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Bài làm

    \(\frac{\left(9,45-3,2\right).4:0,2}{\left(0,297+0,721\right).625}\)

\(=\frac{6,25.20}{1,018.625}\)

\(=\frac{6,25.4}{1,018.125}\)

~ Chỗ này chắc bn cũng biết tính nôits như thế nào nhỉ, bạn tự động não làm nốt nhé. ~
# Học tốt #

16 tháng 8 2017

Ta có:

\(A=\left(x-\frac{1}{2}\right).\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)=\frac{1}{3}\)

\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{1}{3}\)

\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{3}\)

\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{10}\right)=\frac{1}{3}\)

\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\frac{9}{10}=\frac{1}{3}\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}.\frac{10}{9}\Leftrightarrow x=\frac{47}{54}\)

\(B=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{96.101}=\frac{1}{10.x}\)

\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\right)=\frac{1}{10}-\frac{1}{x}\)

\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\)

\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\Leftrightarrow B=\frac{1}{5}.\frac{100}{101}=\frac{1}{10}-\frac{1}{x}\)

\(\Leftrightarrow B=\frac{1}{x}=\frac{1}{10}-\frac{20}{101}=-\frac{99}{1010}\Leftrightarrow x=-\frac{1010}{99}\)

16 tháng 8 2017

c) Sai đề nhé bạn vì không có kết quả nên không tìm được x.

d) \(\left(x-5\right).\left(10-9\frac{40}{41}\right):\left(1-\frac{81}{82}\right):\left(1-\frac{204}{205}\right)=2050\)

\(\Rightarrow\left(x-5\right).\frac{1}{41}.82.205=2050\)

\(\Rightarrow\left(x-5\right).2.205=2050\Leftrightarrow x-5=2050:410=5\Leftrightarrow x=10\)

3 tháng 3 2016

Ta có:1/2-1/3+1/3-1/4+...+1/a-1/(a+1)

=>1/2-1/(a+1)

22 tháng 6 2017

a)\(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)

\(=\frac{13}{3.5}+\frac{13}{5.7}+\frac{13}{7.9}+\frac{13}{9.11}\)

\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=\frac{13}{2}\cdot\frac{8}{33}\)

\(=\frac{52}{33}\)

22 tháng 6 2017

a) Đặt A= 13/15 + 13/35 + 13/63 + 13/99

A = 13/2 ( 2/15 + 2/35 + 2/63 + 2/99)

A= 13/2 ( 2/ 3.5 + 2/5.7 + 2/7.9 + 2/9.11)

A= 13/2 ( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11)

A= 13/2 ( 1/3 - 1/11) 

A= 13/2 . 8/33

A= 52/33  

20 tháng 8 2016

b)\(\left(2016.1017+2017.2018\right).\left(1+\frac{1}{2}:\frac{3}{2}-\frac{4}{3}\right)\)

\(\left(2016.2017+2017.2018\right)\left(1+\frac{1}{3}-\frac{4}{3}\right)\)

\(\left(2016.2017+2017.2018\right).\left(\frac{4}{3}-\frac{4}{3}\right)\)

\(\left(2016.2017+2017.2018\right).0\)

\(=0\)

20 tháng 8 2016

a) \(1001.789+456.128.128-789+912.436\)

\(=\left(1001.789-789\right)+\left(456.2.64.128+912.436\right)\)

\(=789.1000+912.4\left(2048+109\right)\)

\(=789000+912.4.2157\)

\(=8657736\)

12 tháng 6 2018

\(a,\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2017\cdot2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=1-\frac{1}{2018}\)

\(=\frac{2017}{2018}.\)

\(b,\left[x\cdot\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)

\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}:\frac{9}{4}\)

\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{7}{2}\cdot\frac{4}{9}\)

\(\Leftrightarrow\left[x\cdot\frac{5}{3}-1\right]:9=\frac{14}{9}\)

\(\Leftrightarrow x\cdot\frac{5}{3}-1=\frac{14}{9}\cdot9\)

\(\Leftrightarrow x\cdot\frac{5}{3}-1=14\)

\(\Leftrightarrow x\cdot\frac{5}{3}=14+1\)

\(\Leftrightarrow x\cdot\frac{5}{3}=15\)

\(\Leftrightarrow x=15:\frac{5}{3}\)

\(\Leftrightarrow x=15\cdot\frac{3}{5}\)

\(\Leftrightarrow x=9.\)

12 tháng 6 2018

a)\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2017.2018}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\)

\(=\frac{1}{1}-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

b)\(\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:2,25\)

\(\Leftrightarrow\left[x.\frac{5}{3}-1\right]:9=3\frac{1}{2}:\frac{9}{4}=1\frac{5}{9}\)

\(\Rightarrow x.\frac{5}{3}-1=1\frac{5}{9}.9=14\)

\(\Rightarrow x.\frac{5}{3}=14+1=15\)

\(\Rightarrow x=15:\frac{5}{3}=9\)