tìm x,y thoả mãn |x+2y| + ( x^2-4x +4 )=0 28/08/2021 Bởi Allison tìm x,y thoả mãn |x+2y| + ( x^2-4x +4 )=0
$|x+2y| + ( x^2-4x +4 )=0$ $⇔|x+2y| + ( x-2)^2=0$ Mà $|x+2y| ;( x-2)^2≥0∀x$ nên $|x+2y| + ( x-2)^2≥0∀x$ Dấu “=” xảy ra $⇔\left \{ {{|x+2y|=0} \atop {(x-2)^2=0}} \right.⇔$ $\left \{ {{x+2y=0} \atop {x-2=0}} \right.$ $⇔\left \{ {{y=\frac{-x}2} \atop {x=2}} \right.⇔$ $\left \{ {{y=\frac{-2}2=-1} \atop {x=2}} \right.$ Vậy (x;y)=(-1;2) thỏa mãn đề bài. Bình luận
|x+2y|+(x-2)²=0 =>|x+2y|=-(x-2)² TH1:x+2y=-(x-2)² =>x+2y=-( x²-4x +4 ) =>x+2y=-x²+4x-4 =>(x²-3x)+(2y+4)=0 =>x(x-3)+2(y+2)=0 =>(x+2)(x-3)(y+2)=0 =>x= -2;3 y=-2 TH2: x+2y=(x-2)² =>x+2y=x²-4x+4 =>(-x²+5x)+(2y-4)=0 =>-x(x-5)+2(y-2)=0 =>(-x+2)(x-5)(y-2)=0 =>x=2;5 y=2 Bình luận
$|x+2y| + ( x^2-4x +4 )=0$
$⇔|x+2y| + ( x-2)^2=0$
Mà $|x+2y| ;( x-2)^2≥0∀x$ nên $|x+2y| + ( x-2)^2≥0∀x$
Dấu “=” xảy ra $⇔\left \{ {{|x+2y|=0} \atop {(x-2)^2=0}} \right.⇔$ $\left \{ {{x+2y=0} \atop {x-2=0}} \right.$
$⇔\left \{ {{y=\frac{-x}2} \atop {x=2}} \right.⇔$ $\left \{ {{y=\frac{-2}2=-1} \atop {x=2}} \right.$
Vậy (x;y)=(-1;2) thỏa mãn đề bài.
|x+2y|+(x-2)²=0
=>|x+2y|=-(x-2)²
TH1:x+2y=-(x-2)²
=>x+2y=-( x²-4x +4 )
=>x+2y=-x²+4x-4
=>(x²-3x)+(2y+4)=0
=>x(x-3)+2(y+2)=0
=>(x+2)(x-3)(y+2)=0
=>x= -2;3
y=-2
TH2: x+2y=(x-2)²
=>x+2y=x²-4x+4
=>(-x²+5x)+(2y-4)=0
=>-x(x-5)+2(y-2)=0
=>(-x+2)(x-5)(y-2)=0
=>x=2;5
y=2