LEVER

\(\frac{3}{a+b\sqrt{3}}-\frac{2}{a-b\sqrt{3}}=7-20\sqrt{3}\)\(\Leftrightarrow\frac{3\left(a-b\sqrt{3}\right)-2\left(a+b\sqrt{3}\right)}{a^2-3b^2}=7-20\sqrt{3}\)

\(\Leftrightarrow\frac{a-5\sqrt{3}b}{a^2-3b^2}=7-20\sqrt{3}\)\(\Leftrightarrow\frac{a-5\sqrt{3}b}{a^2-3b^2}=\frac{7-20\sqrt{3}}{49-48}\Leftrightarrow\frac{a-5\sqrt{3}b}{a^2-3b^2}=\frac{7-20\sqrt{3}}{7^2-3.4^2}\Leftrightarrow\hept{\begin{cases}a=7\\b=4\end{cases}}\)

Sai đề ?

sr nha này : \(\frac{3}{a+\sqrt{3}}-\frac{2}{a-b\sqrt{3}}=7-20\sqrt{3}\)tìm a,b hữu tỉ

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)

\(A=\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\)\(\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)

\(=\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\right)\)\(:\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)

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

\(=\frac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\)\(.\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)

\(=\frac{2\sqrt{x}+3}{2\sqrt{x}+1}.\frac{5\sqrt{x}}{2\sqrt{x}+3}=\frac{5\sqrt{x}}{2\sqrt{x}+1}\)

\(A\in Z\Leftrightarrow\frac{5\sqrt{x}}{2\sqrt{x}+1}\in Z\Leftrightarrow\frac{10\sqrt{x}}{2\sqrt{x}+1}\in Z\)

\(\Rightarrow\frac{10\sqrt{x}+5-5}{2\sqrt{x}+1}\in Z\Leftrightarrow5-\frac{5}{2\sqrt{x}+1}\in Z\)

\(\Rightarrow\frac{5}{2\sqrt{x}+1}\in Z\Rightarrow2\sqrt{x}+1\inƯ_5\)

Mà \(Ư_5=\left\{\pm1;\pm5\right\}\)

Nhưng \(2\sqrt{x}+1\ge1\)

\(\Rightarrow\orbr{\begin{cases}2\sqrt{x}+1=1\\2\sqrt{x}+1=5\end{cases}\Rightarrow\orbr{\begin{cases}2\sqrt{x}=0\\2\sqrt{x}=4\end{cases}}}\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}}\)

Vậy \(x\in\left\{0;4\right\}\)

Trả lời:

a, \(A=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}-3}{\sqrt{x}-3}-\frac{2x-\sqrt{x}-3}{x-9}\) \(\left(đkxđ:x\ge0;x\ne9\right)\)

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

\(=\frac{x-3\sqrt{x}}{x-9}+\frac{2x+3\sqrt{x}-9}{x-9}-\frac{2x-\sqrt{x}-3}{x-9}\)

\(=\frac{x-3\sqrt{x}+2x+3\sqrt{x}-9-2x+\sqrt{x}+3}{x-9}\)

\(=\frac{x+\sqrt{x}-6}{x-9}\)

\(\frac{3}{a+b\sqrt{3}}-\frac{2}{a-b\sqrt{3}}=720\sqrt{3}\)

<=> \(a-5b\sqrt{3}=720\sqrt{3}\left(a^2-3b^2\right)\)

<=> \(a=\sqrt{3}\left(5b+720a^2-2160b^2\right)\)

Do a ,b là số hữu tỉ 

=> \(a=5b+720a^2-2160b^2=0\)

=> \(\hept{\begin{cases}a=0\\5b-2160b^2=0\end{cases}}\)

Mà a,b không đồng thời bằng 0

=> \(a=0;b=\frac{1}{432}\)

Vậy \(a=0;b=\frac{1}{432}\)