Câu hỏi của Song Lam Diệp

Question

giải $\sqrt{25-x^2}-\sqrt{9-x^2}=2$

Lightning Farron · Accepted Answer

$\sqrt{25-x^2}-\sqrt{9-x^2}=2$

Đk:$-3\le 0\le3$

$\Leftrightarrow\left(\sqrt{25-x^2}-5\right)-\left(\sqrt{9-x^2}-3\right)=0$

$\Leftrightarrow\dfrac{25-x^2-25}{\sqrt{25-x^2}+5}-\dfrac{9-x^2-9}{\sqrt{9-x^2}+3}=0$

$\Leftrightarrow\dfrac{-x^2}{\sqrt{25-x^2}+5}-\dfrac{-x^2}{\sqrt{9-x^2}+3}=0$

$\Leftrightarrow-x^2\left(\dfrac{1}{\sqrt{25-x^2}+5}-\dfrac{1}{\sqrt{9-x^2}+3}\right)=0$

$\Rightarrow x=0$Câu trả lời: $\sqrt{25-x^2}-\sqrt{9-x^2}=2$

Đk:$-3\le 0\le3$

$\Leftrightarrow\left(\sqrt{25-x^2}-5\right)-\left(\sqrt{9-x^2}-3\right)=0$

$\Leftrightarrow\dfrac{25-x^2-25}{\sqrt{25-x^2}+5}-\dfrac{9-x^2-9}{\sqrt{9-x^2}+3}=0$

$\Leftrightarrow\dfrac{-x^2}{\sqrt{25-x^2}+5}-\dfrac{-x^2}{\sqrt{9-x^2}+3}=0$

$\Leftrightarrow-x^2\left(\dfrac{1}{\sqrt{25-x^2}+5}-\dfrac{1}{\sqrt{9-x^2}+3}\right)=0$

$\Rightarrow x=0$

Violympic toán 9