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\(\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{5\sqrt{x}}{\sqrt{x}+3}=\frac{22}{x-9}\left(ĐK:x\ge0;x\ne9\right)\)
\(\Leftrightarrow\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)-5\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}=\frac{22}{x-9}\)
\(\Rightarrow\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)-5\sqrt{x}\left(\sqrt{x}-3\right)=22\)
\(\Leftrightarrow x+5\sqrt{x}+6-5x+15\sqrt{x}=22\)
\(\Leftrightarrow-4x+20\sqrt{x}-16=0\)
\(\Leftrightarrow x-5\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}-4=0\\\sqrt{x}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=16\left(tm\right)\\x=1\left(tm\right)\end{cases}}}\)
Vậy tập nghiệm của phương trình đã cho là : \(S=\left\{1;16\right\}\)
Chúc bạn học tốt !!!
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
Bạn tự tìm điều kiện xác định nhé :)
\(Q=\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\left(\frac{\left(3-\sqrt{x}\right)\left(\sqrt{x}+3\right)+\left(\sqrt{x}-2\right)^2-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\frac{3}{\sqrt{x}+3}:\frac{9-x+x-4\sqrt{x}+4-9+x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{3}{\sqrt{x}+3}:\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{3}{\sqrt{x}+3}.\frac{\sqrt{x}+3}{\sqrt{x}-2}=\frac{3}{\sqrt{x}-2}\)
a/ đề \(=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}-\frac{5}{\sqrt{x}+5}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+5\right)-10\sqrt{x}-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\frac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\frac{x-10\sqrt{x}+25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\frac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\frac{\sqrt{x}-5}{\sqrt{x}+5}\)
b/ đề \(=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-\left(6\sqrt{x}-4\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
c/ đề \(=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-\left(3-11\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)
\(\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{5\sqrt{x}}{\sqrt{x}+3}=\frac{22}{x-9}\left(ĐK:x\ge0;x\ne9\right)\)
\(\Leftrightarrow\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)-5\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}=\frac{22}{x-9}\)
\(\Rightarrow\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)-5\sqrt{x}\left(\sqrt{x}-3\right)=22\)
\(\Leftrightarrow x+5\sqrt{x}+6-5x+15\sqrt{x}=22\)
\(\Leftrightarrow-4x+20\sqrt{x}-16=0\)
\(\Leftrightarrow x-5\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x}-4=0\\\sqrt{x}-1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=16\left(tm\right)\\x=1\left(tm\right)\end{array}\right.\)
Vậy tập nghiệm của pt đã cho là \(S=\left\{1;16\right\}\)