Question

Giải hệ phương trình : \(\left\{{}\begin{matrix}x+y+z+t=4\\x+y-z-t=8\\x-y+z-t=12\\x-y-z+t=16\end{matrix}\right.\)

Answer 1

Lời giải:

HPT \(\Leftrightarrow \left\{\begin{matrix} (x+y)+(z+t)=4(1)\\ (x+y)-(z+t)=8(2)\\ (x-y)+(z-t)=12(3)\\ (x-y)-(z-t)=16(4)\end{matrix}\right.\)

Lấy \((1)+(2)\Rightarrow 2(x+y)=12\Rightarrow x+y=6(5)\)

Lấy \((3)+(4)\Rightarrow 2(x-y)=28\Rightarrow x-y=14(6)\)

Lấy \((5)+(6)\Rightarrow 2x=20\Rightarrow x=10\Rightarrow y=6-10=-4\)

Lấy \((1)-(2)\Rightarrow 2(z+t)=-4\Rightarrow z+t=-2(7)\)

Lấy \((3)-(4)\Rightarrow 2(z-t)=-4\Rightarrow z-t=-2(8)\)

Lấy \((7)+(8)\Rightarrow 2z=-4\Rightarrow z=-2\Rightarrow t=-2-z=0\)

Vậy \((x,y,z,t)=(10,-4,-2,0)\)