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3/ Chu vi hình chữ nhật:
\(\left(\dfrac{1}{4}+\dfrac{3}{10}\right)\cdot2=\dfrac{11}{10}\) (chưa biết đơn vị)
Diện tích hình chữ nhật:
\(\dfrac{1}{4}\cdot\dfrac{3}{10}=\dfrac{11}{20}\) (chưa biết đơn vị)
a, \(x+\dfrac{2}{3}=0,2\)
\(\Rightarrow x+\dfrac{2}{3}=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{5}-\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{-7}{10}\)
b, \(\dfrac{17}{7}-\dfrac{6}{5}x=\dfrac{17}{4}\)
\(\Rightarrow\dfrac{6}{5}x=\dfrac{17}{7}-\dfrac{17}{4}\)
\(\Rightarrow\dfrac{6}{5}x=\dfrac{-51}{28}\)
\(\Rightarrow x=\dfrac{-51}{28}:\dfrac{6}{5}\)
\(\Rightarrow x=\dfrac{-85}{56}\)
1,=0 . [2017/2018+2018/2019]
=>0
2,TH1 x-3=0=>x=3
TH2 y-4=0=>y=4
3, -2/4 = -x/10 = 16/y
=>-1/2 = -x/10 = 16/y
=>-1/2 = -x/10 => -5/10 = -x/10 => x=5
-1/2 = 16/y => 16/-32 = 16/y => y = -32
\(\left(\dfrac{2}{3}+x\right)\left(\dfrac{3}{5}-2x\right)=10\)
\(\Leftrightarrow\dfrac{2}{3}+x;\dfrac{3}{5}-2x\inƯ\left(10\right)\)
\(Ư\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)
\(\Leftrightarrow\dfrac{2}{3}+x=1\Rightarrow x=\dfrac{1}{3}\)
\(\dfrac{3}{5}-2x=10\Rightarrow2x=\dfrac{-47}{5}\Rightarrow x=\dfrac{-47}{10}\)
\(\Leftrightarrow\dfrac{2}{3}+x=-1\Rightarrow x=\dfrac{-5}{3}\)
\(\dfrac{3}{5}-2x=-10\Rightarrow2x=\dfrac{53}{5}\Rightarrow x=\dfrac{53}{10}\)
\(\Leftrightarrow\dfrac{2}{3}+x=2\Rightarrow x=\dfrac{4}{3}\)
\(\dfrac{3}{5}-2x=5\Rightarrow2x=\dfrac{-22}{5}\Rightarrow x=\dfrac{-11}{5}\)
\(\Leftrightarrow\dfrac{2}{3}+x=-2\Rightarrow x=\dfrac{-8}{3}\)
\(\dfrac{3}{5}-2x=-5\Rightarrow2x=\dfrac{28}{5}\Rightarrow x=\dfrac{14}{5}\)
\(\Leftrightarrow\dfrac{2}{3}+x=5\Rightarrow x=\dfrac{13}{3}\)
\(\dfrac{3}{5}-2x=2\Rightarrow2x=\dfrac{-7}{5}\Rightarrow x=\dfrac{-7}{10}\)
\(\Leftrightarrow\dfrac{2}{3}+x=-5\Rightarrow x=\dfrac{-17}{3}\)
\(\dfrac{3}{5}-2x=-2\Rightarrow2x=\dfrac{13}{5}\Rightarrow x=\dfrac{13}{10}\)
\(\Leftrightarrow\dfrac{2}{3}+x=10\Rightarrow x=\dfrac{28}{3}\)
\(\dfrac{3}{5}-2x=1\Rightarrow2x=\dfrac{-2}{5}\Rightarrow x=\dfrac{-1}{5}\)
\(\Leftrightarrow\dfrac{2}{3}+x=-10\Rightarrow x=\dfrac{-32}{3}\)
\(\dfrac{3}{5}-2x=-1\Rightarrow2x=\dfrac{8}{5}\Rightarrow x=\dfrac{4}{5}\)
Mk hơi băn khoăn chỗ này . Tại sao mà trong 1 trường hợp mà x lại khác nhau ?
\(c=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+...+\dfrac{1}{27.28.29.30}\)
\(3C=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\)
\(c=\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\)\(=\)\(\dfrac{1}{6}-\dfrac{1}{24360}\)
\(C=\) \(\dfrac{4059}{24360}\)
Ta có:
\(C=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+...+\dfrac{1}{27.28.29.30}\)
\(\Rightarrow3C=\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{27.28.29.30}\)
\(\Rightarrow3C=\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{27.28.29}-\dfrac{1}{28.29.30}\)
\(\Rightarrow3C=\dfrac{1}{1.2.3}-\dfrac{1}{28.29.30}\)
\(\Rightarrow3C=\dfrac{1353}{8120}\)
\(\Rightarrow C=\dfrac{1353}{\dfrac{8120}{3}}=\dfrac{451}{8120}\)
Vậy \(C=\dfrac{451}{8120}\)
a)A=3\(\dfrac{1}{11}\) x \(\dfrac{27}{46}\) x 1\(\dfrac{6}{17}\) x 2\(\dfrac{4}{9}\)
A=\(\dfrac{34}{11}\) x \(\dfrac{27}{46}\) x \(\dfrac{23}{17}\) x \(\dfrac{22}{9}\)
A=\(\dfrac{34\times27\times23\times22}{11\times46\times17\times9}\)
A=\(\dfrac{2\times3}{1}\)
A=6
mk cảm ơn bạn nhìu nha . bk có thể giải cho mk câu B đc ko