Tìm x,y,z

R

ronaldover7

$(x+y)^2$+$z^2$=(x+1)(y-1)
\Rightarrow $[(x+1)+(y-1)]^2$+$z^2$=(x+1)(y-1)
\Rightarrow $(x+1)^2$+$(y+1)^2$+2(x+1)(y-1)+$z^2$=(x+1)(y-1)
\Rightarrow $(x+1)^2$+$(y+1)^2$+(x+1)(y-1)+$z^2$=0
Mà (x+1)^2+(y+1)^2+(x+1)(y-1) \geq 0(áp dụng $a^2$+$b^2$+ab \geq 0)và $z^2$ \geq 0
\Rightarrow x,y,z
 
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