\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\\\dfrac{\dfrac{2}{3}}{x}+\dfrac{\dfrac{2}{3}}{y}+\dfrac{\dfrac{8}{9}}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\\\dfrac{\dfrac{2}{3}}{x}+\dfrac{\dfrac{14}{9}}{y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{6}\left(1\right)\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\left(2\right)\end{matrix}\right.\)

Nhân cả hai vế (1) cho \(\dfrac{2}{3}\) ta có: \(\left\{{}\begin{matrix}\dfrac{2}{3x}+\dfrac{2}{3y}=\dfrac{5.2}{6.3}\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3x}+\dfrac{2}{3y}=\dfrac{10}{18}\left(3\right)\\\dfrac{2}{3x}+\dfrac{14}{9y}=1\left(4\right)\end{matrix}\right.\)

Lấy (4) trừ (3) ta có:

\(\dfrac{14}{9y}-\dfrac{2}{3y}=1-\dfrac{10}{18}\)\(\Leftrightarrow\dfrac{8}{9y}=\dfrac{4}{9}\)\(\Leftrightarrow y=2\Rightarrow x=\dfrac{1}{\dfrac{5}{6}-\dfrac{1}{2}}=3\)

ĐKXĐ: \(x\notin\left\{0;-9\right\}\)

Ta có: \(\dfrac{1}{x+9}-\dfrac{1}{x}=\dfrac{1}{5}+\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{20x}{20x\left(x+9\right)}-\dfrac{20\left(x+9\right)}{20x\left(x+9\right)}=\dfrac{4x\left(x+9\right)+5x\left(x+9\right)}{20x\left(x+9\right)}\)

Suy ra: \(4x^2+36x+5x^2+45x=20x-20x-180\)

\(\Leftrightarrow9x^2+81x+180=0\)

\(\Leftrightarrow x^2+9x+20=0\)

\(\Leftrightarrow x^2+4x+5x+20=0\)

\(\Leftrightarrow x\left(x+4\right)+5\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\left(nhận\right)\\x=-5\left(nhận\right)\end{matrix}\right.\)

Vậy: S={-4;-5}

\(\dfrac{x}{9}-\dfrac{3}{y}=\dfrac{1}{18}\left(ĐKXĐ:y\ne0\right)\)

\(\Rightarrow\dfrac{xy-27}{9y}=\dfrac{1}{18}\)

\(\Rightarrow18\left(xy-27\right)=9y\)

\(\Rightarrow2\left(xy-27\right)=y\)

\(\Rightarrow2xy-54=y\)

\(\Rightarrow2xy-y=54\Rightarrow y\left(2x-1\right)=54\)

\(\Rightarrow y=\dfrac{54}{2x-1}\)

- Suy ra 54 chia hết cho 2x - 1

\(\Rightarrow2x-1\inƯ\left(54\right)\)

\(\Rightarrow2x-1\in\left\{1;-1;2;-2;3;-3;9;-9;27;-27\right\}\)

Cho 2x - 1 bằng từng giá trị ở trên, ta tìm được :

 \(x\in\left\{1;0;\dfrac{3}{2};-\dfrac{1}{2};2;-1;5;-4;14;-13\right\}\). Mà x không có giá trị ngoài tập số nguyên.

\(\Rightarrow x\in\left\{-13;-4;-1;0;1;2;5;14\right\}\)

Thay các giá trị x trên vừa tìm được vào y :

\(\Rightarrow y\in\left\{54;-54;18;-18;6;-6;2;-2\right\}\)

Vậy : Các số x và y thỏa mãn đề bài là : \(\left(x;y\right)\in\left\{\left(1;54\right),\left(0;-54\right),\left(2;18\right),\left(-1;-18\right),\left(5;6\right),\left(-4;-6\right),\left(14;2\right),\left(-13;-2\right)\right\}\)

\(\dfrac{x-2}{5}=\dfrac{1-x}{6}\\ =>\left(x-2\right)\cdot6=\left(1-x\right)\cdot5\\ =>6x-12=5-5x\\ =>6x+5x=5+12\\ =>11x=17\\ x=\dfrac{17}{11}\)

`[x-2]/5=[1-x]/6`

`=>6(x-2)=5(1-x)`

`=>6x-12=5-5x`

`=>6x+5x=5+12`

`=>11x=17`

`=>x=17/11`

\(\Leftrightarrow3x+6+x^2-3x+2=9\)

\(\Leftrightarrow x^2+8=9\)

hay \(x\in\left\{1;-1\right\}\)

ĐKXĐ:\(x\ne\pm2\)

\(\dfrac{3}{x-2}+\dfrac{x-1}{x+2}=\dfrac{9}{x^2-4}\\ \Leftrightarrow\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{9}{\left(x-2\right)\left(x+2\right)}=0\\ \Leftrightarrow\dfrac{3\left(x+2\right)+\left(x-1\right)\left(x-2\right)-9}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow3\left(x+2\right)+\left(x-1\right)\left(x-2\right)-9=0\\ \Leftrightarrow3x+6+x^2-x-2x+2-9=0\\ \Leftrightarrow x^2-1=0\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

a) \(\dfrac{x}{y}=\dfrac{9}{7}\)⇒\(\dfrac{x}{9}=\dfrac{y}{7}\)

\(\dfrac{y}{z}=\dfrac{7}{3}\)⇒\(\dfrac{y}{7}=\dfrac{z}{3}\)

⇒\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)

Áp dụng tính chất dãy tỉ số bằng nhau,ta có:

\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=-\dfrac{15}{5}=-3\)

⇒\(\left\{{}\begin{matrix}x=-3.9=-27\\y=-3.7=-21\\z=-3.3=-9\end{matrix}\right.\)

c: Ta có: 5x=8y=20z

nên \(\dfrac{x}{\dfrac{1}{5}}=\dfrac{y}{\dfrac{1}{8}}=\dfrac{z}{\dfrac{1}{20}}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{\dfrac{1}{5}}=\dfrac{y}{\dfrac{1}{8}}=\dfrac{z}{\dfrac{1}{20}}=\dfrac{x-y-z}{\dfrac{1}{5}-\dfrac{1}{8}-\dfrac{1}{20}}=\dfrac{3}{\dfrac{1}{40}}=120\)

Do đó: x=24; y=15; z=6

a: \(\Leftrightarrow x+2016=0\)

hay x=-2016

b: \(\Leftrightarrow x-100=0\)

hay x=100

a) 4/9 x ( 3/5 + 8/5 - 2/10 )

= 4/9 x 1

=4/9

b) ( 312 + 325 - 247 ) : 13

= 390 : 13

= 30

a)4/9 x 3/5 + 4/9 x 8/5 - 4/9 x 2/10 

=4/9x(3/5+8/5-2/10)

= 4/9x1

=4/9

b,312 : 13 + 325 : 13 - 247 : 13

=(312+325-247) : 13

=         390 : 3

= 130

\(5\left(7-x\right)=-4\left(8x-2\right)\)

\(\Leftrightarrow35-5x=-32x+8\)

\(\Leftrightarrow-5x+32x=8-35\\ \Leftrightarrow27x=-27\\ \Leftrightarrow x=-1\)