a)ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

Ta có: \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\frac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{15\sqrt{x}-11-\left(3x+9\sqrt{x}-2\sqrt{x}-6\right)-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

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

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

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

\(=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

Ta có: \(A-\frac{2}{3}=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}-\frac{2}{3}\)

\(=\frac{3\left(-5\sqrt{x}+2\right)}{3\left(\sqrt{x}+3\right)}-\frac{2\left(\sqrt{x}+3\right)}{3\left(\sqrt{x}+3\right)}\)

\(=\frac{-15\sqrt{x}+6-2\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}\)

\(=\frac{-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}\)

\(=\frac{-17\sqrt{x}-51+51}{3\left(\sqrt{x}+3\right)}\)

\(=\frac{-17}{3}+\frac{17}{\sqrt{x}+3}\)

Ta có: \(\sqrt{x}+3\ge3\forall x\) thỏa mãn ĐKXĐ

\(\Rightarrow\frac{17}{\sqrt{x}+3}\le\frac{17}{3}\forall x\) thỏa mãn ĐKXĐ

\(\Rightarrow\frac{17}{\sqrt{x}+3}-\frac{17}{3}\le\frac{17}{3}-\frac{17}{3}=0\forall x\) thỏa mãn ĐKXĐ

\(\Rightarrow A-\frac{2}{3}\le0\forall x\) thỏa mãn ĐKXĐ

nên \(A\le\frac{2}{3}\)(đpcm)

c) Ta có: \(C=\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\frac{a-2\sqrt{ab}+b}{a-b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\frac{\sqrt{a}-\sqrt{b}+2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)

\(=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}=1\)

Vậy: Giá trị của C không phụ thuộc vào a,b(đpcm)

ĐKCĐ: \(x\ge0;x\ne9,x\ne4\)

\(A=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\\ \)

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

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

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

b, \(A\inℤ\Leftrightarrow\frac{3}{\sqrt{x}-2}\inℤ\)

Nếu x không là số chính phương thì  \(\sqrt{x}\)là số vô tỉ thì \(\sqrt{x}-2\)là số vô tỉ\(\Rightarrow A=\frac{3}{\sqrt{x}-2}\)là số vô tỉ

Nếu x là số chính phương thì \(\sqrt{x}\)là số nguyên thì \(\sqrt{x}-2\inℤ\Rightarrow\sqrt{x}-2\inƯ\left(3\right)\Rightarrow\sqrt{x}-2\in\left\{\pm1;\pm3\right\}\Rightarrow\sqrt{x}\in\left\{1;3;5\right\}\)\(\Rightarrow x\in\left\{1;9;25\right\}\)

Mà theo ĐKXĐ có x khác 9 => \(x\in\left\{1,25\right\}\)

a. ĐK \(\hept{\begin{cases}a\ge0\\a\ne4\\a\ne9\end{cases}}\)

P=\(\frac{2\sqrt{a}-9-\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)+\left(2\sqrt{a}+1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)

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

\(=\frac{a-\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}-2\right)}=\frac{\sqrt{a}+1}{\sqrt{a}-3}\)

b. P = \(\frac{\sqrt{a}+1}{\sqrt{a}-3}=1+\frac{4}{\sqrt{a}-3}\)

P nguyên \(\sqrt{a}-3\inƯ\left(4\right)\Rightarrow\sqrt{a}-3\in\left\{-4;-2;-1;1;2;4\right\}\)

\(\Rightarrow\sqrt{a}\in\left\{1;2;4;5;7\right\}\Rightarrow a\in\left\{1;4;16;25;49\right\}\)

c. \(P< 1\Rightarrow P-1< 0\Rightarrow\frac{\sqrt{a}+1-\sqrt{a}+3}{\sqrt{a}-3}< 0\Rightarrow\frac{4}{\sqrt{a}-3}< 0\)

\(\Rightarrow0\le a< 9\)và \(a\ne4\)

em phai khong biet

moi hoc lop 6 thoi anh a

ĐKXĐ:

\(\left\{{}\begin{matrix}a\ge0\\\sqrt{a}\ne3\\a\ne9\\\frac{2\sqrt{a}-2}{\sqrt{a}-3}-1\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a\ge0\\a\ne9\\\sqrt{a}+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\ge0\\a\ne9\end{matrix}\right.\)

a,

\(Q=\left(\frac{2\sqrt{a}}{\sqrt{a}+3}-\frac{\sqrt{a}}{\sqrt{a}-3}-\frac{3a+3}{a-9}\right):\left(\frac{2\sqrt{a}-2}{\sqrt{a}-3}-1\right)\)

\(=\frac{2\sqrt{a}\left(\sqrt{a}-3\right)-\sqrt{a}\left(\sqrt{a}+3\right)-3a-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}:\frac{2\sqrt{a}-2-\sqrt{a}+3}{\sqrt{a}-3}\)

\(=\frac{2a-6\sqrt{a}-a-3\sqrt{a}-3a-3}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\cdot\frac{\sqrt{a}-3}{\sqrt{a}+1}\)

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

b,

\(Q< -\frac{1}{2}\Leftrightarrow\frac{-2a-9\sqrt{a}-3}{a+4\sqrt{a}+3}< -\frac{1}{2}\)

\(\Leftrightarrow\frac{2a+9\sqrt{a}+3}{a+4\sqrt{a}+3}>\frac{1}{2}\)

\(\Leftrightarrow4a+18\sqrt{a}+6>a+4\sqrt{a}+3\)

\(\Leftrightarrow3a+14\sqrt{a}+3>0\)

Vậy với mọi thỏa ĐKXĐ thì \(Q< -\frac{1}{2}\)

c,

\(Q=\frac{-2a-9\sqrt{a}-3}{a+4\sqrt{a}+3}=-\frac{\left(a+4\sqrt{a}+3\right)+a+5\sqrt{a}}{a+4\sqrt{a}+3}=-1-\frac{a+5\sqrt{a}}{a+4\sqrt{a}+3}\)

mình nghx đề có vấn đề, số xấu quá

a) A = \(\left(\frac{2\sqrt{a}\left(\sqrt{a}-3\right)}{a-9}+\frac{\sqrt{a}\left(\sqrt{a}+3\right)}{a-9}-\frac{3a-3}{a-9}\right):\left(\frac{2\sqrt{a}-2}{\sqrt{a}-3}-\frac{\sqrt{a}-3}{\sqrt{a}-3}\right)\) (quy đồng lên thôi)

\(=\left(\frac{2a-6\sqrt{a}}{a-9}+\frac{a+3\sqrt{a}}{a-9}-\frac{3a-3}{a-9}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-3}\right)\) (khai triển)

\(=\left(\frac{-3\sqrt{a}+3}{a-9}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-3}\right)\) (rút gọn)

\(=\frac{-3\left(\sqrt{a}-1\right)}{a-9}.\frac{\sqrt{a}-3}{\sqrt{a}+1}=\frac{-3\left(\sqrt{a}-1\right)\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)\left(\sqrt{a}+1\right)}=\frac{-3\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}+1\right)}\)

\(=\frac{-3\left(t-1\right)}{\left(t+3\right)\left(t+1\right)}\left(\text{đặt }\sqrt{a}=t\ge0\right)\)

b) Để A < 1/2 thì \(\frac{-3\left(t-1\right)}{\left(t+3\right)\left(t+1\right)}< \frac{1}{2}\Leftrightarrow-3\left(t-1\right)< \frac{1}{2}\left(t+3\right)\left(t+1\right)\)

\(\Leftrightarrow-3t+3< \frac{1}{2}\left(t^2+4t+3\right)\)

\(\Leftrightarrow-6t+6< t^2+4t+3\)

\(\Leftrightarrow t^2+10t-3>0\)

Giải ra nhưng số xấu quá:(

c) + d) Bí

Sai thì chịu:(

\(a,\left(1+\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\left(1-\frac{a+\sqrt{a}}{1+\sqrt{a}}\right)=\left(1+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)

\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1^2-\sqrt{a}^2=1-a\)

\(b,\left(2-\frac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(2-\frac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)=\left(2-\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}\right)\left(2-\frac{-\sqrt{a}\left(\sqrt{b}-5\right)}{\sqrt{b}-5}\right)\)

\(=\left(2-\sqrt{a}\right)\left(2+\sqrt{a}\right)=2^2-\sqrt{a}^2=2-a\)

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

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

\(d,\left(\frac{a-\sqrt{a}}{\sqrt{a}-1}+2\right)\left(2-\frac{\sqrt{a}+a}{1+\sqrt{a}}\right)=\left(\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}+2\right)\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)

\(=\left(\sqrt{a}+2\right)\left(2-\sqrt{a}\right)=2^2-\sqrt{a}^2=2-a\)