We define the symbol | a B | C D |, which means | a B | C D | = ad BC We specify the symbol | a B| |c d|, It means | a B| |c d| = ad-bc , For example, 1 2| |3 4| =1*4-2*3=-2 If | x + 1 2x| |2 / 3, 1 / 2 | = 1, Find the value of X

We define the symbol | a B | C D |, which means | a B | C D | = ad BC We specify the symbol | a B| |c d|, It means | a B| |c d| = ad-bc , For example, 1 2| |3 4| =1*4-2*3=-2 If | x + 1 2x| |2 / 3, 1 / 2 | = 1, Find the value of X

½×(x+1)-2x×2/3=1
3(x+1)-8x=6
3x+3-8x=6
3x-8x=6-3
-5x=3
x=-0.6

Real numbers a and B satisfy a - (a + b)

||a|-(a+b)|0
b>0
When a = 0, the inequality (1) does not hold
a> 0, a · (a + b) > a ·| a + B|
A + b > | a + B |, and a + B ≤| a + B |, the inequality has no solution
In conclusion, A0 is obtained

Prove (AC BD) &# 178; + (BC + AD) &# 178; = (A & # 178; + B & # 178;) (C & # 178; + D & # 178;)