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a )
Để biểu thức được xác định thì :
\(\left\{{}\begin{matrix}\sqrt{x+1}>0\\\sqrt{x-1}\ge0\end{matrix}\right.\Leftrightarrow x\ge1\)
b )
Để biểu thức được xác định thì :
\(\sqrt{x^2}-1>0\Leftrightarrow x>1\)
Chúc bạn học tốt !
Ta có A=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\) với x≥ 9, x ∈ R
Để A > 0 \(\Leftrightarrow\) \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\) > 0
\(\Leftrightarrow\) \(\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}-2>0\\\sqrt{x}+1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}-2< 0\\\sqrt{x}+1>0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}>2\\\sqrt{x}< -1\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}< 2\\\sqrt{x}>-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>4\\x< 1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 4\\x>1\end{matrix}\right.\end{matrix}\right.\)
Kết hợp với ĐKXĐ\(\Rightarrow\) x ∈ ∅
ĐKXĐ: x≥9, x∈R
Ta có:
A= \(\left[\dfrac{1+\sqrt{x}-\sqrt{x}}{1+\sqrt{x}}\right]\):\(\left[\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{x-2\sqrt{x}-3\sqrt{x}+6}\right]\)
= \(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
=\(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
=\(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
=\(\dfrac{1}{1+\sqrt{x}}\):\(\dfrac{1}{\sqrt{x}-2}\)
=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
a) ĐKXĐ: \(x\ge0;x\ne1\)
b) A= \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}\) + \(\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}\)- \(\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
A= \(\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)}\) - \(\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)- \(\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
= \(\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
= \(\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
= \(\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
= \(\dfrac{-5x+5\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
= \(\dfrac{\left(\sqrt{x}-1\right)\left(-5\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x+3}\right)}\)
= \(\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
c) GTLN (Max)
A= \(\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
= -5+\(\dfrac{17}{\sqrt{x}+3}\)
Ta có: \(\sqrt{x}\)\(\ge\)0 (ĐKXĐ) \(\Rightarrow\) \(\sqrt{x}+3\ge3\)
\(\Rightarrow\) \(\dfrac{1}{\sqrt{x}+3}\le\dfrac{1}{3}\)
\(\Rightarrow\) \(\dfrac{17}{\sqrt{x}+3}\le\dfrac{17}{3}\)
\(\Rightarrow\) \(-5+\dfrac{17}{\sqrt{x}+3}\le-5+\dfrac{17}{3}\)
\(\Leftrightarrow\) A\(\le\dfrac{2}{3}\)
Dấu "=" xảy ra khi \(\sqrt{x}=0\) \(\Rightarrow\) \(x=0\)
Vậy Max A =\(\dfrac{2}{3}\) khi \(x=0\)
Bài 2:
a: ĐKXĐ: 2/3x-1/5>=0
=>2/3x>=1/5
hay x>=3/10
b: ĐKXĐ: \(\dfrac{x+1}{2x-3}>=0\)
=>2x-3>0 hoặc x+1<=0
=>x>3/2 hoặc x<=-1
c: ĐKXĐ: \(\left\{{}\begin{matrix}3x-5>=0\\x-4>0\end{matrix}\right.\Leftrightarrow x>4\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}3-x>=0\\x>=0\\3-x< >4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0\le x\le3\\x< >-1\end{matrix}\right.\Leftrightarrow0\le x\le3\)
b: ĐKXĐ: \(\left\{{}\begin{matrix}x-2>=0\\7-2x>=0\\x-2< >7-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2\le x\le\dfrac{7}{2}\\x< >3\end{matrix}\right.\)
a: \(A=\dfrac{1}{\sqrt{x}-1}-\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+1-x-2-x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{-2x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
Để A>0 thì \(\dfrac{-2x+\sqrt{x}}{\sqrt{x}-1}>0\)
=>\(\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}< 0\)
=>1/2<căn x<1
=>1/4<x<1
b: \(B=\dfrac{2}{A}+\sqrt{x}\)
\(=\dfrac{2\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{-2x+\sqrt{x}}+\sqrt{x}\)
\(=\dfrac{2\left(x\sqrt{x}-1\right)-2x\sqrt{x}+x}{-2x+\sqrt{x}}=\dfrac{x-2}{-2x+\sqrt{x}}=\dfrac{-\left(x-2\right)}{2x-\sqrt{x}}< =0\)
Dấu '=' xảy ra khi x=2
a) ĐKXĐ: \(2-x^2\ge0\Leftrightarrow\left|x\right|< \sqrt{2}\Leftrightarrow-\sqrt{2}\le x\le\sqrt{2}\)
b) ĐKXĐ: \(5x^2-3>0\Leftrightarrow\left|x\right|>\sqrt{\dfrac{3}{5}}\Leftrightarrow x>\sqrt{\dfrac{3}{5}}\) hoặc \(x< -\sqrt{\dfrac{3}{5}}\)
c) ĐKXĐ: \(-\left(2x-1\right)^2=0\Leftrightarrow x=\dfrac{1}{2}\)
d) ĐKXĐ: \(\left(x-1\right)\left(x+2\right)>0\Leftrightarrow x>1\) hoặc \(x< -2\)
Mình nghĩ đề câu a) là \(\frac{1}{1-\sqrt{x^2-3}}\) khi đó
\(1-\sqrt{x^2-3}\ne0\Rightarrow\sqrt{x^2-3}\ne1\Rightarrow x\ne\pm2\)và \(x^2-3\ge0\Leftrightarrow-\sqrt{3}\le x\le\sqrt{3}\)
b)
\(\sqrt{16-x^2}\ge0;\sqrt{2x+1}\ge0;\sqrt{x^2-8x+14}\ge0\)và \(\sqrt{2x+1}\ne0\)
\(\Leftrightarrow-4\le x\le4;x\ge-\frac{1}{2};4-\sqrt{2}\le x\le4+\sqrt{2};x\ne\frac{1}{2}\)
Như vậy \(-\frac{1}{2}< x\le4+\sqrt{2}\)
ĐK:
\(\left\{{}\begin{matrix}x-1\ge0\\2-x>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x< 2\end{matrix}\right.\)
ĐKXĐ: \(1\le x< 2\)