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23 tháng 6 2018

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

\(2.\dfrac{1}{\sqrt{9-12x+4x^2}}=\dfrac{1}{\sqrt{\left(2x-3\right)^2}}=\dfrac{1}{|2x-3|}\)

\(3.\dfrac{1}{\sqrt{x+2\sqrt{x-1}}}=\dfrac{1}{\sqrt{x-1+2\sqrt{x-1}+1}}=\dfrac{1}{\sqrt{\left(\sqrt{x-1}+1\right)^2}}=\dfrac{1}{|\sqrt{x-1}+1|}\)

23 tháng 6 2018

bạn cho mình hỏi tại sao từ \(\sqrt{x-1-2\sqrt{x-1}+1}\) sang \(\sqrt{\left(\sqrt{x-1}-1\right)^2}\)

8 tháng 8 2018

\(a.Để:A=\sqrt{x-3}-\sqrt{\dfrac{1}{4-x}}\) xác định thì :

\(\left\{{}\begin{matrix}x-3\ge0\\4-x>0\end{matrix}\right.\)\(\Leftrightarrow3\le x< 4\)

\(b.Để:B=\dfrac{1}{\sqrt{x-1}}+\dfrac{2}{\sqrt{x^2-4x+4}}=\dfrac{1}{\sqrt{x-1}}+\dfrac{2}{\sqrt{\left(x-2\right)^2}}\)xác định thì :

\(x-1>0\Leftrightarrow x>1\)

17 tháng 8 2017

các biểu thức trong căn pt hết về HĐT rồi phá ra là done

\(P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{2\sqrt{x}+7}{x-4}\right)\)

\(=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\left(\dfrac{x+2\sqrt{x}-x+\sqrt{x}+2-2\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)

\(=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-5}\)

\(=\dfrac{-x+8\sqrt{x}-15+\left(x-4\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)

\(=\dfrac{-x+8\sqrt{x}-15+x\sqrt{x}-2x-4\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)

\(=\dfrac{x\sqrt{x}-3x+4\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)

24 tháng 9 2021

\(ĐK:x\ge0;x\ne4\\ P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\dfrac{x+2\sqrt{x}-x+\sqrt{x}+2-2\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-5}\\ P=\dfrac{\left(3-\sqrt{x}\right)\left(\sqrt{x}-5\right)+\left(x-4\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\\ P=\dfrac{8\sqrt{x}-15-x+x\sqrt{x}-2x-4\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\\ P=\dfrac{x\sqrt{x}-3x+4\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)

14 tháng 8 2018

a/ đkxđ: \(x^2-8x+15>0\)

\(\Leftrightarrow x^2-8x+16-1>0\)

\(\Leftrightarrow\left(x-4\right)^2>1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4>1\Rightarrow x>5\\x-4< -1\Leftrightarrow x< 3\end{matrix}\right.\)

Vậy x < 3 hoặc x > 5

b/ đkxđ: \(2-x^2\ge0\)\(\Leftrightarrow-\sqrt{2}\le x\le\sqrt{2}\)

vậy.........

c/ đkxđ: \(\dfrac{2x-1}{1-x}\ge0\) và 1 - x ≠ 0

=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-1\ge0\\1-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1\le0\\1-x< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x< 1\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\x>1\end{matrix}\right.\end{matrix}\right.\)

=> \(\dfrac{1}{2}\le x< 1\)

Vậy.........

b: \(=\dfrac{\left|x\right|+\left|x-2\right|+1}{2x-1}=\dfrac{x+x-2+1}{2x-1}=\dfrac{2x-1}{2x-1}=1\)

c: \(=\left|x-4\right|+\left|x-6\right|\)

=x-4+6-x=2

NV
1 tháng 3 2019

a/ \(x^2+4x-5>0\Rightarrow\left[{}\begin{matrix}x>1\\x< -5\end{matrix}\right.\)

b/ \(\left\{{}\begin{matrix}2x-1\ge0\\x-\sqrt{2x-1}>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\\left\{{}\begin{matrix}x>0\\x^2>2x-1\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\ne1\end{matrix}\right.\)

c/ \(\left\{{}\begin{matrix}x^2-3\ge0\\1-\sqrt{x^2-3}\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge\sqrt{3}\\x\le-\sqrt{3}\end{matrix}\right.\\x\ne\pm2\end{matrix}\right.\)

d/ \(\left\{{}\begin{matrix}x+\dfrac{1}{x}\ge0\\-2x\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>0\\x\le0\end{matrix}\right.\) \(\Rightarrow\) không tồn tại x thỏa mãn

e/ \(\left\{{}\begin{matrix}3x-1\ge0\\5x-3\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\x\ge\dfrac{3}{5}\end{matrix}\right.\) \(\Rightarrow x\ge\dfrac{3}{5}\)

17 tháng 7 2018

\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)

\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)

\(\Leftrightarrow\sqrt{x}-2< 0\)

\(\Leftrightarrow x< 4\)

Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)

KL............

\(2.\) Tương tự bài 1.

\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)

\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)

9 tháng 9 2021

\(D=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{3\sqrt{x}+1}{x-1}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\left(x\ge0;x\ne1\right)\\ D=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\\ D=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\cdot\dfrac{1}{\sqrt{x}+2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)