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a/ ĐKXĐ: \(x\ge0;x\ne4\)
\(P=\frac{\left(2+\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}-\frac{\left(2-\sqrt{x}\right)^2}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\frac{4}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\)
\(=\frac{4+4\sqrt{x}+x-4+4\sqrt{x}-x+4}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}=\frac{8\sqrt{x}+4}{4-x}\)
\(P=2\Leftrightarrow\frac{8\sqrt{x}+4}{4-x}=2\)
\(\Leftrightarrow4\sqrt{x}+2=4-x\)
\(\Leftrightarrow x+4\sqrt{x}-2=0\Rightarrow\left[{}\begin{matrix}\sqrt{x}=\sqrt{6}-2\\\sqrt{x}=-\sqrt{6}-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=10-4\sqrt{6}\)
Câu c đề thiếu
1) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne\frac{4}{9}\end{matrix}\right.\)
Ta có: \(Q=\frac{-5\sqrt{x}+4}{3\sqrt{x}-2}+\frac{6\sqrt{x}+4}{2\sqrt{x}+3}+\frac{29\sqrt{x}-28}{3\left(6x+5\sqrt{x}-6\right)}\)
\(=\frac{3\left(-5\sqrt{x}+4\right)\left(2\sqrt{x}+3\right)}{3\left(3\sqrt{x}-2\right)\left(2\sqrt{x}+3\right)}+\frac{3\left(6\sqrt{x}+4\right)\left(3\sqrt{x}-2\right)}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}+\frac{29\sqrt{x}-28}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}\)
\(=\frac{3\left(-10x-7\sqrt{x}+12\right)}{3\left(3\sqrt{x}-2\right)\left(2\sqrt{x}+3\right)}+\frac{3\left(18x-8\right)}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}+\frac{29\sqrt{x}-28}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}\)
\(=\frac{-30x-21\sqrt{x}+36+54x-24+29\sqrt{x}-28}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}\)
\(=\frac{24x+8\sqrt{x}-16}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}\)
\(=\frac{8\left(3x+3\sqrt{x}-2\sqrt{x}-2\right)}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}\)
\(=\frac{8\left(\sqrt{x}+1\right)\left(3\sqrt{x}-2\right)}{3\left(2\sqrt{x}+3\right)\left(3\sqrt{x}-2\right)}\)
\(=\frac{8\sqrt{x}+8}{6\sqrt{x}+9}\)
2) Để \(Q>\frac{8}{3}\) thì \(Q-\frac{8}{3}>0\)
\(\Leftrightarrow\frac{8\sqrt{x}+8}{6\sqrt{x}+9}-\frac{8}{3}>0\)
\(\Leftrightarrow\frac{24\sqrt{x}+24}{3\left(6\sqrt{x}+9\right)}-\frac{8\left(6\sqrt{x}+9\right)}{3\left(6\sqrt{x}+9\right)}>0\)
\(\Leftrightarrow\frac{24\sqrt{x}+24-48\sqrt{x}-72}{9\left(2\sqrt{x}+3\right)}>0\)
mà \(9\left(2\sqrt{x}+3\right)>0\forall x\) thỏa mãn ĐKXĐ
nên \(-24\sqrt{x}-48>0\)
\(\Leftrightarrow-24\left(\sqrt{x}+2\right)>0\)
\(\Leftrightarrow\sqrt{x}+2< 0\)(Vô lý)
Vậy: Không có giá trị nào của x thỏa mãn \(Q>\frac{8}{3}\)
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 \(x=0\)
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= \(\left[\dfrac{1+\sqrt{x}-\sqrt{x}}{1+\sqrt{x}}\right]\):\(\left[\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{x-2\sqrt{x}-3\sqrt{x}+6}\right]\)
= \(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
=\(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
=\(\left[\dfrac{1}{1+\sqrt{x}}\right]\):\(\left[\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\)
=\(\dfrac{1}{1+\sqrt{x}}\):\(\dfrac{1}{\sqrt{x}-2}\)
=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
ĐKXĐ
a) \(x\ge0\) và \(x\ne4\)
b)=\(\sqrt{\dfrac{x+2}{x-2}}\)
ĐKXĐ
TH1 \(\left\{{}\begin{matrix}X+2\ge0\\X-2>0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}X\ge-2\\X>2\end{matrix}\right.\) => X > 2
TH2 \(\left\{{}\begin{matrix}x+2\le0\\x-2< 0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\le-2\\x< 2\end{matrix}\right.\) => X \(\le\) -2
vậy ĐKXĐ\(\left[{}\begin{matrix}X>2\\X\le-2\end{matrix}\right.\)
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\y\ge0\\x\ne y\end{matrix}\right.\)
b) Ta có: \(A=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\frac{1}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\frac{\sqrt{x}-\sqrt{y}}{1}\)
\(=x-y\)
c) Ta có: \(x=\sqrt{3+2\sqrt{2}}\)
\(=\sqrt{2+2\cdot\sqrt{2}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(=\left|\sqrt{2}+1\right|\)
\(=\sqrt{2}+1\)(Vì \(\sqrt{2}>1>0\))(nhận)
Ta có: \(y=\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{2-2\cdot\sqrt{2}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\left|\sqrt{2}-1\right|\)
\(=\sqrt{2}-1\)(Vì \(\sqrt{2}>1\))(nhận)
Thay \(x=\sqrt{2}+1\) và \(y=\sqrt{2}-1\) vào biểu thức A=x-y, ta được:
\(A=\sqrt{2}+1-\left(\sqrt{2}-1\right)\)
\(=\sqrt{2}+1-\sqrt{2}+1\)
\(=2\)
Vậy: Khi \(x=\sqrt{3+2\sqrt{2}}\) và \(y=\sqrt{3-2\sqrt{2}}\) thì A=2