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a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
ĐKXĐ:
a/ \(A=\left[\frac{\left(\sqrt{a}+3\right)^2-\left(\sqrt{a}-3\right)^2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\right].\frac{\sqrt{a}-3}{3\sqrt{a}}=\left(\frac{a+6\sqrt{a}+9-a+6\sqrt{a}-9}{\sqrt{a}+3}\right).\frac{1}{3\sqrt{a}}\)
\(=\frac{12\sqrt{a}}{\sqrt{a}+3}.\frac{1}{3\sqrt{a}}=\frac{4}{\sqrt{a}+3}\)
b/ Để \(A\in Z\)thì \(\sqrt{a}+3\inƯ\left(4\right)=\left\{1;-1;2;-2;4;-4\right\}\)
Với \(\sqrt{a}+3=1\Rightarrow\sqrt{a}=-2\) (vô nghiệm)
Với \(\sqrt{a}+3=-1\Rightarrow\sqrt{a}=-4\)(vô nghiệm)
Với \(\sqrt{a}+3=2\Rightarrow\sqrt{a}=-1\) (vô nghiệm)
Với \(\sqrt{a}+3=-2\Rightarrow\sqrt{a}=-5\)(vô nghiệm)
Với \(\sqrt{a}+3=4\Rightarrow\sqrt{a}=1\Rightarrow a=1\)(nhận)
Với \(\sqrt{a}+3=-4\Rightarrow\sqrt{a}=-7\)(vô nghiệm)
Vậy a = 1 thì \(A\in Z\)
c) \(C=\frac{\left(2\sqrt{x}+x\right)\left(\sqrt{x}+1\right)-\left(x\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\frac{x+\sqrt{x}+1-\left(\sqrt{x}+2\right)}{x+\sqrt{x}+1}=\)
\(C=\frac{x\sqrt{x}+2x+x+2\sqrt{x}-x\sqrt{x}+1}{\left(\left(\sqrt{x}\right)^3-1\right)\left(\sqrt{x}+1\right)}\times\frac{x+\sqrt{x}+1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}\times\frac{x+\sqrt{x}+1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\times\frac{1}{x-1}=\)
\(C=\frac{3x+2\sqrt{x}+1}{x-1}\times\frac{1}{x-1}=\frac{3x+2\sqrt{x}+1}{\left(x-1\right)^2}.\)
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
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 => \(x\in\left\{1,25\right\}\)
\(P=\left(\frac{\sqrt{a}}{3+\sqrt{a}}+\frac{a+9}{9-a}\right):\left(\frac{3\sqrt{a}+1}{a-3\sqrt{a}}-\frac{1}{\sqrt{a}}\right)\)
\(P=\left[\frac{\sqrt{a}\left(3-\sqrt{a}\right)}{9-a}+\frac{a+9}{9-a}\right]:\left[\frac{3\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-3\right)}-\frac{1}{\sqrt{a}}\right]\)
\(P=\frac{3\sqrt{a}-a+a+9}{9-a}:\left[\frac{3\sqrt{a}+1-\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-3\right)}\right]\)
\(P=\frac{3\sqrt{a}+9}{9-a}:\frac{2\sqrt{a}+4}{\sqrt{a}\left(\sqrt{a}-3\right)}\)
\(P=\frac{3\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(3-\sqrt{a}\right)}.\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{2\sqrt{a}+4}\)
\(P=\frac{-3\sqrt{a}}{2\sqrt{a}+4}\)
b) theo câu a) \(P=\frac{-3\sqrt{a}}{2\sqrt{a}+4}\) với \(ĐKXĐ:a\ge0;a\ne9\)
theo bài ra \(P< -1\Leftrightarrow\frac{-3\sqrt{a}}{2\sqrt{a}+4}< -1\)
\(\Rightarrow\frac{-3\sqrt{a}}{2\sqrt{a}+4}+1< 0\)
\(\Rightarrow\frac{-3\sqrt{a}+2\sqrt{a}+4}{2\sqrt{a}+4}< 0\)
\(\Rightarrow\frac{4-\sqrt{a}}{2\sqrt{a}+4}< 0\)
\(\Rightarrow4-\sqrt{a}< 0\)
\(\Rightarrow-\sqrt{a}< -4\)
\(\Rightarrow\sqrt{a}>4\)
\(\Rightarrow a>16\)
kết hợp đkxđ a>0 và a\(\ne9\)ta có:
với \(a>16\) \(\Leftrightarrow\hept{\begin{cases}0< a< 16\\a\ne9\end{cases}}\)