Bài 3:
\(\left\{{}\begin{matrix}2x^2+y^2=19\\x^2+9y^2=6xy\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x^2+y^2=19\\x^2-6xy+9y^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x^2+y^2=19\\\left(x-3y\right)^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=3y\\2x^2+y^2=19\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=3y\\2\cdot\left(3y\right)^2+y^2=19\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}19y^2=19\\x=3y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y^2=1\\x=3y\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\left\{{}\begin{matrix}y=1\\x=3\cdot1=3\end{matrix}\right.\\\left\{{}\begin{matrix}y=-1\\x=3\cdot\left(-1\right)=-3\end{matrix}\right.\end{matrix}\right.\)
bài 1:
a: \(P=\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\dfrac{\sqrt{a}-a}{1-\sqrt{a}}\right):\dfrac{1-a}{a+\sqrt{a}}\)
\(=\left(\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}+1}-\dfrac{\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\right):\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\)
\(=\left(a-\sqrt{a}+1-\sqrt{a}\right)\cdot\dfrac{\sqrt{a}}{1-\sqrt{a}}\)
\(=\dfrac{\left(1-\sqrt{a}\right)^2\cdot\sqrt{a}}{1-\sqrt{a}}=\sqrt{a}\left(1-\sqrt{a}\right)\)
b: Để P=-2 thì \(\sqrt{a}\left(1-\sqrt{a}\right)=-2\)
=>\(\sqrt{a}\left(\sqrt{a}-1\right)=2\)
=>\(a-\sqrt{a}-2=0\)
=>\(\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)=0\)
=>\(\sqrt{a}-2=0\)
=>a=4(nhận)
