Toán [toán 9] thi học sinh giỏi

L

leminhnghia1

Giải:

ĐK: $-7 \le x \le 1$

$\sqrt{1-x}+\sqrt{x+7}-2x-10=0$

$\iff 2x+10-\sqrt{1-x}-\sqrt{x+7}=0$

$\iff 2x+6+(2-\sqrt{1-x})+(2-\sqrt{x+7})=0$

$\iff 2(x+3)+\dfrac{x+3}{2+\sqrt{1-x}}-\dfrac{x+3}{2+\sqrt{x+7}}=0$

$\iff (x+3)(2+\dfrac{1}{2+\sqrt{1-x}}-\dfrac{1}{2+\sqrt{x+7}})=0$

$\iff (x+3)(\dfrac{1}{2+\sqrt{1-x}}+\dfrac{2\sqrt{x+7}+3}{2+\sqrt{x+7}})=0$

$\iff x=-3$ (vì phần sau luôn dương)
 
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