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c) Ta có: \(\left\{{}\begin{matrix}\dfrac{x+2}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x+1}+\dfrac{10}{y-2}=25\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y-2}=22\\\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=\dfrac{1}{2}\\\dfrac{1}{x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=1\\y-2=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{5}{2}\end{matrix}\right.\)
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a) \(\hept{\begin{cases}3\left(x+1\right)+2\left(x+2y\right)=4\\4\left(x+1\right)-\left(x+2y\right)=9\end{cases}}\Leftrightarrow\hept{\begin{cases}3\left(x+1\right)+2\left(x+2y\right)=4\\8\left(x+1\right)-2\left(x+2y\right)=18\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}11\left(x+1\right)=22\\3\left(x+1\right)+2\left(x+2y\right)=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\4y+8=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-1\end{cases}}\)
b) ĐK : y khác 0
\(\hept{\begin{cases}x+\frac{1}{y}=-\frac{1}{2}\\2x-\frac{3}{y}=-\frac{7}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}3x+\frac{3}{y}=-\frac{3}{2}\\2x-\frac{3}{y}=-\frac{7}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}5x=-5\\3x+\frac{3}{y}=-\frac{3}{2}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-1\\-3+\frac{3}{y}=-\frac{3}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\\frac{3}{y}=\frac{3}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\left(tm\right)\end{cases}}\)
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ĐKXĐ : \(\left\{{}\begin{matrix}x^3\ge0\\x-1\ge0\\\sqrt{x}-1\ne0\\\sqrt{x-1}\pm\sqrt{x}\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge0\\x\ge1\\x\ne1\\\sqrt{x-1}\ne\pm\sqrt{x}\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x>1\\x-1\ne x\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x>1\\0\ne-1\end{matrix}\right.\) => x > 1
Ta có : \(\frac{1}{\sqrt{x-1}-\sqrt{x}}+\frac{1}{\sqrt{x-1}+\sqrt{x}}+\frac{\sqrt{x^3}-x}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x-1}+\sqrt{x}}{\left(\sqrt{x-1}-\sqrt{x}\right)\left(\sqrt{x-1}+\sqrt{x}\right)}+\frac{\sqrt{x-1}-\sqrt{x}}{\left(\sqrt{x-1}+\sqrt{x}\right)\left(\sqrt{x-1}-\sqrt{x}\right)}+\frac{x\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x-1}+\sqrt{x}}{x-1-x}+\frac{\sqrt{x-1}-\sqrt{x}}{x-1-x}+x\)
\(=\frac{\sqrt{x-1}+\sqrt{x}}{-1}+\frac{\sqrt{x-1}-\sqrt{x}}{-1}+x\)
\(=\frac{\sqrt{x-1}+\sqrt{x}+\sqrt{x-1}-\sqrt{x}}{-1}+x\)
\(=\frac{2\sqrt{x-1}}{-1}+x\)
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Ta có: \(A=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right).\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\) ( ĐK: \(x\ne0,\)\(x\ne9,\)\(x\ge3\))
\(\Leftrightarrow A=\frac{\sqrt{x}.\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{3\sqrt{x}-x+x+9}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{2\sqrt{x}+4}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{3\sqrt{x}-9}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{2\sqrt{x}+4}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{3\left(\sqrt{x}-3\right)}{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)}.\frac{2\sqrt{x}+4}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)
\(\Leftrightarrow A=\frac{3.\left(2\sqrt{x}+4\right)}{\left(9-x\right).\sqrt{x}}\)
\(\Leftrightarrow A=\frac{6\sqrt{x}+12}{9\sqrt{x}-x}\)
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\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{\left(x-1\right)^2}{x^2-1}\right).\frac{x+2003}{x}\)ĐKXĐ: \(x\ne-1;0;1\)
\(A=\frac{\left(x+1\right)^2-\left(x-1\right)^2+\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}.\frac{x+2003}{x}\)
\(A=\frac{\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}.\frac{x+2003}{x}\)
\(A=\frac{x+1}{x-1}.\frac{x+2003}{x}\)
\(A=\frac{x^2+2004x+2003}{x^2-x}\)
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c,chia cả tử và mẫu cho x,sau đó đặt 3x+2/x=t
các câu còn lại hiện chưa giải đc vì chưa có giấy nháp,lúc nào rảnh mình chỉ cho cách làm
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Lời giải:
ĐK: $x\geq 0$
a)
Khi \(x=\frac{\sqrt{7-4\sqrt{3}}}{2}=\frac{\sqrt{4+3-2\sqrt{4.3}}}{2}=\frac{\sqrt{(2-\sqrt{3})^2}}{2}=\frac{2-\sqrt{3}}{2}=\frac{4-2\sqrt{3}}{4}=\frac{(\sqrt{3}-1)^2}{2^2}\)
\(\Rightarrow \sqrt{x}=\frac{\sqrt{3}-1}{2}\)
\(\Rightarrow \left\{\begin{matrix} 4\sqrt{x}=2(\sqrt{3}-1)\\ (\sqrt{x}+1)^2=\frac{4+2\sqrt{3}}{4}\end{matrix}\right.\) \(\Rightarrow P=-20+12\sqrt{3}\)
b)
\(P=\frac{4\sqrt{x}}{(\sqrt{x}+1)^2}=\frac{1}{2}\)\(\Leftrightarrow 8\sqrt{x}=x+1+2\sqrt{x}\)
\(\Leftrightarrow x-6\sqrt{x}+1=0\)
\(\Leftrightarrow (\sqrt{x}-3)^2=8\Rightarrow \sqrt{x}-3=\pm 2\sqrt{2}\)
\(\Rightarrow \sqrt{x}=3-2\sqrt{2}\Rightarrow x=17\pm 12\sqrt{2}\)
(đều thỏa mãn)
Lời giải:
ĐK: $x\geq 0$
a)
Khi \(x=\frac{\sqrt{7-4\sqrt{3}}}{2}=\frac{\sqrt{4+3-2\sqrt{4.3}}}{2}=\frac{\sqrt{(2-\sqrt{3})^2}}{2}=\frac{2-\sqrt{3}}{2}=\frac{4-2\sqrt{3}}{4}=\frac{(\sqrt{3}-1)^2}{2^2}\)
\(\Rightarrow \sqrt{x}=\frac{\sqrt{3}-1}{2}\)
\(\Rightarrow \left\{\begin{matrix} 4\sqrt{x}=2(\sqrt{3}-1)\\ (\sqrt{x}+1)^2=\frac{4+2\sqrt{3}}{4}\end{matrix}\right.\) \(\Rightarrow P=-20+12\sqrt{3}\)
b)
\(P=\frac{4\sqrt{x}}{(\sqrt{x}+1)^2}=\frac{1}{2}\)\(\Leftrightarrow 8\sqrt{x}=x+1+2\sqrt{x}\)
\(\Leftrightarrow x-6\sqrt{x}+1=0\)
\(\Leftrightarrow (\sqrt{x}-3)^2=8\Rightarrow \sqrt{x}-3=\pm 2\sqrt{2}\)
\(\Rightarrow \sqrt{x}=3-2\sqrt{2}\Rightarrow x=17\pm 12\sqrt{2}\)
(đều thỏa mãn)