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\(Q=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
b.\(Q< 1\)
\(\Leftrightarrow x-\sqrt{x}-2< x-5\sqrt{x}+6\)
\(\Leftrightarrow4\sqrt{x}-8< 0\)
\(\Leftrightarrow0\le x< 4\)
Vay de Q<1 thi \(0\le0< 4\)
a, ĐK \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)
\(P=\frac{x-1}{\sqrt{x}}:\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)
Ta thấy \(P=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}>0\forall x>0,x\ne1\)
b, P=\(\frac{x+2\sqrt{x}+1}{\sqrt{x}-1}=\frac{\frac{2}{2+\sqrt{3}}+2\sqrt{\frac{2}{2+\sqrt{3}}}+1}{\sqrt{\frac{2}{2+\sqrt{3}}}-1}\)
=\(\frac{\frac{4}{\left(\sqrt{3}+1\right)^2}+2.\sqrt{\left(\frac{2}{\left(\sqrt{3}+1\right)^2}\right)}+1}{\sqrt{\left(\frac{2}{2+\sqrt{3}}\right)^2}-1}=\frac{\frac{4}{\left(\sqrt{3}+1\right)^2}+2.\frac{2}{\sqrt{3}+1}+1}{\frac{2}{\sqrt{3}+1}-1}\)
\(=\frac{12+6\sqrt{3}}{1-3}=-6-3\sqrt{3}\)
Để \(A=\frac{5}{\sqrt{x}-1}\)nguyên thì \(5⋮\sqrt{x}-1\)
\(\Rightarrow\sqrt{x}-1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{2;0;6;-4\right\}\)
\(\Leftrightarrow x\in\left\{4;0;36\right\}\)( thỏa )
Để \(B=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3+4⋮\sqrt{x}-3\)
Vì \(\sqrt{x}-3⋮\sqrt{x}-3\)
\(\Rightarrow4⋮\sqrt{x}-3\)
\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{4;2;5;1;7;-1\right\}\)
\(\Leftrightarrow x\in\left\{16;4;25;1;49\right\}\)
Máy kia tương tự đi ăn cơm đây :>
Mk lm nốt câu C
\(C=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+2}{\sqrt{x}-1}=1+\frac{3}{\sqrt{x}-1}\)
Để C nguyên\(\Leftrightarrow\sqrt{x}-1\inƯ_{\left(3\right)}=\left\{\pm3;\pm1\right\}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=3\left(tm\right)\\\sqrt{x}-1=-3\left(l\right)\\\sqrt{x}-1=-1\left(tm\right)\\\sqrt{x}-1=1\left(tm\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=16\\x=0\\x=4\end{matrix}\right.\)