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Trước tiên ta chứng minh:
\(x\sqrt{x}-3\sqrt{x}+3>0\)
\(\Leftrightarrow\left(x\sqrt{x}-2x+\sqrt{x}\right)+\left(2x-4\sqrt{x}+2\right)+1>0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)^2+2\left(\sqrt{x}-1\right)^2+1>0\)(đúng )
\(\Rightarrow A=\frac{\sqrt{x}}{x\sqrt{x}-3\sqrt{x}+3}\ge0\)
Ta chứng minh:
\(A=\frac{\sqrt{x}}{x\sqrt{x}-3\sqrt{x}+3}< 2\)
\(\Leftrightarrow2x\sqrt{x}-6\sqrt{x}+6-\sqrt{x}>0\)
\(\Leftrightarrow2x\sqrt{x}-7\sqrt{x}+6>0\)
\(\Leftrightarrow\left(2x\sqrt{x}-4x+2\right)+\left(4x-\frac{2.2.7}{4}\sqrt{x}+\frac{49}{16}\right)+\frac{47}{16}>0\)
\(\Leftrightarrow2\sqrt{x}\left(\sqrt{x}-1\right)^2+\left(2\sqrt{x}-\frac{7}{2}\right)^2+\frac{47}{16}>0\)(đúng )
Từ đây ta được: \(0\le A< 1\)
\(\Rightarrow A=\left\{0;1\right\}\)
Thế A vô tìm x nha. Cái nào thỏa mãn thì lụm không thì bỏ nha.
Thuy Duong Nguyen đánh đề cẩn thận hơn bạn nhé
Lời giải :
a) ĐKXĐ : \(x\ne1\)
\(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(A=\frac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}+3\right)\left(2-3\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\frac{15\sqrt{x}-11-3x+6-7\sqrt{x}-2x-\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\frac{\left(\sqrt{x}-1\right)\left(-5\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(A=\frac{2-5\sqrt{x}}{\sqrt{x}+3}\)
b) \(x=3-2\sqrt{2}=2-2\sqrt{2}+1=\left(\sqrt{2}-1\right)^2\)
\(\Leftrightarrow\sqrt{x}=\sqrt{2}-1\)
Khi đó \(A=\frac{2-5\left(\sqrt{2}-1\right)}{\sqrt{2}-1+3}\)
\(A=\frac{2-5\sqrt{2}+5}{\sqrt{2}+2}=\frac{7-5\sqrt{2}}{\sqrt{2}+2}\)
c) \(A=\frac{1}{2}\)
\(\Leftrightarrow\frac{2-5\sqrt{x}}{\sqrt{x}+3}=\frac{1}{2}\)
\(\Leftrightarrow2\left(2-5\sqrt{x}\right)=\sqrt{x}+3\)
\(\Leftrightarrow4-10\sqrt{x}-\sqrt{x}-3=0\)
\(\Leftrightarrow1-11\sqrt{x}=0\)
\(\Leftrightarrow11\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x}=\frac{1}{11}\)
\(\Leftrightarrow x=\frac{1}{121}\)( thỏa )
d) A nguyên \(\Leftrightarrow2-5\sqrt{x}⋮\sqrt{x}+3\)
\(\Leftrightarrow-5\left(\sqrt{x}+3\right)+17⋮\sqrt{x}+3\)
Vì \(-5\left(\sqrt{x}+3\right)⋮\sqrt{x}+3\)
\(\Rightarrow17⋮\sqrt{x}+3\)
\(\Rightarrow\sqrt{x}+3\inƯ\left(17\right)=\left\{17\right\}\)( vì \(\sqrt{x}+3\ge3\))
\(\Leftrightarrow\sqrt{x}=14\)
\(\Leftrightarrow x=196\)( thỏa )
Vậy....
\(a,ĐKXĐ:\orbr{\begin{cases}x+2\sqrt{x}+3\ne0\\\sqrt{x}+3\ne0\end{cases}}\)
\(\Leftrightarrow\orbr{ }\sqrt{x}\ne-3\)
Rút gọn: p/s: sau phân số thứ 2 ở mẫu ko có x à? Bạn chép đề sai?
Mình giải câu a thôi nha b,c,d tương tự
a/ để \(\frac{2}{x-1}\)nguyên thì x - 1 phải là ước nguyên của 2 hay (x - 1) = (-1, 1, -2, 2)
=> x = (0, 2, -1; 3)
\(M=\left(\frac{\sqrt{x}+3}{\sqrt{x}-3}-\frac{\sqrt{x}-3}{\sqrt{x}+3}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+3}-1\right)\) ĐKXĐ : \(x\ge0;x\ne-3;x\ne3\)
\(M=\frac{\left(\sqrt{x}+3\right)^2-\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-3}\)
\(M=\frac{x+6\sqrt{x}+9-x+6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{-3}\)
\(M=\frac{12\sqrt{x}}{\sqrt{x}+3}.\frac{1}{-3}\)
\(M=\frac{-4\sqrt{x}}{\sqrt{x}+3}\)
\(P=\left(\frac{3x+3}{x-9}-\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{3-\sqrt{x}}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right).ĐKXĐ:x\ge0,x\ne9\)
\(=\left(\frac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\frac{3x+3-2x+6\sqrt{x}-x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(=\frac{3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3}{\sqrt{x}+3}\)
\(b,x=20-6\sqrt{11}=11-2.3\sqrt{11}+9\)
\(=\left(\sqrt{11}-3\right)^2\)
\(P=\frac{3}{\sqrt{x}+3}=\frac{3}{\sqrt{\left(\sqrt{11}-3\right)^2}+3}=\frac{3}{\sqrt{11}-3+3}=\frac{3\sqrt{11}}{11}\)
\(c,P>\frac{1}{2}\Rightarrow\frac{3}{\sqrt{x}+3}>\frac{1}{2}\)
\(\Leftrightarrow\frac{3}{\sqrt{x}+3}-\frac{1}{2}>0\)
\(\Leftrightarrow\frac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)
\(\Leftrightarrow\frac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)\(\Leftrightarrow\frac{3-\sqrt{x}}{2\left(\sqrt{x}+3\right)}>0\)
vì \(2\left(\sqrt{x}+3\right)>0\) (nếu x=0 =>pt vô nghiệm)
\(\Rightarrow3-\sqrt{x}>0\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\)
Kết hợp ĐKXĐ: \(0< x< 9\)
\(a,\sqrt{4-4x+x^2}+\sqrt{\frac{2}{x^2+6x+9}}=\sqrt{\left(x-2\right)^2}+\sqrt{\frac{2}{\left(x+3\right)^2}}\)
\(đkxđ\Leftrightarrow\hept{\begin{cases}x+2\ge0\\x+3>0\end{cases}\Rightarrow\hept{\begin{cases}x\ge-2\\x>-3\end{cases}\Rightarrow}x\ge-2}\)
\(b,\frac{5\sqrt{x}}{\sqrt{x}-3}+\frac{2}{\sqrt{x}}\)
\(đkxđ\Leftrightarrow\hept{\begin{cases}x>0\\\sqrt{x}-3\ne0\end{cases}\Rightarrow\hept{\begin{cases}x>0\\\sqrt{x}\ne\sqrt{9}\end{cases}\Rightarrow}\hept{\begin{cases}x>0\\x\ne9\end{cases}}}\)
\(c,\sqrt{3-\sqrt{x}}\)
\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ge0\\3-\sqrt{x}\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\ge0\\\sqrt{x}\le3\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x>0\\\sqrt{x}\le9\end{cases}\Rightarrow\hept{\begin{cases}x>0\\x\le3\end{cases}}}\)
\(\Rightarrow0< x\le3\)