y=(2x+3)/(x+1)(x
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y=(2x+3)/(x+1)(xy=(2x+3)/(x+1)(xy=(2x+3)/(x+1)(x分离常数法.y=(2x+3)/(x+1)=[2(x+1)+1]/(x+1)=2+1/(x+1)因为x∈(
y=(2x+3)/(x+1)(x
y=(2x+3)/(x+1)(x<=0且x不等于-1)值域?
y=(2x+3)/(x+1)(x
分离常数法.
y=(2x+3)/(x+1)=[2(x+1)+1]/(x+1)=2+1/(x+1)
因为 x∈(-∞,-1)∪(-1,0]
所以 x+1∈(-∞,0)∪(0,1]
1/(x+1)∈(-∞,0)∪[1,+∞)
从而 y=2+1/(x+1)∈(-∞,2)∪[3,+∞)
即值域为(-∞,2)∪[3,+∞)
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