The four numbers a, B, C and D are arranged in two rows and two columns, and a vertical line is added on both sides to mark them as | a B |, which is defined as | a B | = AD + BC, | C D | C D| The four numbers a, B, C and D are arranged in two rows and two columns, and a vertical line is added on each side to mark them as | a B |. Define | a B | = AD + BC, | C D | C D |. The above marks are called second-order determinants. If | x + 1 X-1 | X-1 x + 1 | = 6, then x + 1 | = 6=

The four numbers a, B, C and D are arranged in two rows and two columns, and a vertical line is added on both sides to mark them as | a B |, which is defined as | a B | = AD + BC, | C D | C D| The four numbers a, B, C and D are arranged in two rows and two columns, and a vertical line is added on each side to mark them as | a B |. Define | a B | = AD + BC, | C D | C D |. The above marks are called second-order determinants. If | x + 1 X-1 | X-1 x + 1 | = 6, then x + 1 | = 6=

(x+1)(x+1)-(x-1)(x-1)=6
4x=6
x=1.5
LZ are you sure it's AD + BC? This thing you describe is called determinant in high number, and the method is ad BC
If AD + BC is determined, then
(x+1)(x+1)+(x-1)(x-1)=6
x=√2
If the explanation is not clear enough,

The four numbers a.b.c.d are arranged in two rows and two columns, and a vertical line is added on both sides to mark them as | a B |, which is defined as | a B | = ad BC,
The four numbers a, B, C and D are arranged in two rows and two columns, and a vertical line is added on each side to mark them as | a B |, which is defined as | a B | = ad BC,
|c d| | c d|
The above notation is called a determinant of order 2
|x+1 x-1|
|X-1 x + 1 | = 6, then x=

（x＋1）²-（x-1）²=6
x²＋2x＋1-x²＋2x-1=6
4x=6
x=3/2

The four numbers a, B, C, D are arranged in two rows and two columns, and a vertical line is added on each side to mark them as | a B C B |, which defines | a B C D | = ad BC,
If | x + 2x-1|
|X + 1 x - 2 | = 5x, find the value of X

(x + 2) (X-2) - (x + 1) (x-1) = 5x, the solution is x = - 3 / 5