a) \(7x=63\Leftrightarrow x=9\)

b) \(x=682\)

c) \(x=80\)

d) \(1043\)

a x * 7=63

x=63:7

x=9

b 8 * x=640

x=640:8

x=80

c x=341 * 2

x=682

d x:7=149

x=149 * 7

x =1043

a. X* 7 = 56 + 7 

    X* 7 = 63

     X = 63:7=9                  

b. 8 * X = 720 - 80 

    8 * X = 640

    X= 640 :8=80

 c. X chia 2 = 341'

     X =341* 2=682            

d. X chia 7 = 114 + 35

   X chia 7 = 149

    X  = 149* 7=1043

`- 9 . ( x - 6 ) + 20 = 56`

`=> -9 . ( x - 6 ) = 56-20`

`=> -9 . ( x - 6 ) =36`

`=>x-6=36:(-9)`

`=>x-6=-4`

`=>x=-4+6`

`=>x=2`

`-------------`

` ( 245 - x ) + 7^2=149`

`=>( 245 - x ) +49=149`

`=> 245 - x =149-49`

`=> 245 - x =100`

`=>x=245-100`

`=>x=145`

`------------`

`(2^x-3).7=35`

`=> 2^x-3=35:7`

`=> 2^x-3=5`

`=>2^x=5+3`

`=>2^x=8`

`=>2^x=2^3`

`=>x=3`

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

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

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

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

\(\Rightarrow2xy-54=y\)

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

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

- 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 : \(\left(x;y\right)\in\left\{\left(1;54\right),\left(0;-54\right),\left(2;18\right),\left(-1;-18\right),\left(5;6\right),\left(-4;-6\right),\left(14;2\right),\left(-13;-2\right)\right\}\)

Ta có: \(\left(x^2+2x+3\right)^2-9\left(x^2+2x+3\right)+18=0\)

\(\Leftrightarrow\left(x^2+2x+3\right)^2-3\left(x^2+2x+3\right)-6\left(x^2+2x+3\right)+18=0\)

\(\Leftrightarrow\left(x^2+2x+3\right)\left(x^2+2x+3-3\right)-6\left(x^2+2x+3-3\right)=0\)

\(\Leftrightarrow\left(x^2+2x+3-6\right)\left(x^2+2x\right)=0\)

\(\Leftrightarrow\left(x^2+2x-3\right)\cdot x\cdot\left(x+2\right)=0\)

\(\Leftrightarrow\left(x^2+2x+1-4\right)\cdot x\cdot\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+1-2\right)\left(x+1+2\right)\cdot x\cdot\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+3\right)\cdot x\cdot\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\\x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\\x=0\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{1;-3;0;-2\right\}\)

=>\(5\cdot\dfrac{3\sqrt{x-3}}{5}-7\cdot\dfrac{2\sqrt{x-3}}{3}-7\cdot\sqrt{x^2-9}+18\cdot\sqrt{\dfrac{9}{81}\left(x^2-9\right)}=0\)

=>\(3\cdot\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}=7\cdot\sqrt{x^2-9}-18\cdot\dfrac{3}{9}\cdot\sqrt{x^2-9}\)

=>\(-\dfrac{5}{3}\sqrt{x-3}=\sqrt{x^2-9}\)

=>\(\sqrt{x-3}\left(\sqrt{x+3}+\dfrac{5}{3}\right)=0\)

=>x-3=0

=>x=3

Áp dụng t/c dãy tỉ số bằng nhau:

a.

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{-21}{7}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.\left(-3\right)=-6\\y=5.\left(-3\right)=-15\end{matrix}\right.\)

b.

\(5x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x-y}{3-5}=\dfrac{10}{-2}=-5\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.\left(-5\right)=-15\\y=5.\left(-5\right)=-25\end{matrix}\right.\)

c.

\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{3x}{15}=\dfrac{-2y}{-4}=\dfrac{3x-2y}{15-4}=\dfrac{44}{11}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=5.4=20\\y=2.4=8\end{matrix}\right.\)

d.

\(\dfrac{x}{3}=\dfrac{y}{16}=\dfrac{3x}{9}=\dfrac{-y}{-16}=\dfrac{3x-y}{9-16}=\dfrac{35}{-7}=-5\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.\left(-5\right)=-15\\y=16.\left(-5\right)=-80\end{matrix}\right.\)

\(\frac{x}{27}-\frac{2}{9}=\frac{6}{18}\)

\(\frac{x}{27}=\frac{6}{18}+\frac{2}{9}\)

\(\frac{x}{27}=\frac{6}{18}+\frac{4}{18}\)

\(\frac{x}{27}=\frac{5}{9}\)

\(\Rightarrow\text{ }9x=27\cdot5\)

\(9x=135\)

\(x=135\text{ : }9\)

\(x=15\)

\(a,A⋮3\Leftrightarrow x⋮3\\ b,A⋮9\Leftrightarrow x:9dư3\)