Toán Lớp 9: B= $\frac{x+9}{√x+1}$ (ĐK: x$\geq$0)
So sánh B với 2
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Toán Lớp 9: B= $\frac{x+9}{√x+1}$ (ĐK: x$\geq$0)
So sánh B với 2
Comments ( 1 )
ĐKXĐ:x>=0
Ta có: B-2=\frac{x+9}{\sqrt{x}+1}-2
=\frac{x+9-2(\sqrt{x}+1)}{\sqrt{x}+1}
=\frac{x+9-2\sqrt{x}-2}{\sqrt{x}+1}
=\frac{x-2\sqrt{x}+7}{\sqrt{x}+1}
Mà x>=0<=>\sqrt{x}>=0<=>\sqrt{x}+1>=1>0
Và x-2\sqrt{x}+7=x-2\sqrt{x}+1+6=(\sqrt{x}-1)^2+6
Mà (\sqrt{x}-1)^2>=0<=>(\sqrt{x}-1)^2+6>=6>0
=>\frac{x-2\sqrt{x}+7}{\sqrt{x}+1}>0