Question

[Nguyễn Trang]

Hỗ trợ mình bài này ạ. Đây là toàn nang cao lớp 9

Answer 1

[Nguyễn Việt Lâm]

Ta có bài toán quen thuộc sau: 

Nếu \(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\) thì \(x+y=0\)

Do đó từ giả thiết ta chỉ cần chứng minh được \(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\) thì bài toán được giải quyết.

Đặt \(\left\{{}\begin{matrix}x+\sqrt{x^2+1}=a>0\\y+\sqrt{y^2+1}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2+1}=a-x\\\sqrt{y^2+1}=b-y\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^2+1=a^2+x^2-2ax\\y^2+1=b^2+y^2-2by\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2ax=a^2-1\\2by=b^2-1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{a^2-1}{2a}\\y=\dfrac{b^2-1}{2b}\end{matrix}\right.\)

Thế vào giả thiết:

\(\left(\dfrac{a^2-1}{2a}+\sqrt{1+\left(\dfrac{b^2-1}{2b}\right)^2}\right)\left(\dfrac{b^2-1}{2b}+\sqrt{1+\left(\dfrac{a^2-1}{2a}\right)^2}\right)=1\)

\(\Leftrightarrow\left(\dfrac{a^2-1}{2a}+\sqrt{\dfrac{\left(b^2+1\right)^2}{\left(2b\right)^2}}\right)\left(\dfrac{b^2-1}{2b}+\sqrt{\dfrac{\left(a^2+1\right)^2}{\left(2a\right)^2}}\right)=1\)

\(\Leftrightarrow\left(\dfrac{a+b}{2}+\dfrac{a-b}{2ab}\right)\left(\dfrac{a+b}{2}-\dfrac{a-b}{2ab}\right)=1\) 

\(\Leftrightarrow\left(\dfrac{a+b}{2}\right)^2-\left(\dfrac{a-b}{2ab}\right)^2=1\) (1)

Chú ý rằng: \(1=\dfrac{4ab}{4ab}=\dfrac{\left(a+b\right)^2-\left(a-b\right)^2}{4ab}\)

Do đó (1) tương đương:

\(\left(\dfrac{a+b}{2}\right)^2-\dfrac{\left(a-b\right)^2}{\left(2ab\right)^2}=\dfrac{\left(a+b\right)^2}{4ab}-\dfrac{\left(a-b\right)^2}{4ab}\)

\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{4}\left(1-\dfrac{1}{ab}\right)+\dfrac{\left(a-b\right)^2}{4ab}\left(1-\dfrac{1}{ab}\right)=0\)

\(\Leftrightarrow\left[\dfrac{\left(a+b\right)^2}{4}+\dfrac{\left(a-b\right)^2}{4ab}\right]\left(1-\dfrac{1}{ab}\right)=0\)

Do \(a;b>0\Rightarrow\dfrac{\left(a+b\right)^2}{4}+\dfrac{\left(a-b\right)^2}{4ab}>0\)

\(\Rightarrow1-\dfrac{1}{ab}=0\Leftrightarrow ab=1\)

Hay \(\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)

\(\Rightarrow x+y=0\Rightarrow P=100\)