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/

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

x \(\in\) = { 0; 40;80;120;160}

`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`

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

`a)`\(A=167\times12-204:x\)

         \(A=2004-204:x\)

`x=5,1xx3,2+5,1xx6,7+51xx0,01`

`x=5,1xx3,2+5,1xx6,7+5,1xx0,1`

`x=5,1xx(3,2+6,7+0,1)`

`x=5,1xx10`

`x=51`

\(A=2004-204:51\)

\(A=2004-4\)

\(A=2000\)

`b)`\(A=2004-204:x=2002\)

\(\Rightarrow-204:x=-2\)

\(\Rightarrow x=102\)

Ta có :\(6=2.3;4=2^2\)

\(BCNN\left(6,4\right)=2^2.3=12\)

⇒x ∈ BC ( 6 , 4 ) = B ( 12 ) = { 0;12;24;36;48;60;...}

mà 16 ≤ x ≤50

⇒x∈{24;46;48}

vậyx ∈ {24;46;48}

a) Ta có : 24:x; 36:x; 150:x và x lớn nhất

⇒x là ƯCLN ( 24;36;150)

24=2³.3

36=2².3²

150=2.3.5²

UCLN (24;36;160)=2 . 3 = 6

\(71\dfrac{23}{45}+\left(43\dfrac{22}{45}+1\dfrac{1}{6}\right)\)

\(=\dfrac{3218}{45}+\left(\dfrac{1957}{45}+\dfrac{7}{6}\right)\)

\(=\dfrac{3218}{45}+\dfrac{1957}{45}+\dfrac{7}{6}\)

\(=115+\dfrac{7}{6}\) \(=\dfrac{697}{6}\)