\(\dfrac{9}{4}-\dfrac{1}{4}:y-\dfrac{1}{2}=\dfrac{2}{3}\\ =>\dfrac{1}{4}:y=\dfrac{9}{4}-\dfrac{1}{2}-\dfrac{2}{3}\\ =>\dfrac{1}{4}:y=\dfrac{27}{12}-\dfrac{6}{12}-\dfrac{8}{12}\\ =>\dfrac{1}{4}:y=\dfrac{13}{12}\\ =>y=\dfrac{1}{4}:\dfrac{13}{12}\\ =>y=\dfrac{3}{13}\)

`9/4- 1/4:y - 1/2 = 2/3`

`<=> 9/4 - 1/4:y - 2/4 = 2/3`

`<=> 7/4 - 1/4:y= 2/3`

`<=> 1/4:y = 7/4 - 2/3`

`<=> 1/4:y = 21/12 - 8/12`

`<=> 1/4:y = 13/12`

`<=> y = 1/4 : 13/12`

`<=> y = 1/4 . 12/13`

`<=> y = 3/13`

Vậy ..

\(5,1\times y-y:10=36,25\times4\)

\(5,1\times y-y\times0,1=145\)

\(\left(5,1-0,1\right)\times y=145\)

\(5\times y=145\)

\(y=145:5\)

\(y=29\)

`5,1 xx y - y : 10 = 36,25 xx 4`

`=>5,1 xx y - y xx 0,1 = 36 x 4 + 0,25 x 4`

`=>y xx (5,1 - 0,1) = 144+1`

`=>y xx 5 = 145`

`=>y=145:5`

`=>y=29`

Vậy: ... 

`x(x-5)=0`

TH1: `x=0`

TH2: `x-5=0`

`=>x=0+5`

`=>x=5`

Vậy: `x∈{0;5}`

Bổ sung cho @HT.Phong

Kết luận: \(x\) \(\in\) {0; 5}

\(\Leftrightarrow\left(6x^2+2xy-8x\right)+\left(3xy+y^2-4y\right)+\left(3x+y-4\right)=1\)

\(\Leftrightarrow2x\left(3x+y-4\right)+y\left(3x+y-4\right)+\left(3x+y-4\right)=1\)

\(\Leftrightarrow\left(3x+y-4\right)\left(2x+y+1\right)=1\)

Ta có bảng sau:

Vậy \(\left(x;y\right)=\left(5;-12\right);\left(5;-10\right)\)

Thanks very much bro!

\(124\times\left(56+67\right)-24\times\left(67+56\right)\\ =124\times123-24\times123\\ =\left(124-24\right)\times123\\ =100\times123\\ =12300\)

  124 x (56 + 67) - 24 x (67 + 56)

= (56 + 67) x (125 - 24)

= 123 x 100 

= 12300

Do 729 chia hết cho 3 \(\Rightarrow2x^2\) chia hết  cho 3 \(\Rightarrow x\) chia hết cho 3

\(\Rightarrow x=3x_1\)

\(\Rightarrow2\left(3x_1\right)^2+3y^2=729\)

\(\Rightarrow6x_1^2+y^2=243\)

Tương tự, 243 chia hết cho 3 \(\Rightarrow y=3y_1\)

\(\Rightarrow6x_1^2+9y_1^2=243\)

\(\Rightarrow2x_1^2+3y_1^2=81\)

Lý luận tương tự ta có \(\left\{{}\begin{matrix}x_1=3x_2\\y_1=3y_2\end{matrix}\right.\)

\(\Rightarrow2x_2^2+3y_2^2=1\) (1)

(1) ko có nghiệm nguyên nên pt đã cho ko có nghiệm nguyên

\(\dfrac{2}{3}\times\dfrac{8}{17}+\dfrac{2}{3}\times\dfrac{5}{17}+\dfrac{2}{3}\times\dfrac{4}{17}\)

\(=\dfrac{2}{3}\times\left(\dfrac{8}{17}+\dfrac{5}{17}+\dfrac{4}{17}\right)\)

\(=\dfrac{2}{3}\times1\)

\(=\dfrac{2}{3}\)

`x ∈ ƯC (150,35)`

Ta có: 

`150 = 2 . 3 . 5^2`

`35 = 5 . 7 `

`=> UCLN(150;35) = 5`

Hay `x ∈ U(5) = {-5;-1;1;5}`

Vậy ....

`x ∈ BC (25,30,40)`

Ta có: 

`25 = 5^2`

`30 = 2.3.5`

`40 = 2^3 . 5`

`=> BCNN = 5^2 . 2^3 . 3 = 600`

`=> x ∈ B(600) = {0;600;1200;....}`

Vậy ....