\(\left(3-\frac{9}{10}-\left|x+2\right|\right):\left(\frac{19}{10}-1-\frac{2}{5}\right)+\frac{4}{5}=1\)

\(\left(\frac{21}{10}-\left|x+2\right|\right):\frac{1}{2}+\frac{4}{5}=1\)

\(\left(\frac{21}{10}-\left|x+2\right|\right):\frac{1}{2}=\frac{1}{5}\)

\(\frac{21}{10}-\left|x+2\right|=\frac{1}{10}\)

=> |x+2| = 2

TH1: x + 2 = 2 => x = 0

TH2: x + 2 = -2 => x = -4

KL:...

a)\(\frac{X}{5}=\frac{5}{6}+\frac{-19}{30}\)

   \(\frac{X}{5}=\frac{1}{5}\)

Vậy \(X=1\)

b)\(X-\frac{41}{5}=\frac{-2}{3}\)

                   \(X=\frac{-2}{3}+\frac{41}{5}\)

                   \(X=\frac{113}{15}\)

Vậy \(X=\frac{113}{15}\)

c)\(\frac{31}{5}-X=\frac{11}{3}+\frac{7}{10}\)

    \(\frac{31}{5}-X=\frac{131}{30}\)

                   \(X=\frac{31}{5}-\frac{131}{30}\)

                    \(X=\frac{11}{6}\)

Vậy \(X=\frac{11}{6}\)

d)\(\frac{9}{X}=\frac{2}{5}+\frac{-7}{20}\)

   \(\frac{9}{X}=\frac{1}{20}\)

     \(X=9:\frac{1}{20}\)

     \(X=180\)

Vậy \(X=180\)

Hc tốt

câu b bài 2:

\(\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\)

\(=\dfrac{1}{5}\)

câu a bài 2:

\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{10\cdot11\cdot12}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{12}\)

\(=1-\dfrac{1}{12}=\dfrac{11}{12}\)

( 1 - 3/10 - x ) : ( 19/10 - 1 - 2/ 5 ) + 4/5 = 1 

=> (7/10 - x) : (9/10 - 2/5) = 1 - 4/5

=> (7/10 - x) : 1/2              = 3/5

=> (7/10 - x)                      = 3/5 : 1/2 

=> 7/10 - x                         = 6/5

=> x                                   = 7/10 - 6/5

=> x                                   = -1/2

ko ghi lại đề

=>(7/10-x):1/2+4/5=1

=>(7/10-x):1/2=1/5

=>7/10-x=1/10

=>x=3/5

Câu hỏi tương tự nha !!!!!

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