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A=\([\)\(\frac{2}{7}\)\(\times\)(\(\frac{1}{4}-\frac{1}{3}\))\(]\)\(\div\)\([\)(\(\frac{2}{7}\times\)(\(\frac{3}{9}-\frac{2}{5}\))\(]\)
=(\(\frac{2}{7}\times\)\(\frac{-1}{12}\))\(\div(\)\(\frac{2}{7}\times\)\(\frac{-1}{15}\))
=\(\frac{-1}{42}\)\(\div\)\(\frac{-2}{35}\)
=\(\frac{-1}{42}\)\(\times\)\(\frac{35}{-2}\)
=\(\frac{5}{12}\)
a) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2-\frac{11}{5}:\frac{-11}{5}=\left(-\frac{1}{10}\right)^2+1=1\frac{1}{100}\)
b) \(\left(-\frac{5}{7}\right)^2+8.\left(0,5\right)^2+\left(-1\right)^{2010}=\frac{25}{49}+2+1=3\frac{25}{49}\)
c) \(\frac{9999^2}{3333^2}+\left(0,5\right)^2.\left(-2\right)^4-\left(-\frac{4}{3}\right)^2=9+1-\frac{16}{9}=8\frac{2}{9}\)
d) \(\left|-\frac{2}{5}+\frac{1}{7}\right|:\frac{-3}{35}+\frac{-3}{7}.\frac{7}{5}=\frac{9}{35}.\frac{35}{-3}-\frac{3}{5}=-3\frac{3}{5}\)
e) \(\frac{1}{2}-\left(-0,4\right)+\frac{1}{3}+\frac{1}{5}-\frac{-1}{6}+\frac{-4}{35}+\frac{1}{41}\)
\(=\frac{1}{2}+\frac{2}{5}+\frac{1}{3}+\frac{1}{5}+\frac{1}{6}-\frac{4}{35}+\frac{1}{41}=1\frac{732}{1435}\)
\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
\(=\frac{1\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}{4.\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}=1\)
\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}\)
=1
\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.\frac{5}{12}.....\frac{30}{62}.\frac{31}{64}=2^x\)
=>\(\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}.\frac{4}{2.5}.\frac{5}{2.6}....\frac{30}{2.31}.\frac{31}{2.32}=2^x\)
=>\(\frac{1.2.3.4.5....30.31}{2.2.2.3.2.4.2.5.2.6...2.31.2.32}=2^x\)
=>\(\frac{2.3.4.5...30.31}{2^{31}.32.\left(2.3.4.5...31\right)}=2^x\)
=>\(\frac{1}{2^{31}.2^5}=2^x\)
=>\(\frac{1}{2^{36}}=2^x\)
=> x=36
Vậy x=36
Chúc bn học tốt nhé!
a) \(\frac{3}{5}.x-\frac{1}{5}=\frac{4}{5}\)
\(\Leftrightarrow\frac{3}{5}.x=\frac{4}{5}+\frac{1}{5}\)
\(\Leftrightarrow\frac{3}{5}.x=1\)
\(\Leftrightarrow x=1:\frac{3}{5}\)
\(\Leftrightarrow x=\frac{5}{3}\)
Vậy : \(x=\frac{5}{3}\)
b) \(\frac{4}{7}+\frac{5}{7}:x=1\)
\(\Leftrightarrow\frac{5}{7}:x=1-\frac{4}{7}\)
\(\Leftrightarrow\frac{5}{7}:x=\frac{3}{7}\)
\(\Leftrightarrow x=\frac{5}{7}:\frac{3}{7}\)
\(\Leftrightarrow x=\frac{5}{3}\)
Vậy : \(x=\frac{5}{3}\)
c) \(-\frac{12}{7}.\left(\frac{3}{4}-x\right).\frac{1}{4}=-1\)
\(\Leftrightarrow\frac{-12.1}{7.4}.\left(\frac{3}{4}-x\right)=-1\)
\(\Leftrightarrow-\frac{3}{7}.\left(\frac{3}{4}-x\right)=-1\)
\(\Leftrightarrow\frac{3}{4}-x=-1:\left(-\frac{3}{7}\right)\)
\(\Leftrightarrow\frac{3}{4}-x=\frac{7}{3}\)
\(\Leftrightarrow x=\frac{3}{4}-\frac{7}{3}=-\frac{19}{12}\)
Vậy : \(x=-\frac{19}{12}\)
d) \(x:\frac{17}{8}=-\frac{2}{5}.-\frac{9}{17}+3\)
\(\Leftrightarrow x:\frac{17}{8}=\frac{273}{85}\)
\(\Leftrightarrow x=\frac{273}{85}.\frac{17}{8}\)
\(\Leftrightarrow x=\frac{273}{40}\)
Vậy : \(x=\frac{273}{40}\)
\(\)
\(=\frac{11}{-5}\cdot\frac{-9}{11}\cdot\frac{15}{-14}\cdot\frac{2}{5}+-\frac{2}{77}\cdot\frac{5}{-3}\)
\(=\frac{9}{5}\cdot-\frac{15}{14}\cdot\frac{2}{5}+\frac{10}{231}\)
\(=-\frac{841}{1155}\)
Bài làm :
a)\(=-\frac{3}{5}+\frac{28}{5}\times\frac{9}{14}=-\frac{3}{5}+\frac{18}{5}=3\)
b)\(=\frac{55}{126}+\frac{5}{42}+\frac{4}{9}=1\)
c)\(=-\frac{51}{13}-\frac{27}{13}=-6\)
d)\(=\frac{7}{3}-11\frac{1}{4}\times\frac{2}{15}=\frac{7}{3}-\frac{3}{2}=\frac{5}{6}\)
e)\(=1\times\frac{8}{3}\times0,25=\frac{2}{3}\)
gợi ý:
Bước 1: tính bên trái và bên phải của x
Bước 2:lập luận ra để bài khó thành cơ bản
Bước 3:bn chỉ cần ghi kết quả
chú ý: Bài này lớp 6 hk nhiều r mak
chúc thành công
Nhân vô rồi chuyển dấu lên và nhóm nhân -1ra ngoài rồi trg ngoặc là dãy có quy luật giải dãy đó r nhân phá ngoặc