Câu hỏi của Bảo Hân

Question

Cho biểu thức: P= $\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{\left(1-x\right)^2}{2}$a) Rút gọn A với x ≥ 0; x ≠ 1b) Tìm x để A dươngc) Tìm GTLN của Amn giải chi t...

Nguyễn Lê Phước Thịnh · Accepted Answer

a: $A=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(x-1\right)^2}{2}$$=-\sqrt{x}\left(\sqrt{x}-1\right)$b:Để A>0 thì $\sqrt{x}-1< 0$=>00 thì $\sqrt{x}-1< 0$=>0

Hoàng Phú Thiện · Answer

a) $P=\left[\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right].\dfrac{\left(1-x\right)^2}{2}$$=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\left(1-x\right)^2}{2}$$=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\left(x-1\right)^2}{2}$$=\dfrac{-2\sqrt{x}\left(x-1\right)^2}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}$$=\dfrac{-2\sqrt{x}.\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)^2}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}$$=-\sqrt{x}.\left(\sqrt{x}-1\right)$b) Yêu cầu bài toán: $-\sqrt{x}.\left(\sqrt{x}-1\right)>0\Leftrightarrow\sqrt{x}-1< 0$ (do $-\sqrt{x}< 0$) $\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1$ Câu trả lời: a) $P=\left[\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right].\dfrac{\left(1-x\right)^2}{2}$$=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\left(1-x\right)^2}{2}$$=\dfrac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\left(x-1\right)^2}{2}$$=\dfrac{-2\sqrt{x}\left(x-1\right)^2}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}$$=\dfrac{-2\sqrt{x}.\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)^2}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}$$=-\sqrt{x}.\left(\sqrt{x}-1\right)$b) Yêu cầu bài toán: $-\sqrt{x}.\left(\sqrt{x}-1\right)>0\Leftrightarrow\sqrt{x}-1< 0$ (do $-\sqrt{x}< 0$) $\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1$

Trần Tuấn Hoàng · Answer

a) $ĐKXĐ:x\ge0;x\ne1$$P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{\left(1-x\right)^2}{2}$$=\left[\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right].\dfrac{\left(1-x\right)^2}{2}$$=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\left(1-x\right)^2}{2}$$=\dfrac{\left(x-\sqrt{x}-2\right)-\left(x+\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\left(1-x\right)^2}{2}$$=\dfrac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\left[\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\right]^2}{2}$$=\dfrac{-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2$$=-\sqrt{x}\left(\sqrt{x}-1\right)$b) $A>0$$\Rightarrow-\sqrt{x}\left(\sqrt{x}-1\right)>0$$\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)< 0$$\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}< 0\\sqrt{x}-1>0\end{matrix}\right.\\left\{{}\begin{matrix}\sqrt{x}>0\\sqrt{x}-1< 0\end{matrix}\right.\end{matrix}\right.$$\Leftrightarrow0< x< 1$c) $A=-\sqrt{x}\left(\sqrt{x}-1\right)=-x+\sqrt{x}=-\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}$- Dấu "=" xảy ra khi $\left(\sqrt{x}-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{4}$- Vậy $MaxA=\dfrac{1}{4}$ Câu trả lời: a) $ĐKXĐ:x\ge0;x\ne1$$P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{\left(1-x\right)^2}{2}$$=\left[\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right].\dfrac{\left(1-x\right)^2}{2}$$=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\left(1-x\right)^2}{2}$$=\dfrac{\left(x-\sqrt{x}-2\right)-\left(x+\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\left(1-x\right)^2}{2}$$=\dfrac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\dfrac{\left[\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\right]^2}{2}$$=\dfrac{-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2$$=-\sqrt{x}\left(\sqrt{x}-1\right)$b) $A>0$$\Rightarrow-\sqrt{x}\left(\sqrt{x}-1\right)>0$$\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)< 0$$\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}< 0\\sqrt{x}-1>0\end{matrix}\right.\\left\{{}\begin{matrix}\sqrt{x}>0\\sqrt{x}-1< 0\end{matrix}\right.\end{matrix}\right.$$\Leftrightarrow0< x< 1$c) $A=-\sqrt{x}\left(\sqrt{x}-1\right)=-x+\sqrt{x}=-\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{1}{4}=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}$- Dấu "=" xảy ra khi $\left(\sqrt{x}-\dfrac{1}{2}\right)^2=0\Leftrightarrow x=\dfrac{1}{4}$- Vậy $MaxA=\dfrac{1}{4}$

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