Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(\left\{{}\begin{matrix}\left(\sqrt{3}+1\right)x+\left(\sqrt{3}-1\right)y=\sqrt{3}\\2\sqrt{3}x-2y=3\sqrt{3}+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(\sqrt{3}+1\right)^2\cdot x+\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)y=\sqrt{3}\left(\sqrt{3}+1\right)\\2\sqrt{3}x-2y=3\sqrt{3}+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(4+2\sqrt{3}\right)+2y=3+\sqrt{3}\\2\sqrt{3}\cdot x-2y=3\sqrt{3}+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\left(4+2\sqrt{3}+2\sqrt{3}\right)=3+\sqrt{3}+3\sqrt{3}+1\\2\sqrt{3}\cdot x-2y=3\sqrt{3}+1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\2y=2\sqrt{3}-3\sqrt{3}-1=-\sqrt{3}-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=1\\y=\dfrac{-\sqrt{3}-1}{2}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}x\sqrt{3}+y\sqrt{2}=1\\x\sqrt{2}+y\sqrt{3}=\sqrt{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\sqrt{6}+2y=\sqrt{2}\\x\sqrt{6}+3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2y-3y=\sqrt{2}-3\\x\sqrt{3}+y\sqrt{2}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-y=\sqrt{2}-3\\x\sqrt{3}=1-y\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3-\sqrt{2}\\x\sqrt{3}=1-\sqrt{2}\left(3-\sqrt{2}\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=3-\sqrt{2}\\x\sqrt{3}=1-3\sqrt{2}+2=3-3\sqrt{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=3-\sqrt{2}\\x=\sqrt{3}-\sqrt{6}\end{matrix}\right.\)
c: \(\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=\left(x+1\right)\left(y-3\right)\\\left(x-5\right)\left(y+4\right)=\left(x-4\right)\left(y+1\right)\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}xy-2y-y+2=xy-3x+y-3\\xy+4x-5y-20=xy+x-4y-4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2x-y+2=-3x+y-3\\4x-5y-20=x-4y-4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-2x-y+3x-y=-3-2=-5\\4x-5y-x+4y=-4+20\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2y=-5\\3x-y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-6y=-15\\3x-y=16\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-5y=-15-16=-31\\x-2y=-5\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{31}{5}\\x=-5+2y=-5+\dfrac{62}{5}=\dfrac{37}{5}\end{matrix}\right.\)
a) \(ĐK:y-2x+1\ge0;4x+y+5\ge0;x+2y-2\ge0,x\le1\)
Th1: \(\hept{\begin{cases}y-2x+1=0\\3-3x=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}0=0\\-1=\sqrt{10}-1\end{cases}}\)(không thỏa mãn)
Th2: \(x,y\ne1\)
\(2x^2-y^2+xy-5x+y+2=\sqrt{y-2x+1}-\sqrt{3-3x}\)\(\Leftrightarrow\left(x+y-2\right)\left(2x-y-1\right)=\frac{x+y-2}{\sqrt{y-2x+1}+\sqrt{3-3x}}\)\(\Leftrightarrow\left(x+y-2\right)\left(\frac{1}{\sqrt{y-2x+1}+\sqrt{3-3x}}+y-2x+1\right)=0\)
Dễ thấy \(\frac{1}{\sqrt{y-2x+1}+\sqrt{3-3x}}+y-2x+1>0\)nên x + y - 2 = 0
Thay y = 2 - x vào phương trình \(x^2-y-1=\sqrt{4x+y+5}-\sqrt{x+2y-2}\), ta được: \(x^2+x-3=\sqrt{3x+7}-\sqrt{2-x}\)\(\Leftrightarrow x^2+x-2=\sqrt{3x+7}-1+2-\sqrt{2-x}\)\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=\frac{3\left(x+2\right)}{\sqrt{3x+7}+1}+\frac{x+2}{2+\sqrt{2-x}}\)\(\Leftrightarrow\left(x+2\right)\left(\frac{3}{\sqrt{3x+7}+1}+\frac{1}{2+\sqrt{2-x}}+1-x\right)=0\)
Vì \(x\le1\)nên\(\frac{3}{\sqrt{3x+7}+1}+\frac{1}{2+\sqrt{2-x}}+1-x>0\)suy ra x = -2 nên y = 4
Vậy nghiệm của hệ phương trình là (x;y) = (-2;4)
b) \(\hept{\begin{cases}x^2+y^2=5\\x^3+2y^3=10x-10y\end{cases}}\Leftrightarrow\hept{\begin{cases}2\left(x^2+y^2\right)=10\left(1\right)\\x^3+2y^3=10\left(x-y\right)\left(2\right)\end{cases}}\)
Thay (1) vào (2), ta được: \(x^3+2y^3=2\left(x^2+y^2\right)\left(x-y\right)\Leftrightarrow\left(2y-x\right)\left(x^2+2y^2\right)=0\)
* Th1: \(x^2+2y^2=0\)(*)
Mà \(x^2\ge0\forall x;2y^2\ge0\forall y\Rightarrow x^2+2y^2\ge0\)nên (*) xảy ra khi x = y = 0 nhưng cặp nghiệm này không thỏa mãn hệ
* Th2: 2y - x = 0 suy ra x = 2y thay vào (1), ta được: \(y^2=1\Rightarrow y=\pm1\Rightarrow x=\pm2\)
Vậy hệ có 2 nghiệm \(\left(x,y\right)\in\left\{\left(2;1\right);\left(-2;-1\right)\right\}\)