{x+y=5,y+z=4,z+x=3
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{x+y=5,y+z=4,z+x=3
{x+y=5,y+z=4,z+x=3
{x+y=5,y+z=4,z+x=3
1式+2式+3式=2x+2y+2z=12
x+y+z=6再减去1式得
z=1x=2 y=3
x+y=5,y+z=4,z+x=3
x+y+z = 5+4+3 /2 = 6
z= 1 x = 2 y = 3
你好
{x+y=5, (1)
y+z=4, (2)
z+x=3 (3)
(1)+(2)+(3)
2(x+y+z)=12
x+y+z=6 (4)
(4)-(1)
z=1
(4)-(2)
x=2
(4)-(3)
y=3
方程组的解为
{x=2...
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你好
{x+y=5, (1)
y+z=4, (2)
z+x=3 (3)
(1)+(2)+(3)
2(x+y+z)=12
x+y+z=6 (4)
(4)-(1)
z=1
(4)-(2)
x=2
(4)-(3)
y=3
方程组的解为
{x=2
y=3
z=1
数学辅导团为您解答,不理解请追问,理解请及时选为满意回答!(*^__^*)谢谢!
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