a) \(19\times\left(52+38\right)-9\times\left(52+38\right)=\)

\(=\)     \(19\times90-9\times90\) 

\(=90\times\left(19-9\right)\) 

\(=90\times10\) 

\(=900\) 

b) \(\dfrac{2}{5}\div\dfrac{3}{7}\times\dfrac{3}{7}\div\dfrac{2}{5}+2019=\)

\(=\left(\dfrac{2}{5}\div\dfrac{2}{5}\right)\times\left(\dfrac{3}{7}\div\dfrac{3}{7}\right)+2019\) 

\(=1\times1+2019\) 

\(=2019\)

37 x 18 - 9 x 14 + 100

=37*2*9-9*14+100

=74*9-9*14+100

=9*(74-14)+100

=9*60+100

=540+100

=640

15 x 2 + 15 x 3 + 15 x 5

=15*(2+3+5)

=15*10

=150

52 + 37 + 48 + 63

=(52+48)+(37+63)

=100+100

=200

37 x 18 - 9 x 14 + 100 = 37 x 2 x 9 - 9 x 14 + 100

= 74 x 9 - 9 x 14 + 100

= (74 - 14) x 9 +100

= 60 x 9 + 100

= 540 + 100

= 550

15 x 2 + 15 x 3 + 15 x 5 = 15 x ( 2 + 3 +5 )

= 15 x 10

= 150

52 + 37 + 48 + 63 = (52 + 48) + (37 + 63)

= 100 + 100

= 200

Duyệt!!!

19 x ( 52 + 38 ) - 9 x ( 52 + 38 )

= 19 x 90 - 9 x 90

= ( 19 - 9 ) x 90

= 10 x 90

= 900

\(\dfrac{2}{5}:\dfrac{3}{7}\times\dfrac{3}{7}:\dfrac{2}{5}+2019=\left(\dfrac{2}{5}:\dfrac{2}{5}\right)\times\left(\dfrac{3}{7}:\dfrac{3}{7}\right)+2019=1\times1+2019=1+2019=2020\)

x = 23                                         

x + 3 + x + 5 + x + 7 - x - 9 = 52

2x + 6 = 52

2x = 46

x = 23

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c) \(\left|x\right|=3,5\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-3,5\end{matrix}\right.\)

d) \(\left|x\right|=-2,7\Rightarrow x\in\varnothing\) 

l) \(\left|x+\dfrac{3}{4}\right|-5=-2\Rightarrow\left|x+\dfrac{3}{4}\right|=3\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=3\\x+\dfrac{3}{4}=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3-\dfrac{3}{4}\\x=-3-\dfrac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{15}{4}\end{matrix}\right.\)

Đính chính câu l \(x=-\dfrac{15}{4}\) không phải \(x=\dfrac{15}{4}\)

a) (x + 1/5)² = 9/25

=> (x + 1/5)² = (3/5)²

=> \(\orbr{\begin{cases}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{4}{5}\end{cases}}\)

Vậy ...

\(a,\text{ }\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)

\(\left(x+\frac{1}{5}\right)^2=\left(\pm\frac{3}{5}\right)^2\)

\(x+\frac{1}{5}=\pm\frac{3}{5}\)

\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{-3}{5}\\x+\frac{1}{5}=\frac{3}{5}\end{cases}}\)                               \(\Rightarrow\orbr{\begin{cases}x=\frac{-3}{5}-\frac{1}{5}\\x=\frac{3}{5}-\frac{1}{5}\end{cases}}\)                   \(\Rightarrow\orbr{\begin{cases}x=-\frac{4}{5}\\x=\frac{2}{5}\end{cases}}\)

                                \(\Rightarrow\text{ }x\in\left\{-\frac{4}{5}\text{ ; }\frac{2}{5}\right\}\)

a: \(x-\frac13x+\frac15x=-\frac{26}{5}\)

=>\(x\left(1-\frac13+\frac15\right)=-\frac{26}{5}\)

=>\(x\left(\frac23+\frac15\right)=-\frac{26}{5}\)

=>\(x\cdot\left(\frac{10}{15}+\frac{3}{15}\right)=-\frac{26}{5}\)

=>\(x\cdot\frac{13}{15}=-\frac{26}{5}\)

=>\(x=-\frac{26}{5}:\frac{13}{15}=-\frac{26}{5}\cdot\frac{15}{13}=-2\cdot3=-6\)

b: \(\frac23\left(x+\frac95\right)-\frac{3}{10}\left(5x-\frac13\right)=\frac{7}{15}\)

=>\(\frac23x+\frac{18}{15}-\frac32x+\frac{1}{10}=\frac{7}{15}\)

=>\(\frac{-5}{6}x+\frac65+\frac{1}{10}=\frac{7}{15}\)

=>\(-\frac56x+\frac{13}{10}=\frac{7}{15}\)

=>\(-\frac56x=\frac{7}{15}-\frac{13}{10}=\frac{14}{30}-\frac{39}{30}=\frac{-25}{30}=-\frac56\)

=>x=1