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\(1\frac{13}{15}.0,75-\left(\frac{8}{15}+25\%\right).\frac{24}{47}-3\frac{12}{13}:3\)
\(=\frac{28}{15}.\frac{3}{4}-\left(\frac{8}{15}+\frac{1}{4}\right).\frac{24}{47}-\frac{51}{13}:3\)
\(=\frac{7}{5}-\frac{47}{60}.\frac{24}{47}-\frac{17}{13}\)
\(=\frac{7}{5}-\frac{2}{5}-\frac{17}{13}\)
\(=\frac{-4}{13}\)
\(4\frac{1}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Leftrightarrow\frac{13}{3}.\frac{-1}{3}\le x\le\frac{2}{3}.\frac{-11}{12}\)
\(\Leftrightarrow\frac{-13}{9}\le x\le\frac{-11}{18}\)
\(\Leftrightarrow x=-1\)
a) Ta có:
\(x-\left\{\left[-x-\left(x+3\right)\right]-\left[\left(x+2018\right)-\left(x+2019\right)\right]+21\right\}\)
\(=x-\left\{\left[-x-x-3\right]-\left[x+2018-x-2019\right]+21\right\}\)
\(=x-\left\{\left[-2x-3\right]-\left[2018-2019\right]+21\right\}\)
\(=x+2x+-3+1-21\)
\(=3x-23\)
=> \(3x-23=2020\)
\(3x=2020+23=2043\)
=> \(x=2043:3=681\)
Nhầm
\(=x-\left\{-2x-3+1+21\right\}\\ =x+2x+3-1-21\)
\(=3x-17\\ =>3x-17=2020\\ 3x=2020+17=2037\\ x=2037:3=679\)
a) \(x=\frac{9}{10}\)
b) \(x=\frac{-4}{3}\)
c) \(x=\frac{1}{42}\)
d) \(x=\frac{-47}{10}\)
ko có thời gian nên mình chỉ cho đáp án thôi nhé
thông cảm cho mình ngen
đúng thì k đấy
chúc bạn học giỏi
a,-3/5.2/7+-3/7.3/5+-3/7
=-3/7.2/5+(-3/7).3/5+(-3/7)
=-3/7(2/5+3/5+1)
=-3/7.2
=-6/7
\(A=\left(1-\frac{1}{2010}\right)\left(1-\frac{2}{2010}\right)...\left(1-\frac{2010}{2010}\right)\left(1-\frac{2011}{2010}\right)\)
\(=\left(1-\frac{1}{2010}\right)\left(1-\frac{2}{2010}\right)...0\left(1-\frac{2011}{2010}\right)\)
\(=0\)
b)=(2/3 +2/7 - 2/28)/(-3/3 -3/7 + 3/28)
=[2(1/3+1/7-1/28)]/[(-3)(1/3+1/7-1/28)]
=2/-3
=-2/3
1.\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...........\frac{49}{50}=\frac{1}{50}\)
1/ \(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{49}{50}=\frac{1.\left(2.3.4...49\right)}{50.\left(2.3.4....49\right)}=\frac{1}{50}\)
2/ Chưa học dạng này:v
=1/1.2.3+1/2.3.4+1/3.4.5+...+1/10.11.12
=2/2.(1/1.2.3+1/2.3.4+1/3.4.5+...1/10.11.12
=1/2.(2/1.2.3+2/2.3.4+2/3.4.5+...+2/10.11.12)
=1/2.(1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+...1/10.11-1/11.12)
=1/2.(1/1.2-1/11.12)
=1/2.65/132
=65/264
2/Sửa đề thế này mới làm được: \(\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{10.11.12}\)
Nhận xét: \(\frac{1}{2.3.4}=\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right);...;\frac{1}{10.11.12}=\frac{1}{2}\left(\frac{1}{10.11}-\frac{1}{11.12}\right)\)
Suy ra \(A=\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{10.11.12}\)
\(=\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{10.11}-\frac{1}{11.12}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{11.12}\right)=\frac{7}{88}\)
(1-\(\frac{1}{2}\)) .( 1-\(\frac{1}{3}\)) . ( 1 - \(\frac{1}{4}\)) .........( 1 - \(\frac{1}{50}\))
+ Đầu tiên mình sẽ cho là 1 = \(\frac{2}{2}\)
=( \(\frac{2}{2}\)- \(\frac{1}{2}\)) . ( \(\frac{2}{2}\)- \(\frac{1}{3}\)) . ( \(\frac{2}{2}\)- \(\frac{1}{4}\)) ........( \(\frac{2}{2}\)- \(\frac{1}{50}\))
= \(\frac{1}{2}\). \(\frac{2}{3}\). \(\frac{3}{4}\). ...... . \(\frac{49}{50}\)
= \(\frac{1}{50}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{50}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{49}{50}\)
\(=\frac{1.2.3....49}{2.3.4......50}\)
\(=\frac{1}{50}\)