Câu hỏi của Diệu Anh

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Rút gọn$C=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}$ với $x>0,x\ne1$ - tìm GTNN của C- tìm x để N= $\dfrac{2\sqrt{x}}{C}$ nhận giá tr...

Nguyễn Lê Phước Thịnh · Accepted Answer

*Rút gọnTa có: $C=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}$$=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}$$=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2$$=x-\sqrt{x}+1$Ta có: $C=x-\sqrt{x}+1$$=x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}$$=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x$ thỏa mãn ĐKXĐDấu '=' xảy ra khi $\sqrt{x}=\dfrac{1}{2}$hay $x=\dfrac{1}{4}$Câu trả lời: *Rút gọnTa có: $C=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}$$=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}$$=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2$$=x-\sqrt{x}+1$Ta có: $C=x-\sqrt{x}+1$$=x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}$$=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x$ thỏa mãn ĐKXĐDấu '=' xảy ra khi $\sqrt{x}=\dfrac{1}{2}$hay $x=\dfrac{1}{4}$

Lê Thị Thục Hiền · Answer

$C=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\left(x>0;x\ne1\right)$$=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}$$=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2$$=x-\sqrt{x}+1$$=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}$Dấu "=" xảy ra khi $\sqrt{x}-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{4}$Vậy $C_{min}=\dfrac{3}{4}$$N=\dfrac{2\sqrt{x}}{C}=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=\dfrac{2}{\sqrt{x}+\dfrac{1}{\sqrt{x}}-1}$Áp dụng AM-GM có: $\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2$Dấu "=" xảy ra khi x=1 (ktm đk)Suy ra dấu bằng ko xảy ra $\Rightarrow\sqrt{x}+\dfrac{1}{\sqrt{x}}-1>2-1=1$$\Rightarrow\dfrac{2}{\sqrt{x}+\dfrac{1}{\sqrt{x}}-1}< 2$ $\Rightarrow N< 2$ mà $N>0$,$N$ nguyên$\Rightarrow N=1\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=1$$\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{3+\sqrt{5}}{2}\\sqrt{x}=\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7+3\sqrt{5}}{2}\x=\dfrac{7-3\sqrt{5}}{2}\end{matrix}\right.$ (tm)Vậy...Câu trả lời: $C=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\left(x>0;x\ne1\right)$$=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}$$=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2$$=x-\sqrt{x}+1$$=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}$Dấu "=" xảy ra khi $\sqrt{x}-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{4}$Vậy $C_{min}=\dfrac{3}{4}$$N=\dfrac{2\sqrt{x}}{C}=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=\dfrac{2}{\sqrt{x}+\dfrac{1}{\sqrt{x}}-1}$Áp dụng AM-GM có: $\sqrt{x}+\dfrac{1}{\sqrt{x}}\ge2$Dấu "=" xảy ra khi x=1 (ktm đk)Suy ra dấu bằng ko xảy ra $\Rightarrow\sqrt{x}+\dfrac{1}{\sqrt{x}}-1>2-1=1$$\Rightarrow\dfrac{2}{\sqrt{x}+\dfrac{1}{\sqrt{x}}-1}< 2$ $\Rightarrow N< 2$ mà $N>0$,$N$ nguyên$\Rightarrow N=1\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=1$$\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{3+\sqrt{5}}{2}\\sqrt{x}=\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7+3\sqrt{5}}{2}\x=\dfrac{7-3\sqrt{5}}{2}\end{matrix}\right.$ (tm)Vậy...

hâyztohehe · Answer

$\Rightarrow C=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1$ * $\Rightarrow C=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}$ Dấu = xảy ra $\Leftrightarrow x=\dfrac{1}{2}$* Ta có $N=\dfrac{2\sqrt{x}}{C}=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}>0\left(1\right)$ Xét $N-2=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}-2=\dfrac{2\sqrt{x}-2x+2\sqrt{x}-2}{x-\sqrt{x}+1}=\dfrac{-2x+4\sqrt{x}-2}{x-\sqrt{x}+1}=\dfrac{-2\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+1}< 0\left(dox\ne1\right)\Rightarrow N< 2\left(2\right)$ Từ (1) và (2) $\Rightarrow0< N< 2$. Mà N nguyên nên N=1 $\Rightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=1\Rightarrow2\sqrt{x}=x-\sqrt{x}+1\Leftrightarrow x-3\sqrt{x}+1=0$$\Delta=9-4=5\Rightarrow$ pt có 2 nghiệm phân biệt: $x_1=\dfrac{\sqrt{5}+3}{2}\left(TM\right);x_2=\dfrac{3-\sqrt{5}}{2}\left(TM\right)$Câu trả lời: $\Rightarrow C=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1$ * $\Rightarrow C=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}$ Dấu = xảy ra $\Leftrightarrow x=\dfrac{1}{2}$* Ta có $N=\dfrac{2\sqrt{x}}{C}=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}>0\left(1\right)$ Xét $N-2=\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}-2=\dfrac{2\sqrt{x}-2x+2\sqrt{x}-2}{x-\sqrt{x}+1}=\dfrac{-2x+4\sqrt{x}-2}{x-\sqrt{x}+1}=\dfrac{-2\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+1}< 0\left(dox\ne1\right)\Rightarrow N< 2\left(2\right)$ Từ (1) và (2) $\Rightarrow0< N< 2$. Mà N nguyên nên N=1 $\Rightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}+1}=1\Rightarrow2\sqrt{x}=x-\sqrt{x}+1\Leftrightarrow x-3\sqrt{x}+1=0$$\Delta=9-4=5\Rightarrow$ pt có 2 nghiệm phân biệt: $x_1=\dfrac{\sqrt{5}+3}{2}\left(TM\right);x_2=\dfrac{3-\sqrt{5}}{2}\left(TM\right)$

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