A=|x-1|+|x+2012|

A=|1-x|+|x+2012|>=|1-x+x+2012|=2013

dấu = xảy ra <=>1-x>=0  =>x<1                     =>-2012<x<1

                            x+2012>0   =>x>-2012

                            1-x<0        =>x>1                   =>1<x<-2012(vô lý)

                             x+2012<0   =>x<-2012

vật min A =2013<=> -2012<x<1

A=|x-1|+|x+2012|

A=|1-x|+|x+2012|  >=    |1-x+x+2012|=2013

dấu = xảy ra khi

+)1-x>0 => x>1                   =>-2012<x<1

x+2012>0 => x>-2012

+)1-x<0 =>x>1                    =>1<x<-2012(vô lý) 

x+2012<0 => x<-2012

vậy A min =2013 <=> -2012<x<1

\(\text{nhanh lên}\)

thôi ko cần nx :>

\(9.\left(x-1\right)^2-\frac{4}{9}:\frac{2}{9}=\frac{1}{4}\)

\(9.\left(x-1\right)^2-2=\frac{1}{4}\)

\(9.\left(x-1\right)^2=\frac{9}{4}\)

\(\left(x-1\right)^2=\left(\frac{9}{2}\right)^2\)

\(x-1=\frac{9}{2}\)

\(x=\frac{11}{2}\)

\(9\left(x-1\right)^2-\frac{4}{9}:\frac{2}{9}=\frac{1}{4}\)

\(\Leftrightarrow9\left(x-1\right)^2-2=\frac{1}{4}\)

\(\Leftrightarrow9\left(x-1\right)^2=\frac{9}{4}\)

\(\Leftrightarrow\left(x-1\right)^2=\frac{1}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=\frac{1}{2}\\x-1=-\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{2}\end{cases}}\)

vậy....

\(D=\frac{2^{16}\cdot9^4}{6^6\cdot8^3}\)

\(=\frac{2^{16}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^6\cdot\left(2^3\right)^3}\)

\(=\frac{2^{16}\cdot3^8}{2^6\cdot3^6\cdot2^9}\)

\(=\frac{2^{16}\cdot3^8}{3^6\cdot2^{15}}\)

\(=\frac{2\cdot3^2}{1\cdot1}\)

\(=2\cdot9=18\)

\(D=\frac{2^{16}\cdot9^4}{6^6\cdot8^3}\)

\(=\frac{2^{16}\cdot3^8}{\left(2\cdot3\right)^6\cdot2^9}\)

\(=\frac{2^{16}\cdot3^8}{2^{15}\cdot3^6}\)

\(=2\cdot3^2\)

\(=18\)

5^3/8

hay 5^3/8

Mai thi rồi giúp mình với🙁🙁🙁