Câu hỏi của Linh Lê

Question

Cho biểu thức :

$P=\left(\dfrac{1+\sqrt{1-x}}{1-x+\sqrt{1-x}}+\dfrac{1-\sqrt{1+x}}{1+x-\sqrt{1+x}}\right)^2.\dfrac{x^2-1}{2}+1$

Tìm x để $P$ ≤ $\dfrac{\sqrt{2}}{2}$

Nguyễn Việt Lâm · Accepted Answer

ĐKXĐ: $-1< x< 1$

$P=\left(\dfrac{1+\sqrt{1-x}}{\sqrt{1-x}\left(\sqrt{1-x}+1\right)}-\dfrac{\sqrt{1+x}-1}{\sqrt{1+x}\left(\sqrt{1+x}-1\right)}\right)^2\dfrac{x^2-1}{2}+1$

$P=\left(\dfrac{1}{\sqrt{1-x}}-\dfrac{1}{\sqrt{1+x}}\right)^2\dfrac{x^2-1}{2}+1$

$P=\left(\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1-x^2}}\right)^2\dfrac{x^2-1}{2}+1$

$P=\dfrac{2-2\sqrt{1-x^2}}{1-x^2}\times\dfrac{\left(x^2-1\right)}{2}+1=\sqrt{1-x^2}$

$P\le\dfrac{\sqrt{2}}{2}\Rightarrow\sqrt{1-x^2}\le\dfrac{\sqrt{2}}{2}\Rightarrow1-x^2\le\dfrac{1}{2}\Rightarrow x^2\ge\dfrac{1}{2}$

$\Rightarrow\left[{}\begin{matrix}x\ge\dfrac{\sqrt{2}}{2}\x\le\dfrac{-\sqrt{2}}{2}\end{matrix}\right.$

Kết hợp với ĐKXĐ ban đầu ta được $\left[{}\begin{matrix}-1< x\le\dfrac{-\sqrt{2}}{2}\\dfrac{\sqrt{2}}{2}\le x< 1\end{matrix}\right.$Câu trả lời: ĐKXĐ: $-1< x< 1$

$P=\left(\dfrac{1+\sqrt{1-x}}{\sqrt{1-x}\left(\sqrt{1-x}+1\right)}-\dfrac{\sqrt{1+x}-1}{\sqrt{1+x}\left(\sqrt{1+x}-1\right)}\right)^2\dfrac{x^2-1}{2}+1$

$P=\left(\dfrac{1}{\sqrt{1-x}}-\dfrac{1}{\sqrt{1+x}}\right)^2\dfrac{x^2-1}{2}+1$

$P=\left(\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1-x^2}}\right)^2\dfrac{x^2-1}{2}+1$

$P=\dfrac{2-2\sqrt{1-x^2}}{1-x^2}\times\dfrac{\left(x^2-1\right)}{2}+1=\sqrt{1-x^2}$

$P\le\dfrac{\sqrt{2}}{2}\Rightarrow\sqrt{1-x^2}\le\dfrac{\sqrt{2}}{2}\Rightarrow1-x^2\le\dfrac{1}{2}\Rightarrow x^2\ge\dfrac{1}{2}$

$\Rightarrow\left[{}\begin{matrix}x\ge\dfrac{\sqrt{2}}{2}\x\le\dfrac{-\sqrt{2}}{2}\end{matrix}\right.$

Kết hợp với ĐKXĐ ban đầu ta được $\left[{}\begin{matrix}-1< x\le\dfrac{-\sqrt{2}}{2}\\dfrac{\sqrt{2}}{2}\le x< 1\end{matrix}\right.$

Duc Nguyendinh · Answer

Hình như đề bài sai hay sao vậyCâu trả lời: Hình như đề bài sai hay sao vậy

Rimuru tempest · Answer

ĐK $-1\le x\le1$

$P=\left(\dfrac{1+\sqrt{1-x}}{\left(\sqrt{1-x}\right)^2+\sqrt{1-x}}+\dfrac{1-\sqrt{1+x}}{\left(\sqrt{1+x}\right)^2-\sqrt{1+x}}\right).\dfrac{x^2-1}{2}+1$

$P=\left(\dfrac{1+\sqrt{1-x}}{\left(\sqrt{1-x}+1\right).\sqrt{1-x}}+\dfrac{1-\sqrt{1+x}}{-\sqrt{1+x}\left(1-\sqrt{1+x}\right)}\right).\dfrac{x^2-1}{2}+1$

$P=\left(\dfrac{1}{\sqrt{1-x}}-\dfrac{1}{\sqrt{1+x}}\right).\dfrac{x^2-1}{2}+1$

$P=\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{\left(1-x\right)\left(1+x\right)}}.\dfrac{\left[-\left(1-x\right)\left(1+x\right)\right]}{2}+1$

$P=\dfrac{-\sqrt{\left(1-x\right)\left(1+x\right)}\left(\sqrt{1+x}-\sqrt{1-x}\right)}{2}+1$

$P=\dfrac{-\sqrt{\left(1-x\right)\left(1+x\right)}\left(\sqrt{1+x}-\sqrt{1-x}\right)+2}{2}$

$P\le\dfrac{\sqrt{2}}{2}$

$\Leftrightarrow\dfrac{-\sqrt{\left(1-x\right)\left(1+x\right)}\left(\sqrt{1+x}-\sqrt{1-x}\right)+2}{2}\le\dfrac{\sqrt{2}}{2}$

$\Leftrightarrow\sqrt{\left(1-x\right)\left(1+x\right)}\left(\sqrt{1+x}-\sqrt{1-x}\right)-2\ge-\sqrt{2}$

$\Leftrightarrow\sqrt{\left(1-x\right)\left(1+x\right)}\left(\sqrt{1+x}-\sqrt{1-x}\right)\ge2-\sqrt{2}$

Đặt $t=\sqrt{1+x}-\sqrt{1-x}\Leftrightarrow\sqrt{\left(1-x\right)\left(1+x\right)}=\dfrac{2-t^2}{2}$

$\Leftrightarrow\dfrac{2-t^2}{2}.t\ge2-\sqrt{2}$

$\Leftrightarrow2t-t^3\ge4-2\sqrt{2}$

$\Leftrightarrow t^3-2t+4-2\sqrt{2}\le0$

giải tìm đc t rồi thế lên tìm đc x , tại mình ko có máy tính nên tới đây là ko làm đc :)))Câu trả lời: ĐK $-1\le x\le1$