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 \(Q>\frac{8}{3}\)

\(ĐK:x\ge1,y\ge2,z\ge3\)

\(PT\Leftrightarrow\sqrt{x-1}+\frac{1}{\sqrt{x-1}}+\sqrt{y-2}+\frac{1}{\sqrt{y-2}}+\sqrt{z-3}+\frac{1}{\sqrt{z-3}}=6\)

Theo bđt AM-GM thì \(VT\ge6\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\sqrt{x-1}=\frac{1}{\sqrt{x-1}}=1\\\sqrt{y-2}=\frac{1}{\sqrt{y-2}}=1\\\sqrt{z-3}=\frac{1}{\sqrt{z-3}}=1\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=1\\y=3\\z=4\end{cases}}\)

a) Ta có: \(A=\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}}{x-4\sqrt{x}+3}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{\sqrt{x}-1}\)

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

\(=\dfrac{4\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{1}{2\left(\sqrt{x}-1\right)}\)

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

a)= \(\left(3+\sqrt{5}\right)\left(\sqrt{\left(3-\sqrt{5}\right)^2}\right)\)=\(\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)=9-5=4\)

b)= \(\frac{2\left(3-\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}+\frac{\sqrt{2^2.7}}{2}-2\)=\(\frac{2\left(3-\sqrt{7}\right)}{9-7}+\sqrt{7}-2\)=1

c) =\(\frac{3}{3\left(\sqrt{7}-2\right)}-\frac{3}{3\left(\sqrt{7}+2\right)}\)=\(\frac{1}{\sqrt{7}-2}-\frac{1}{\sqrt{7}+2}=\frac{\sqrt{7}+2-\left(\sqrt{7}-2\right)}{\left(\sqrt{7}+2\right)\left(\sqrt{7}-2\right)}\)=\(\frac{4}{7-4}=\frac{4}{3}\)

d) =\(\left(\sqrt{3}+1\right)\sqrt{\frac{\left(14-6\sqrt{3}\right)^{ }\left(5-\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}}\)=\(\left(1+\sqrt{3}\right)\sqrt{\frac{\left(88-44\sqrt{3}\right)}{25-3}}\)=\(\left(1+\sqrt{3}\right)\sqrt{\frac{22\left(4-2\sqrt{3}\right)}{22}}\)=\(\left(1+\sqrt{3}\right)\sqrt{\left(\sqrt{3}-1\right)^2}=\left(1+\sqrt{3}\right)\left(\sqrt{3}-1\right)\)=3-1 = 2

e) = \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{7\sqrt{x}-3}{x-9}+\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{3-\sqrt{x}}\)= \(\frac{x-4\sqrt{x}+3}{x-9}+\frac{7\sqrt{x}-3}{x-9}+\sqrt{x}\)= \(\frac{x+3\sqrt{x}}{x-9}+\sqrt{x}=\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\sqrt{x}\)= \(\frac{\sqrt{x}}{\sqrt{x}-3}+\sqrt{x}=\frac{x-2\sqrt{x}}{\sqrt{x}-3}\)

a) \(A=\frac{-\sqrt{x}+2+4}{\sqrt{x}-2}=-1+\frac{4}{\sqrt{x}-2}\)

Để \(A\in Z\Leftrightarrow\sqrt{x}-2\in\left\{-4;-2;-1;1;2;4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{-2;0;1;3;4;6\right\}\)

Mà \(x\in Z;\sqrt{x}\ge0\Rightarrow x\in\left\{0;1;9;16;36\right\}\)

b)\(A=\frac{4\sqrt{x}-2+3}{2\sqrt{x}-1}=2+\frac{3}{2\sqrt{x}-1}\)

Để \(A\in Z\Leftrightarrow2\sqrt{x}-1\in\left\{-3;-1;1;3\right\}\)

\(\Leftrightarrow2\sqrt{x}\in\left\{-2;0;2;4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{-1;0;1;2\right\}\Leftrightarrow x\in\left\{0;1;4\right\}\)

a) A= \(\frac{-\sqrt{x}+6}{\sqrt{x}-2}=\frac{-\sqrt{x}+2+4}{\sqrt{x}-2}=\frac{-\left(\sqrt{x}-2\right)+4}{\sqrt{x}-2}=\frac{4}{\sqrt{x}-2}-1\)

⇒ \(\sqrt{x}-2\inƯ\left(4\right)\) ⇒ x = 36

Đề hay nhỉ:) bấm vào câu hỏi thì ko thấy đề hai câu đầu, thoát ra ngoài trang hỏi đáp thì lại thấy;) công nhận người hỏi đăng hay:)

a) \(A=\frac{-\sqrt{x}+6}{\sqrt{x}-2}=\frac{-\sqrt{x}+2+4}{\sqrt{x}-2}=-1+\frac{4}{\sqrt{x}-2}\)

Do x thuộc Z nên để A nguyên \(\sqrt{x}-2\inƯ\left(4\right)=\left\{-2;-1;1;2;4\right\}\) (ko xét -4 vì \(\sqrt{x}-2\ge-2>-4\))

Rồi làm tiếp.

Sai thì em chịu

a) Ta có: \(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)

\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\dfrac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)

\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-1\right):\left(\dfrac{25-x-\left(x-9\right)+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)

\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}+5}-\dfrac{\sqrt{x}+5}{\sqrt{x}+5}\right):\left(\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)

\(=\dfrac{\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}:\dfrac{x+9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-5}{\sqrt{x}+5}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{x+9}\)

\(=\dfrac{-5\left(\sqrt{x}-3\right)}{x+9}\)