Toán Lớp 9: giải pt:
2 căn x – 1 < căn 2 + 2
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Toán Lớp 9: giải pt:
2 căn x – 1 < căn 2 + 2
Comments ( 2 )
2 $\sqrt{x}$ -1 < $\sqrt{2}$ +2
⇔2 $\sqrt{x}$< $\sqrt{2}$ +1
⇔4x< 3+ $\sqrt{2}$
⇔x< (3+$\sqrt{2}$) /4
vậy S={x | x< $\frac{3+$\sqrt{2}}{4}$}
ĐK:x>=0
2\sqrt{x}-1<\sqrt{x}+2
<=>2\sqrt{x}-\sqrt{x}<2+1
<=>\sqrt{x}<3
<=>x<9
Kết hợp ĐK:0<=x<9