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#)Giải :

     \(\frac{2}{9}.\frac{5}{17}+\frac{2}{9}.\frac{13}{17}-\frac{2}{9}.\frac{1}{17}+2018\frac{7}{9}\)

\(=\frac{2}{9}\left(\frac{5}{17}+\frac{13}{17}-\frac{1}{17}\right)+2018\frac{7}{9}\)

\(=\frac{2}{9}.1+2018\frac{7}{9}\)

\(=\frac{2}{9}+2018\frac{7}{9}\)

\(=2018\frac{9}{9}=2019\)

             #~Will~be~Pens~#

\(\frac{2}{9}\times\frac{5}{17}+\frac{2}{9}\times\frac{13}{17}-\frac{2}{7}\times\frac{1}{17}+2018\frac{7}{9}\)

\(=\frac{2}{9}\times\left(\frac{5}{17}+\frac{13}{17}-\frac{1}{17}\right)+2018\frac{7}{9}\)

\(=\frac{2}{9}\times1+2018\frac{7}{9}\)

\(=\frac{2}{9}+2018\frac{7}{9}\)

\(=\frac{2}{9}+2018+\frac{7}{9}\)

\(=\left(\frac{2}{9}+\frac{7}{9}\right)+2018\)

\(=1+2018=2019\)

~ Hok tốt ~

dễ mà bro 

a)x+5= 9

x=9-5

x=4

b) -3x=24

x=24:-3

x=-8

c) [x]-9=15

[x]=15+9

[x]=24

x=24

d)[x-9]=-2+17

[x-9]=15

[x]=15+9

[x]=24

x=24

a) \(x+5=-2+11\)                                b)\(-3x=5+29\)

\(\Rightarrow x=-2+11-5\)                                   \(\Rightarrow-3x=24\)

\(\Rightarrow x=4\)                                                       \(\Rightarrow-8\)        

c)\(\left|x\right|-9=-2+17\)                         d)\(\left|x-9\right|=-2+17\)

     \(\left|x\right|=15+9\)                                    \(\left|x-9\right|=15\)

      \(\left|x\right|=24\)                                          \(x-9=15\) hoặc \(x-9=15\)

x = 24 hoặc - 24                                    \(x=24\) hoặc \(x=-6\)

\(\dfrac{9}{17}\times\dfrac{21}{13}+\dfrac{9}{17}\times\dfrac{5}{13}-\dfrac{9}{17}\times2\)

\(=\dfrac{9}{17}\times\left(\dfrac{21}{13}+\dfrac{5}{13}-2\right)\)

\(=\dfrac{9}{17}\times\left(\dfrac{26}{13}-2\right)=\dfrac{9}{17}\times\left(2-2\right)\)

\(=\dfrac{9}{17}\times0=0\)

= 9/17 x ( 21/13 + 5/13 - 2 )

= 9/17 x 0

= 0

=9/17 x (21/13 + 5/13 - 2)

=9/17 x (26/13 - 2)

=9/17 x (13 - 2)

=9/17 x 11

= 99/17

Tick cho mik nhé!!!

a) x = 4

b) x = -8

c) | x | - 9 = -2 + 17

| x | = 15 + 9

| x | = 24

x = 24 hoặc x = -24

d) |x – 9| = -2 + 17

|x – 9| = 15

x – 9 = 15 hoặc x – 9 = -15

x = 24 hoặc x = -6

\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)

\(\Rightarrow x\left(\frac{1}{2}-\frac{2}{3}\right)=\frac{7}{12}\Rightarrow x\cdot\left(\frac{3}{6}-\frac{4}{6}\right)=\frac{7}{12}\)

\(\Rightarrow x\cdot\left(-1\right)=\frac{7}{12}\Rightarrow x=\frac{7}{12}:\left(-1\right)=\frac{7}{-12}\)

\(c,\frac{\left(x-5\right)}{12}\cdot\frac{9}{29}=\frac{-6}{29}\Rightarrow\frac{\left(x-5\right)}{12}=\frac{-6}{29}:\frac{9}{26}\)

\(\frac{\Rightarrow\left(x-5\right)}{12}=\frac{-6}{9}=\frac{-2}{3}\Rightarrow x-5=-\frac{2}{3}\cdot12\)

\(\Rightarrow x-5=\frac{-24}{3}=-8\Rightarrow x=-8+5=-3\)

\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\)

\(\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)

\(\Rightarrow-\frac{1}{6}x=\frac{7}{12}\)

\(\Rightarrow x=-\frac{7}{2}\)

\(c,\frac{x-5}{12}\cdot\frac{9}{29}=-\frac{6}{29}\)

\(\Rightarrow\frac{x-5}{12}=-\frac{2}{3}\)

\(\Rightarrow x-5=12.\left(-\frac{2}{3}\right)\)

\(\Rightarrow x-5=-8\)

\(\Rightarrow x=-3\)

b)\(\frac{-8}{17}+\frac{5}{17}< \frac{x}{17}< \frac{-6}{17}+\frac{9}{17}\)\(\left(x\in Z\right)\)

\(\Rightarrow\frac{-3}{17}< \frac{x}{17}< \frac{3}{17}\)

\(\Rightarrow-3< x< 3\)

\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)

Vậy \(x\in\left\{-2;-1;0;1;2\right\}\)

\(\left|x\right|-9=-2+17\)

\(\left|x\right|-9=15\)

\(\left|x\right|=24\)

=> \(\orbr{\begin{cases}x=24\\x=-24\end{cases}}\)

\(\left|x-9\right|=-2+17\)

\(\left|x-9\right|=15\)

=> \(\orbr{\begin{cases}x-9=15\\x-9=-15\end{cases}}\)

=> \(\orbr{\begin{cases}x=24\\x=-6\end{cases}}\)

\(\left(2^{17}+17^2\right)\left(9^{15}-15^9\right)\left(4^2-2^4\right)\)

\(=\left(2^{17}+17^2\right)\left(9^{15}-15^9\right)\left(16-16\right)\)

\(=\left(2^{17}+17^2\right)\left(9^{15}-15^9\right).0\)

\(=0\)

p/s: chúc bạn học tốt

ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}u=v\\v=\sqrt{17-x^2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u+v+uv=9\\u^2+v^2=17\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}uv=9-\left(u+v\right)\\\left(u+v\right)^2-2uv=17\end{matrix}\right.\)

\(\Rightarrow\left(u+v\right)^2+2\left(u+v\right)-35=0\)

\(\Rightarrow\left[{}\begin{matrix}u+v=5\Rightarrow uv=4\\u+v=-7\Rightarrow uv=16\end{matrix}\right.\)

\(\Rightarrow...\)