`1`, `a, |2x+3|+|4y^2-9|=0`
Dấu "`=`" xảy ra
`<=>`$\left \{ {{|2x+3|=0} \atop {|4y^2-9|=0}} \right.$
`<=>`$\left \{ {{2x+3=0} \atop {4y^2-9=0}} \right.$
`<=>` $\left \{ {{2x=-3} \atop {4y^2=9}} \right.$
`<=>` $\left \{ {{\frac{-3}{2}} \atop {2^2. y^2=3^2}} \right.$
`<=>`$\left \{ {{\frac{-3}{2}} \atop {(2.y)^2=3^2}} \right.$
`<=>` $\left \{ {{\frac{-3}{2}} \atop {2.y=±3}} \right.$
`<=>`$\left \{ {{\frac{-3}{2}} \atop {±\frac{3}{2}}} \right.$
Vậy `x=-3/2; y=±3/2`
`b, |x-2007|+|y-2008| <=0 (1)`
Ta có: `|x-2007| >=0` với `∀ y`
`|y-2008| >=0` với `∀ y`
`=> |x-2007|+|y-2008|>=0` với `∀ x,y (2)`
Từ `(1)` và `(2) => |x-2007|+|y-2008|=0`
Dấu "`=`" xảy ra
`<=>` $\left \{ {{|x-2007|=0} \atop {|y-2008|=0}} \right.$
`<=>` $\left \{ {{x-2007=0} \atop {y-2008}} \right.$
`<=>` $\left \{ {{x=2007} \atop {y=2008}} \right.$
Vậy `x=2007; y=2008`
#Học tốt!
