}.\frac{4\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}{1\left(3+\frac{1}{3}-\frac{3}{7}-\frac{3}{53}\right)}:\frac{4\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}{5\left(1+\frac{1}{17}+\frac{1}{19}+\frac{1}{2003}\right)}\)
\(\Rightarrow B=-\frac{6}{5}.4:\frac{4}{5}\)
\(\Rightarrow B=-\frac{24}{5}:\frac{4}{5}\)
\(\Rightarrow B=-\frac{24}{5}.\frac{5}{4}\)
\(\Rightarrow B=-6\)
Vậy \(B=-6\)
~ Ủng hộ nhé
KC
23 tháng 4 2018
\(\frac{\frac{4}{17}+\frac{4}{19}-\frac{4}{2111}}{\frac{5}{17}+\frac{5}{19}-\frac{5}{2111}}-\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{-\frac{5}{123}+\frac{5}{19}-\frac{5}{371}+1}\)
\(=\frac{4.\left(\frac{1}{17}+\frac{1}{19}-\frac{1}{2111}\right)}{5.\left(\frac{1}{17}+\frac{1}{19}-\frac{1}{2111}\right)}+\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{5.\left(\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}\right)}=\frac{4}{5}+\frac{1}{5}=1\)
29 tháng 6 2018
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
12 tháng 5 2022
Sửa đề; \(\dfrac{1}{5}\cdot\dfrac{4\left(3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}\right)}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2003}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2003}}\)
\(=\dfrac{1}{5}\cdot4:\dfrac{4}{5}=\dfrac{4}{5}\cdot\dfrac{5}{4}=1\)
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) \(\frac{\frac{4}{17}-\frac{4}{177}-\frac{4}{1779}}{\frac{5}{17}-\frac{5}{177}-\frac{5}{1779}}\)
\(=\frac{4\left(\frac{1}{17}-\frac{1}{177}-\frac{1}{1779}\right)}{5\left(\frac{1}{17}-\frac{1}{177}-\frac{1}{1779}\right)}\)
\(=\frac{4}{5}\)