Tìm x,y,z sao cho a) x ²+ 3y ² + 2z ² – 2x+ 12y+ 4z+ 15= 0 b) 3x ² + y ² + z ² + 2x – 2y+ 2xy+3 = 0 07/08/2021 Bởi Alexandra Tìm x,y,z sao cho a) x ²+ 3y ² + 2z ² – 2x+ 12y+ 4z+ 15= 0 b) 3x ² + y ² + z ² + 2x – 2y+ 2xy+3 = 0
Đáp án: a.$(x,y,z)=(1,-2,-1)$ b.$(x,y,z)=(1,0,0)$ Giải thích các bước giải: a.$x^2+3y^2+2z^2-2x+12y+4z+15=0$ $\rightarrow (x^2-2x+1)+(3y^2+12y+12)+(2z^2+4z+2)=0$ $\rightarrow (x-1)^2+3(y^2+4y+4)+2(z^2+2z+1)=0$ $\rightarrow (x-1)^2+3(y+2)^2+2(z+1)^2=0$ $\rightarrow \begin{cases}x-1=0\\y+2=0\\z+1=0\end{cases}\rightarrow (x,y,z)=(1,-2,-1)$ b.$3x^2+y^2+z^2+2x-2y+2xy+3=0$ $\rightarrow (y^2+2y(x-1)+x^2-2x+1)+(2x^2-4x+2)+z^2=0$ $\rightarrow (y^2+2y(x-1)+(x-1)^2)+2(x^2-2x+1)+z^2=0$ $\rightarrow (y+x-1)^2+2(x-1)^2+z^2=0$ $\rightarrow \begin{cases}y+x-1=0\\x-1=0\\z=0\end{cases}\rightarrow (x,y,z)=(1,0,0)$ Bình luận
Đáp án:
a.$(x,y,z)=(1,-2,-1)$
b.$(x,y,z)=(1,0,0)$
Giải thích các bước giải:
a.$x^2+3y^2+2z^2-2x+12y+4z+15=0$
$\rightarrow (x^2-2x+1)+(3y^2+12y+12)+(2z^2+4z+2)=0$
$\rightarrow (x-1)^2+3(y^2+4y+4)+2(z^2+2z+1)=0$
$\rightarrow (x-1)^2+3(y+2)^2+2(z+1)^2=0$
$\rightarrow \begin{cases}x-1=0\\y+2=0\\z+1=0\end{cases}\rightarrow (x,y,z)=(1,-2,-1)$
b.$3x^2+y^2+z^2+2x-2y+2xy+3=0$
$\rightarrow (y^2+2y(x-1)+x^2-2x+1)+(2x^2-4x+2)+z^2=0$
$\rightarrow (y^2+2y(x-1)+(x-1)^2)+2(x^2-2x+1)+z^2=0$
$\rightarrow (y+x-1)^2+2(x-1)^2+z^2=0$
$\rightarrow \begin{cases}y+x-1=0\\x-1=0\\z=0\end{cases}\rightarrow (x,y,z)=(1,0,0)$