Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Em thử nhá!
b) ĐK: \(x\ge\frac{1}{3}\)
Tách pt thành: \(\left(x^2-2x+1\right)+\left(3x+1\right)-2\sqrt{3x+1}.2+4=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(\sqrt{3x+1}-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\\sqrt{3x+1}=2\end{matrix}\right.\Leftrightarrow x=1\left(TMĐK\right)\)
Vậy....
b)Đặt $S=x+y,P=xy$ thì được:
\(\left\{ \begin{align} & S+P=2+3\sqrt{2} \\ & {{S}^{2}}-2P=6 \\ \end{align} \right.\Rightarrow {{S}^{2}}+2S+1=11+6\sqrt{2}={{\left( 3+\sqrt{2} \right)}^{2}}\)
\(\begin{array}{l} \Rightarrow \left\{ \begin{array}{l} S = 2 + \sqrt 2 \\ P = 2\sqrt 2 \end{array} \right. \Rightarrow \left( {x;y} \right) \in \left\{ {\left( {2;\sqrt 2 } \right),\left( {\sqrt 2 ;2} \right)} \right\}\\ \left\{ \begin{array}{l} S = - 4 - \sqrt 2 \\ P = 6 + 4\sqrt 2 \end{array} \right.\left( {VN} \right) \end{array} \)
\( c)\left\{ \begin{array}{l} 2{x^2} + xy + 3{y^2} - 2y - 4 = 0\\ 3{x^2} + 5{y^2} + 4x - 12 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} 2\left( {2{x^2} + xy + 3{y^2} - 2y - 4} \right) - \left( {3{x^2} + 5{y^2} + 4x - 12} \right) = 0\\ 3{x^2} + 5{y^2} + 4x - 12 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} {x^2} + 2xy + {y^2} - 4x - 4y + 4 = 0\\ 3{x^2} + 5{y^2} + 4x - 12 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} {\left( {x + y - 2} \right)^2} = 0\\ 3{x^2} + 5{y^2} + 4x - 12 = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x + y - 2 = 0\\ 3{x^2} + 5{y^2} + 4x - 12 = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = 1\\ y = 1 \end{array} \right. \)
a)Trừ theo vế của \(pt\left(2\right)\) cho \(pt\left(1\right)\):
\(\left(5x+3y\right)-\left(3x+2y\right)=-4-1\)
\(\Leftrightarrow2x+y=-5\). Khi đó
\(3x+2y=1\Leftrightarrow2\left(2x+y\right)-x=1\)
\(\Leftrightarrow2\cdot\left(-5\right)-x=1\)\(\Leftrightarrow x=-11\)
\(\Rightarrow3x+2y=1\Rightarrow y=\dfrac{1-3x}{2}=\dfrac{1-3\cdot\left(-11\right)}{2}=17\)
Vậy nghiệm hpt \(\left(x;y\right)=\left(-11;17\right)\)
b)\(2x^2+2\sqrt{3}x-3=0\)
\(\Delta=\left(2\sqrt{3}\right)^2-\left(4\cdot2\cdot\left(-3\right)\right)=36\)
\(\Rightarrow x_{1,2}=\dfrac{-2\sqrt{3}\pm\sqrt{36}}{4}\)
c)\(9x^4+8x^2-1=0\)
\(\Leftrightarrow9x^4-x^2+9x^2-1=0\)
\(\Leftrightarrow x^2\left(9x^2-1\right)+\left(9x^2-1\right)=0\)
\(\Leftrightarrow\left(9x^2-1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)\left(x^2+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\3x+1=0\\x^2+1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\pm\dfrac{1}{3}\\x^2+1>0\left(loai\right)\end{matrix}\right.\)
a) \(x^4-30x^2+31x-30=0\)
\(\Leftrightarrow\left(x^4+x\right)+\left(-30x^2+30x-30\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x^2+x-30\right)\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+6\right)\left(x^2-x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-6\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x+y+z=2\left(1\right)\\2xy-z^2=4 \left(2\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+z^2+2xy+2yz+2xz=4\\2xy-z^2=4\end{matrix}\right.\)
\(\Rightarrow x^2+y^2+z^2+2xy+2yz+2xz=2xy-z^2\)
\(\Leftrightarrow x^2+y^2+2z^2+2yz+2xz=0\)
\(\Leftrightarrow\left(x+z\right)^2+\left(y+z\right)^2=0\)
\(\Rightarrow x=y=-z\) thay vào (1) ta được : \(-z-z+z=2\Rightarrow z=-2\)
\(\Rightarrow x=y=2\)
Vậy \(x=y=2;z=-2\)