a)Ta thấy:

\(-\left|\frac{1}{3}x+2\right|\le0\)

\(\Rightarrow5-\left|\frac{1}{3}x+2\right|\le5-0=5\)

\(\Rightarrow B\le5\)

Dấu "=" xảy ra khi x=-6

Vậy MaxB=5<=>x=-6

b)Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\).Ta có:

\(\left|\frac{1}{2}x-3\right|+\left|\frac{1}{2}x+5\right|\ge\left|\frac{1}{2}x-3+5-\frac{1}{2}x\right|=2\)

\(\Rightarrow C\ge2\)

Dấu "=" xảy ra khi \(\orbr{\begin{cases}x=6\\x=-10\end{cases}}\)

Vậy MinC=2<=>x=6 hoặc -10

Để a/b , a+c/b+d thi a(b+d)< b (a+c)<=> ab+ad < ab +bc <=>ab < bc <=> a/b < c/d

Để a+c/b+d < c/d thì (a+c).đ < (b+d).c <=> ab+cd < bc + cd <=> ad  < bc <=> a/b < c/d

\(\frac{\text{A=1,11+0,19-13.2}}{2,06+0,54}\) -(\(\frac{1}{2}\)

  Ta có 

1+c+ac = abc +c +ca = c(1+a+ab) 

=> 
___1___ = _____1____ = ___1____ 
(1+c+ac) abc + c + ca c(1+a+ab) 

___1___ = ___a_____ = ___a___ 
(1+b+bc) a + ab + abc (1+a+ab) 

=> S = 

___1____ + ___a____ + ___1____ 
(1+ a + ab) (1 +a + ab) c(1+a+ab) 

=> S= 

c + ac +1 
---------------- 
c(1 +a+ ab) 

mà c + ac + 1 = c(1+a+ab) (xem phía trên) 

=> S = 

c(1+a+ab) 
-------------- 
c(1+a+ab) 


=> S=1

\(x^4\ge0\)

\(3x^2\ge0\)

\(\Rightarrow x^4+3x^2\ge0\)

\(\Rightarrow x^4+3x^2-4\ge-4\)

\(\Rightarrow\)GTNN của \(x^4+3x^2-4\)là -4