Question

**Bài 5.** Tìm điều kiện xác định và rút gọn các biểu thức sau:

a) \( A = \frac{\sqrt{x}}{x - 5\sqrt{x}} - \frac{10}{x - 25} \).

b) \( B = \frac{\sqrt{x}}{\sqrt{x} + 3} - \frac{3\sqrt{x}}{\sqrt{x} - 3} + \frac{2x + 36}{x - 9} \).

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

d) \( D = \left( \frac{\sqrt{x}}{\sqrt{x - 1}} + \frac{2\sqrt{x} - 1}{\sqrt{x + 1}} + \frac{6}{1 - x} \right) : \left( 3 - \frac{2}{\sqrt{x - 1}} \right) \).

e) \( E = \left( \frac{\sqrt{x}}{\sqrt{x + 3}} + \frac{2\sqrt{x - 1}}{\sqrt{x - 3}} - \frac{2x - \sqrt{x - 3}}{x - 9} \right) : \frac{3\sqrt{x}}{\sqrt{x - 3}} \).

f) \( F = \frac{\sqrt{x + \frac{xy + \sqrt{y}^2}{\sqrt{x} + \sqrt{y}}}}{\sqrt{x - \sqrt{y}}} : \left( \sqrt{x^3 - \sqrt{y^3}} + \sqrt{xy} \right) \).

Answer 1

bài 5:

câu a: đkxđ:( \(x>0;x\ne25)\)

\(a.A=\frac{\sqrt{x}}{x-5\sqrt{x}}-\frac{10}{x-25}=\frac{\sqrt{x}}{\sqrt{x}\cdot\left(\sqrt{x}-5\right)}-\frac{10}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

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

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

câu b: đkxđ: \(\left(x\ge0;x\ne9\right)\)

\(B=\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{3\sqrt{x}}{\sqrt{x}-3}+\frac{2x+36}{x-9}\) \(\)

\(=\frac{\sqrt{x}\cdot\left(\sqrt{x}-3\right)-3\sqrt{x}\cdot\left(\sqrt{x}+3\right)+2x+36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x-3\sqrt{x}-3x-9\sqrt{x}+2x+36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{-12\sqrt{x}+36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{-12\cdot\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{-12}{\sqrt{x}+3}\)

câu c: đkxđ: \(\left(x>0;x\ne\frac12;x\ne1\right)\)

đặt t = \(\sqrt{x}\) , ta có:

\(C=\left(\frac{t}{t-1}+\frac{t+1}{t}\right)\cdot\frac{t^2}{1-2t^2}\)

\(=\frac{t^2+\left(t+1\right)\left(t-1\right)}{t\cdot\left(t-1\right)}\cdot\frac{t^2}{1-2t^2}\)

\(=\frac{2t^2-1}{t\cdot\left(t-1\right)}\cdot\frac{t^2}{1-2t^2}=\frac{t\cdot\left(2t^2-1\right)}{\left(t-1\right)\left(1-2t^2\right)}\)

\(=\frac{t\cdot\left(2t^2-1\right)}{\left(t-1\right)\left\lbrack-\left(2t^2-1\right)\right\rbrack}=\frac{-t}{t-1}=-\frac{\sqrt{x}}{\sqrt{x}-1}\)

câu d: đkxđ: \(\left(x\ge0;x\ne1;x\ne\frac{25}{9}\right)\)

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

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

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

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

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

câu e: đkxđ: \(\left(x>0;x\ne9\right)\)

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

\(=\frac{\left(x-3\sqrt{x}\right)+\left(2x+5\sqrt{x}-3\right)-\left(2x-\sqrt{x}-3\right)}{x-9}:\frac{3\sqrt{x}}{\sqrt{x}-3}\)

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

\(=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{3\left(x-9\right)}=\frac{x-9}{3\cdot\left(x-9\right)}=\frac13\)

câu f: đkxđ: \(\) \(\left(x;y\ge0;x\ne y\right)\)

\(F=\left(\sqrt{x}+\frac{\sqrt{xy}+\sqrt{y^2}}{\sqrt{x}+\sqrt{y}}\right):\left(\frac{\sqrt{x^2}-\sqrt{y^2}}{\sqrt{x}-\sqrt{y}}+\sqrt{xy}\right)\)

\(=\left(\sqrt{x}+\frac{\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right):\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}+\sqrt{xy}\right)\)

\(=\frac{x+2\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}:\left(\sqrt{x}+\sqrt{y}+\sqrt{xy}\right)\)

\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\sqrt{x}+\sqrt{y}}:\left(\sqrt{x}+\sqrt{y}+\sqrt{xy}\right)\)

\(=\left(\sqrt{x}+\sqrt{y}\right):\left(\sqrt{x}+\sqrt{y}+\sqrt{xy}\right)=\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{xy}}\)