`(67xx36+67xx18):54`

\(=\left[67\times\left(36+18\right)\right]:54\)

\(=\left(67\times54\right):54\)

\(=\dfrac{67}{54}\times1\)

\(=\dfrac{67}{54}\)

67 x 36 + 67 x 18) : 54

= 67x ( 36+18) : 54

= 67x 54 :54

= 67x 1

= 67

-69 + 53 + 46 + (-94) + (-14) + 78=0

`-69 + 53 + 46 + (-94) + (-14) + 78`

`=-69 + 53 + 46 - 94 - 14 + 78`

`=-16 + 46 - 94 - 14 + 78`

`=30  - 94 - 14 + 78`

`=-64 - 14 + 78`

`=-78 + 78`

`=0`

Cách giải: Dùng tương tự thuật toán lập ma phương lẻ

https://itstudent93.wordpress.com/2015/11/12/thuat-toan-giai-ma-phuong-ma-tran-ki-ao/

\(\sqrt{3,6\times16,9}=\sqrt{60,84}=7,8\)

7,8

x=40;80;120;160;0;-40;-80;-120;......

x \(\in\) = { 0; 40;80;120;160}

2x + 7 ⋮ x +2 ⇔ 2(x + 2) + 3 ⋮ x + 2 ⇔ 3 ⋮ x +2

 ⇔ x+2 ϵ {-3; -1; 1;3} ⇔ x ϵ { -5; -3;  -1; 1}

\(\dfrac{2x+7}{x+2}=\dfrac{2\left(x+2\right)+3}{x+2}=2+\dfrac{3}{x+2}\)

Để \(2x+7⋮x+2\) thì \(x+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

`@x+2=1->x=-1`

`@x+2=-1->x=-3`

`@x+2=3->x=1`

`@x+2=-3->x=-5`

Vậy \(x\in\left\{-1;-3;1;-5\right\}\) thì \(2x+7⋮x+2\)

`16 2/7 : (-2,5) -28 2/7 - 2/5`

`= 144/7 : (-5/2) - 198/7 - 2/5`

`=144/7 xx (-2/5)- 198/7 - 2/5`

`=-228/35 - 198/7 - 2/5`

`=-228/35 - 990/35 - 14/35`

`=(-228-990-14)/35`

`=-176/5`

\(x\in\left\{-6;-3;-2;-1;1;2;3;6\right\}\)