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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 : \(x> 0 ; x \neq 1 \)
P = \(\dfrac{3x+3\sqrt{x} - 3}{\sqrt{x^2} +2\sqrt{x} - \sqrt{x} - 2}\) \(- \dfrac{\sqrt{x}+1}{\sqrt{x}+2} + \dfrac{\sqrt{x}-2}{\sqrt{x}} . \dfrac{1-( 1 -\sqrt{x})}{1-\sqrt{x}}\)
= \(\dfrac{3x+3\sqrt{x} - 3 }{\sqrt{x}(\sqrt{x}+2)-(\sqrt{x} - 2)}\) \(- \dfrac{\sqrt{x}+1}{\sqrt{x}+2} + \dfrac{\sqrt{x}-2}{\sqrt{x}}. \dfrac{ 1-1+\sqrt{x}}{1-\sqrt{x}}\)
= \(\dfrac{3x+3\sqrt{x} - 3 }{(\sqrt{x}+2)(\sqrt{x}-1)}\) \(- \dfrac{\sqrt{x}+1}{\sqrt{x}+2} + \dfrac{\sqrt{x}-2}{(\sqrt{x}-1)} \)
= \(\dfrac{3x+3\sqrt{x}-3-(\sqrt{x}-1)(\sqrt{x}-1)-(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x}-1)}\)
= \(\dfrac{3x+3\sqrt{x}-3-(\sqrt{x^2}- 1^2) - (\sqrt {x^2}-2^2)}{(\sqrt{x}+2)(\sqrt{x}-1)}\)
= \(\dfrac{3x+3\sqrt{x} - 3 - x+1-x+4}{(\sqrt{x}+2)(\sqrt{x}-1)} \)
= \(\dfrac{x+3\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x} - 1)}\)
= \(\dfrac{\sqrt{x^2}+2\sqrt{x} +\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x} - 1)} \)
= \(\dfrac{\sqrt{x}(\sqrt{x}+2)+(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x} - 1)} \)
= \(\dfrac{(\sqrt{x}+2)(\sqrt{x}+1)}{(\sqrt{x}+2)(\sqrt{x} - 1)} \)
= \(\dfrac{\sqrt{x}+1}{\sqrt{x} - 1} \)
c, Để P = \(\sqrt{x}\) \(\Leftrightarrow\) \(\dfrac{\sqrt{x}+1}{\sqrt{x} - 1} \) = \(\sqrt{x} \)
\(\Rightarrow\) \(\sqrt{x}+1= \sqrt{x}(\sqrt{x}-1)\)
\(\Leftrightarrow\) \(\sqrt{x}+1 = \sqrt{x^2} - \sqrt{x}\)
\(\Leftrightarrow\) \( \sqrt{x^2} -\sqrt{x} - \sqrt{x} - 1 = 0\)
\(\Leftrightarrow\) \(\sqrt{x^2} - 2\sqrt{x} +1-1-1=0\)
\(\Leftrightarrow\) \((\sqrt{x}-1)^2 - (\sqrt{2})^2 \) = 0
\(\Leftrightarrow\) \((\sqrt{x} - 1 - \sqrt{2})(\sqrt{x} - 1+\sqrt{2})\)
\(\Leftrightarrow\) \(\begin{cases} \sqrt{x} - 1 - \sqrt{2}=0 \\ \sqrt{x} - 1 +\sqrt{2}=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} \sqrt{x} = 1 +\sqrt{2} \\ \sqrt{x} = 1 - \sqrt{2} \end{cases} \) \(\Leftrightarrow\)\(\begin{cases} x = 1+\sqrt{2} = 3+2\sqrt{2} \\ \sqrt{x} = 1-\sqrt{2} < 0 ( LOẠI ) \end{cases} \)
P/s : mk không biết làm phần b
a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\\x\ne\frac{9}{4}\end{matrix}\right.\)
Ta có: \(Q=\frac{\sqrt{x}+2}{-\sqrt{x}+2}+\frac{3\sqrt{x}-4}{2\sqrt{x}-3}+\frac{-7\sqrt{x}+10}{-2x+7\sqrt{x}-6}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(2\sqrt{x}-3\right)}{\left(2-\sqrt{x}\right)\left(2\sqrt{x}-3\right)}+\frac{\left(3\sqrt{x}-4\right)\left(2-\sqrt{x}\right)}{\left(2-\sqrt{x}\right)\left(2\sqrt{x}-3\right)}+\frac{-7\sqrt{x}+10}{\left(2-\sqrt{x}\right)\left(2\sqrt{x}-3\right)}\)
\(=\frac{2x+\sqrt{x}-6-3x+10\sqrt{x}-8-7\sqrt{x}+10}{\left(2-\sqrt{x}\right)\left(2\sqrt{x}-3\right)}\)
\(=\frac{-x+4\sqrt{x}-4}{\left(2-\sqrt{x}\right)\left(2\sqrt{x}-3\right)}\)
\(=\frac{-\left(2-\sqrt{x}\right)^2}{\left(2-\sqrt{x}\right)\left(2\sqrt{x}-3\right)}\)
\(=\frac{\sqrt{x}-2}{2\sqrt{x}-3}\)
b) Để Q<-4 thì Q+4<0
\(\Leftrightarrow\frac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{2\sqrt{x}-3}+\frac{4\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)
\(\Leftrightarrow\frac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)
Trường hợp 1: \(\left\{{}\begin{matrix}9\sqrt{x}-14>0\\2\sqrt{x}-3< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9\sqrt{x}>14\\2\sqrt{x}< 3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}>\frac{14}{9}\\\sqrt{x}< \frac{3}{2}\end{matrix}\right.\)
⇔Loại vì \(\frac{14}{9}>\frac{3}{2}\)
Trường hợp 2: \(\left\{{}\begin{matrix}9\sqrt{x}-14< 0\\2\sqrt{x}-3>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9\sqrt{x}< 14\\2\sqrt{x}>3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}< \frac{14}{9}\\\sqrt{x}>\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< \frac{196}{81}\\x>\frac{9}{4}\end{matrix}\right.\Leftrightarrow\frac{9}{4}< x< \frac{196}{81}\)
Kết hợp ĐKXĐ, ta được:
\(\frac{9}{4}< x< \frac{196}{81}\)
Vậy: Để Q<-4 thì \(\frac{9}{4}< x< \frac{196}{81}\)
1.a)ĐKXĐ:\(\left\{{}\begin{matrix}x-1\ge0\\1-\sqrt{x-1}\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x\ne2\end{matrix}\right.\)
b)ĐKXĐ:\(\left\{{}\begin{matrix}x^2-2x+1\ge0\\\sqrt{x^2-2x+1}\ne0\end{matrix}\right.\Leftrightarrow x^2-2x+1>0\Leftrightarrow\left(x-1\right)^2>0\) luôn đúng với mọi x \(\ne\)1
Vậy biểu thức xác định khi \(x\ne1\)
2.\(B=\frac{\sqrt{8-\sqrt{15}}}{\sqrt{30}-\sqrt{2}}=\frac{\sqrt{16-2\sqrt{15}}}{\sqrt{60}-2}=\frac{\sqrt{15-2\sqrt{15}+1}}{2\sqrt{15}-2}=\frac{\sqrt{\left(\sqrt{15}-1\right)^2}}{2\left(\sqrt{15}-1\right)}=\frac{\sqrt{15}-1}{2\left(\sqrt{15}-1\right)}=\frac{1}{2}\)
3.a)ĐKXĐ:\(x\ge0\)
b)\(Q=\left(\frac{1}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\right):\frac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}+1}-\frac{1}{\sqrt{x}\left(x+1\right)}\right):\frac{x-\sqrt{x}+1}{\sqrt{x}^3+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}\left(x+1\right)}.\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
c)\(Q=\frac{\sqrt{x}-1}{\sqrt{x}}=1-\frac{1}{\sqrt{x}}\)
Để \(Q\in Z\) thì
\(1⋮\sqrt{x}\)
\(\Rightarrow\sqrt{x}\in\left\{-1;1\right\}\)(loại -1 vì \(\sqrt{x}\ge0\))
\(\Rightarrow x\in\left\{1\right\}\)