-12 .( x - 5 ) + 7 .( 3 - x ) = 5

-12x - ( -60 ) + 21 - 7x = 5

-12x + 60 + 21 - 7x = 5

( -12x - 7x ) + ( 60 + 21 ) = 5

-19x + 81 = 5

-19x = 5 - 81

-19x = -76

x = ( -76 ) : ( -19 )

x = 4

\(-5< x< 1\)

\(\Rightarrow\left\{-4;-3;-2;-1;0\right\}\)

Hok tốt

\(-5< x< 1\)

\(\Rightarrow x\in\left\{-4;-3;-2;-1;0\right\}\)

Hok tốt

`48/[x+4]+48/[x-4]=5`           `ĐK: x \ne +-4`

`<=>[48(x-4)+48(x+4)]/[(x-4)(x+4)]=[5(x+4)(x-4)]/[(x-4)(x+4)]`

   `=>48x-192+48x+192=5x^2-80`

`<=>5x^2-96x-80=0`

`<=>5x^2-100+4x-80=0`

`<=>5x(x-20)+4(x-20)=0`

`<=>(x-20)(5x+4)=0`

`<=>` $\left[\begin{matrix} x=20\\ x=\dfrac{-4}{5}\end{matrix}\right.$   (t/m)

Vậy `S={-4/5;20}`

ĐK : \(x\ne\pm4\)

\(\Leftrightarrow\cdot\dfrac{48\left(x+4\right)+48\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}=\dfrac{5\left(x+4\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}\)

\(\Leftrightarrow48x+192+48x-192==5x^2-80\)

\(\Leftrightarrow96x=5x^2-80\)

\(\Leftrightarrow5x^2-96x-80=0\)

\(\Leftrightarrow5x^2+4x-100-80=0\)

\(\Leftrightarrow4\left(x-20\right)+5x\left(x-20\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-20=0\\5x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=20\\x=-\dfrac{4}{5}\end{matrix}\right.\)

\(a.\)\(A=|x|+|2014-x|\ge|x+2014-x|=2014\)

Dấu '=' xảy ra khi\(x\left(2014-x\right)>0\)

TH1:\(\hept{\begin{cases}x>0\\2014-x>0\end{cases}\Leftrightarrow0< x< 2014\left(n\right)}\)

TH2:\(\hept{\begin{cases}x< 0\\2014-x< 0\end{cases}\left(l\right)}\)

Vậy \(A_{min}=2014\)khi\(0< x< 2014\)

\(b.\)\(|x^2+|x-1||=x^2+2\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+|x-1|=-x^2-2\\x^2+|x-1|=x^2+2\end{cases}\Leftrightarrow\orbr{\begin{cases}|x-1|=-2x^2-2\left(l\right)\\|x-1|=2\left(n\right)\end{cases}}}\)

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

V...

Thay x=2 và y=1 vào (d), ta được:

\(2\left(m^2-2m+3\right)+4=1\)

=>\(2m^2-4m+6+4-1=0\)

=>\(2m^2-4m+9=0\)

=>\(m^2-2m+\dfrac{9}{2}=0\)

=>\(m^2-2m+1+\dfrac{7}{2}=0\)

=>\(\left(m-1\right)^2+\dfrac{7}{2}=0\)(vô lý)

Vậy: \(m\in\varnothing\)