b. + Vì \(|6-2x|\ge0\)\(\forall x\)

\(\Rightarrow\)\(|6-2x|-5\ge0-5\)\(\forall x\)

\(\Rightarrow\)B\(\ge\)-5 \(\forall x\)

Vậy GTNN của B= -5 \(\Leftrightarrow\)6-2x=0

                                    \(\Leftrightarrow\)2x=6

                                   \(\Leftrightarrow\)x=3

+ Vì -\(|6-2x|\le0\forall x\)

\(\Rightarrow\)\(|6-2x|-5\le0+5\forall x\)

\(\Rightarrow B\le5\forall x\)

Vậy GTLN của B= 5 \(\Leftrightarrow6-2x=0\)

                                \(\Leftrightarrow2x=1\)

                                \(\Leftrightarrow x=\frac{1}{2}\)

c,+ Vì \(|x+1|\ge0\forall x\)

\(\Rightarrow\)\(3-|x+1|\ge3-0\forall x\)

\(\Rightarrow C\ge3\forall x\)

Vậy GTNN của C=3 \(\Leftrightarrow x+1=0\)

                                 \(\Leftrightarrow x=-1\)

+ Vì \(-|x+1|\le0\forall x\)

\(\Rightarrow3-|x+1|\le3+0\forall x\)

\(\Rightarrow C\le3\forall x\)

Vậy GTLN của \(C=3\Leftrightarrow x+1=0\)

                                     \(\Leftrightarrow x=-1\)

Mình chỉ làm vậy thôi nhé!

THANKS  BẠN NHIỀU NHA

b: \(\left(-0,25\right)^3:\left(-0,25\right)=\left(-0,25\right)^{3-1}=\left(-0,25\right)^2=\left(-\frac14\right)^2=\frac{1}{16}\)

c: \(\left\lbrack\left(-0,5\right)^2\right\rbrack^2=\left\lbrack\left(\frac12\right)^2\right\rbrack^2=\left(\frac12\right)^4=\frac{1}{16}\)

d: \(a^5\cdot a^2=a^{5+2}=a^7\)

f: \(\frac{3^2}{\left(0,375\right)^2}=\left(\frac{3}{0,375}\right)^2=8^2=64\)

e: \(\left(\frac15\right)^5\cdot5^5=\left(\frac15\cdot5\right)^5=1^5=1\)

g: \(\frac{120^3}{40^3}=\left(\frac{120}{40}\right)^3=3^3=27\)

h: \(\left(0,125\right)^3\cdot512=\left(\frac18\right)^3\cdot8^3=1\)

if a<b,bcz of a^b=b^c so b>c c<d d>e e<f f>g g<a bcz of g<a and a<b so g<b (not possible)

Same with a>b ,so a=b.

Do again multiple time ,we get a=b=c=d=e=f so bcs f^g=g^a,so f^g=g^f so g=f.

So totally ,we get a=b=c=d=e=f=g.