已知|a-3|+(b+1)的平方=0,代数式2b-a+m⼀2的值比1⼀2b-a+m的值大1,求m的值

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2024年12月05日 13:59
有5个网友回答
网友(1):

由|a-3|+(b+1)的平方=0
可得,a=3,闭袭b=-1
代入“2b-a+m/2的值比轿者兄1/2b-a+m的值大嫌宏1”
-2-3+m/2=-1/2-3+m+1
解得,m=-5

网友(2):

由已知得知察尺
a-3=0
b+1=0
解得a=3
b=-1
由题意搭高得
2b-a+m/2-1/2b-a+m=1
将a=3,b=-1代入得
后面就不用再说没野了吧?

网友(3):

0的平方等于0,所以a=3,b=-1
代数式2b-a+m/2的值比1/2b-a+m的值大扰段槐1
a,b代人燃隐为 -2-3+m/2+1/缓友2+3-m=1 -1/2m=5/2 m=-5

网友(4):

|a-3|+(悄盯森b+1)的平方=0
因为|a-3|>=0,(则氏b+1)的平方>=0
只有:启亩a-3=0,b+1=0
a=3,b=-1
2b-a+m/2-(1/2b-a+m)=1
3b/2-m/2=1
m=-5

网友(5):

|a-3|+(b+1)的裤漏平方=0,
|a-3|=0,(b+1)^2=0 b=1或-1
2b-a+m/2-(1/2b-a+m)=1
-3/2b-m/族困2=1
m/2=3/胡穗烂2b-1
m=3b-2=1或-5

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