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a: Ta có: \(D=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+1\)

\(=x+\sqrt{x}-2\sqrt{x}-1+1\)

\(=x-\sqrt{x}\)

b: Để D=12 thì D-12=0

\(\Leftrightarrow\sqrt{x}-4=0\)

hay x=16

15 tháng 10 2023

 

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{1;\dfrac{25}{9};\dfrac{9}{4}\right\}\end{matrix}\right.\)

a: \(C=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}-\dfrac{5}{2\sqrt{x}-3}\right):\left(3-\dfrac{2}{\sqrt{x}-1}\right)\)

\(=\dfrac{2\sqrt{x}-5\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}:\dfrac{3\sqrt{x}-3-2}{\sqrt{x}-1}\)

\(=\dfrac{2\sqrt{x}-5\sqrt{x}+5}{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-1}{3\sqrt{x}-5}\)

\(=-\dfrac{1}{2\sqrt{x}-3}\)

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

Khi \(x=4+2\sqrt{3}\) thì \(C=-\dfrac{1}{2\left(\sqrt{3}+1\right)-3}=\dfrac{-1}{2\sqrt{3}-1}=\dfrac{-2\sqrt{3}-1}{11}\)

c: C=-1

=>\(2\sqrt{x}-3=1\)

=>\(\sqrt{x}=2\)

=>x=4(nhận)

d: C>0

=>\(2\sqrt{x}-3< 0\)

=>\(\sqrt{x}< \dfrac{3}{2}\)

=>\(0< =x< \dfrac{9}{4}\)

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< =x< \dfrac{9}{4}\\x< >1\end{matrix}\right.\)

 

10 tháng 7 2021

a) \(B=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\left(x\ge0,x\ne1\right)\)

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

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

\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{x+\sqrt{x}+1}{x-1}=\dfrac{1}{x-1}\)

 

10 tháng 7 2021

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23 tháng 10 2021

a) ĐKXĐ: \(x>0\)

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

\(=x+\sqrt{x}-2\sqrt{x}-1+1=x-\sqrt{x}\)

\(A=x-\sqrt{x}=2\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=0\)

\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)(do \(\sqrt{x}+1\ge1>0\))

b) \(A=x-\sqrt{x}=\sqrt{x}\left(\sqrt{x}-1\right)>0\)(do \(x>1\))

\(\Leftrightarrow A=x-\sqrt{x}=\left|A\right|\)

c) \(A=x-\sqrt{x}=\left(x-\sqrt{x}+\dfrac{1}{4}\right)-\dfrac{1}{4}\)

\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

\(minA=-\dfrac{1}{4}\Leftrightarrow\sqrt[]{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)

23 tháng 10 2021

\(a,A=\dfrac{x\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+1\left(x>0\right)\\ A=\dfrac{x\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-2\sqrt{x}-1+1\\ A=x+\sqrt{x}-2\sqrt{x}=x-\sqrt{x}\\ A=2\Leftrightarrow x-\sqrt{x}-2=0\\ \Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow\sqrt{x}=2\left(\sqrt{x}>0\right)\\ \Leftrightarrow x=4\left(tm\right)\)

\(b,x>1\Leftrightarrow\sqrt{x}-1>0\\ \Leftrightarrow\left|A\right|=\left|x-\sqrt{x}\right|=\left|\sqrt{x}\left(\sqrt{x}-1\right)\right|=\sqrt{x}\left(\sqrt{x}-1\right)=A\left(\sqrt{x}>0\right)\)

\(c,A=x-\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\\ A_{min}=-\dfrac{1}{4}\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)

AH
Akai Haruma
Giáo viên
4 tháng 7 2021

$A=2x-\sqrt{x}=2(x-\frac{1}{2}\sqrt{x}+\frac{1}{4^2})-\frac{1}{8}$

$=2(\sqrt{x}-\frac{1}{4})^2-\frac{1}{8}$

$\geq \frac{-1}{8}$

Vậy $A_{\min}=-\frac{1}{8}$. Giá trị này đạt tại $x=\frac{1}{16}$

 

AH
Akai Haruma
Giáo viên
4 tháng 7 2021

$B=x+\sqrt{x}$

Vì $x\geq 0$ nên $B\geq 0+\sqrt{0}=0$

Vậy $B_{\min}=0$. Giá trị này đạt tại $x=0$

 

27 tháng 8 2021

a, \(x+1\ge0\Leftrightarrow x\ge-1\)

b, \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)

c, \(\left\{{}\begin{matrix}x+1\ge0\\x-2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ge2\end{matrix}\right.\Leftrightarrow x\ge2\)

27 tháng 8 2021

d, \(\left\{{}\begin{matrix}2-3x\ge0\\1-2x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{2}{3}\\x\le\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow x\le\dfrac{1}{2}\)

e, \(\left\{{}\begin{matrix}\sqrt{3}-2x\ge0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{\sqrt{3}}{2}\\x\ne1\end{matrix}\right.\Leftrightarrow x\le\dfrac{\sqrt{3}}{2}\)

25 tháng 5 2021

ĐKXĐ: \(3-2x\ge0\Leftrightarrow x\le\dfrac{3}{2}\)

25 tháng 5 2021

b) ĐKXĐ: \(-1\le x\le3\)

c) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\ne1\\x\ne3\end{matrix}\right.\).

d) ĐKXĐ: \(x< \dfrac{3}{5}\).

18 tháng 9 2021

a) \(D=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

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

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)

\(=\dfrac{-3\sqrt{x}+3}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}-1}=\dfrac{-3}{\sqrt{x}+3}\)

b) \(D=-\dfrac{3}{\sqrt{x}+3}< -\dfrac{1}{4}\)

\(\Leftrightarrow12>\sqrt{x}+3\Leftrightarrow\sqrt{x}< 9\) 

\(\Leftrightarrow0\le x< 81\) và \(x\ne9\)

a) D=\(\left(\dfrac{2\sqrt{x}.\left(\sqrt{x}-3\right)+\sqrt{x}.\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-3\right)}\right)\) \(:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(\Leftrightarrow D=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}\) \(.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(\Leftrightarrow D=\dfrac{-3-3\sqrt{x}}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}+1}\)

\(\Leftrightarrow D=\dfrac{-3.\left(\sqrt{x}+1\right)}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}+1}\)

\(\Leftrightarrow D=\dfrac{-3}{\sqrt{x}+3}\)

b) Để D\(< \dfrac{-1}{4}\) \(\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}< \dfrac{-1}{4}\) 

\(\Leftrightarrow12>\sqrt{x}+3\Leftrightarrow9>\sqrt{x}\Leftrightarrow81>x\ge0\)