Bài 12:
a: ĐKXĐ: x>0; x<>1
\(L=\frac{3x-3\sqrt{x}-3}{x+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{\sqrt{x}}\left(\frac{1}{1-\sqrt{x}}-1\right)\)
\(=\frac{3x-3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{\sqrt{x}}\cdot\frac{1-1+\sqrt{x}}{1-\sqrt{x}}\)
\(=\frac{3x-3\sqrt{x}-3-\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\frac{3x-3\sqrt{x}-3-\left(x-1\right)-\left(x-4\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{3x-3\sqrt{x}-3-x+1-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{x-3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-2}{\sqrt{x}+2}\)
c: Để L là số nguyên thì \(\sqrt{x}-2\) ⋮\(\sqrt{x}+2\)
=>\(\sqrt{x}+2-4\) ⋮\(\sqrt{x}+2\)
=>-4⋮\(\sqrt{x}+2\)
=>\(\sqrt{x}+2\in\left\lbrace2;4\right\rbrace\)
=>\(\sqrt{x}\in\left\lbrace0;2\right\rbrace\)
=>x∈{0;4}
Kết hợp ĐKXĐ, ta được: x=4
Bài 11:
a: ĐKXĐ: a>=0; a<>1
\(K=\left(1+\frac{\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\cdot\sqrt{a}-a+\sqrt{a}-1}\right)\)
\(=\frac{a+\sqrt{a}+1}{a+1}:\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\right)\)
\(=\frac{a+\sqrt{a}+1}{a+1}:\left(\frac{a+1}{\left(\sqrt{a}-1\right)\left(a+1\right)}-\frac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\right)\)
\(=\frac{a+\sqrt{a}+1}{a+1}:\frac{a+1-2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}=\frac{a+\sqrt{a}+1}{a+1}\cdot\frac{\left(a+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)^2}=\frac{a+\sqrt{a}+1}{\sqrt{a}-1}\)
e: Để K là số nguyên thì \(a+\sqrt{a}+1\) ⋮\(\sqrt{a}-1\)
=>\(a-\sqrt{a}+2\sqrt{a}-2+3\) ⋮\(\sqrt{a}-1\)
=>3⋮\(\sqrt{a}-1\)
=>\(\sqrt{a}-1\in\left\lbrace1;-1;3\right\rbrace\)
=>\(\sqrt{a}\in\left\lbrace2;0;4\right\rbrace\)
=>a∈{0;4;16}
