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Bài 1:
\(P=\left(\dfrac{x-\sqrt{x}-2+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{x-\sqrt{x}+2-x-\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{-2\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\cdot\dfrac{1}{\sqrt{x}-1}=\dfrac{-2}{\sqrt{x}+1}\)
\(\left\{{}\begin{matrix}x+y=m-1\\x-y=m+3\end{matrix}\right.\)
\(\Rightarrow x+y+x-y=m-1+m+3\)
\(\Rightarrow2x=2m+2\Rightarrow x=m+1\)
\(\Rightarrow x_0=m+1\) (1)
\(\left\{{}\begin{matrix}x+y=m-1\\x-y=m+3\end{matrix}\right.\)
\(\Rightarrow x+y-\left(x-y\right)=m-1-\left(m+3\right)\)
\(\Rightarrow2y=-4\Rightarrow y=-2\Rightarrow y_0=-2\Rightarrow y_0^2=4\) (2)
-Từ (1) và (2) suy ra:
\(m+1=4\Rightarrow m=3\)
b) \(x^2+2\sqrt{3}x-6=0\)
\(\Leftrightarrow\) \(x^2+2\sqrt{3}x+3-9=0\)
\(\Leftrightarrow\) \(\left(x+\sqrt{3}\right)^2-9=0\)
\(\Leftrightarrow\) \(\left(x+\sqrt{3}-3\right).\left(x+\sqrt{3}+3\right)=0\)
\(\Leftrightarrow\) \(\left[\begin{array}{} x+\sqrt{3}-3=0 \\ x+\sqrt{3}+3=0 \end{array} \right.\)\(\Leftrightarrow\) \(\left[\begin{array}{} x= 3-\sqrt{3} \\ x= -3-\sqrt{3} \end{array} \right.\)
Vậy phương trình có tập nghiệm là S={\(3-\sqrt{3};-3-\sqrt{3}\)}
b)Đặt $S=x+y,P=xy$ thì được:
\(\left\{ \begin{align} & S+P=2+3\sqrt{2} \\ & {{S}^{2}}-2P=6 \\ \end{align} \right.\Rightarrow {{S}^{2}}+2S+1=11+6\sqrt{2}={{\left( 3+\sqrt{2} \right)}^{2}}\)
\(\begin{array}{l} \Rightarrow \left\{ \begin{array}{l} S = 2 + \sqrt 2 \\ P = 2\sqrt 2 \end{array} \right. \Rightarrow \left( {x;y} \right) \in \left\{ {\left( {2;\sqrt 2 } \right),\left( {\sqrt 2 ;2} \right)} \right\}\\ \left\{ \begin{array}{l} S = - 4 - \sqrt 2 \\ P = 6 + 4\sqrt 2 \end{array} \right.\left( {VN} \right) \end{array} \)
\( c)\left\{ \begin{array}{l} 2{x^2} + xy + 3{y^2} - 2y - 4 = 0\\ 3{x^2} + 5{y^2} + 4x - 12 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} 2\left( {2{x^2} + xy + 3{y^2} - 2y - 4} \right) - \left( {3{x^2} + 5{y^2} + 4x - 12} \right) = 0\\ 3{x^2} + 5{y^2} + 4x - 12 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} {x^2} + 2xy + {y^2} - 4x - 4y + 4 = 0\\ 3{x^2} + 5{y^2} + 4x - 12 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} {\left( {x + y - 2} \right)^2} = 0\\ 3{x^2} + 5{y^2} + 4x - 12 = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x + y - 2 = 0\\ 3{x^2} + 5{y^2} + 4x - 12 = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 1\\ y = 1 \end{array} \right. \)