Question

Giải pt và hpt:

a) hpt:

\(\left\{{}\begin{matrix}x+y+xy=11\\x^{2\:}+y^{2\:}-x-y=8\end{matrix}\right.\)

b) pt:\(\left(x+8\right)\sqrt{2x^2-4x\: }=2x^2-2x+12\: \)

Answer 1

a) hpt \(\Leftrightarrow\left\{{}\begin{matrix}x+y+xy=11\\\left(x+y\right)^2-2xy-\left(x+y\right)=8\end{matrix}\right.\)

Đặt S=x+y; P =xy, ta có hệ :

\(\left\{{}\begin{matrix}S+P=11\\S^2-S-2P=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}P=11-S\\S^2-S-2\left(11-S\right)=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}P=11-S\\S^2+S-30=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}P=11-S\\\left[{}\begin{matrix}S=5\\S=-6\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy=11-\left(x+y\right)\\\left[{}\begin{matrix}x+y=5\\x+y=-6\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\curlyvee\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\\\text{hệ vô nghiệm}\end{matrix}\right.\)

Vậy...

Answer 2

b)ĐK : \(\left\{{}\begin{matrix}x\le0\\x\ge2\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{2x^2-4x}-4=\frac{2x^2-2x+12}{x+8}-4\)

\(\Leftrightarrow\frac{2x^2-4x-16}{\sqrt{2x^2-4x}+4}=\frac{2x^2-6x-20}{x+8}\)

\(\Leftrightarrow\frac{\left(x-4\right)\left(x+2\right)}{\sqrt{2x^2-4x}+4}=\frac{\left(x-5\right)\left(x+2\right)}{x+8}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\\frac{x-4}{\sqrt{2x^2-4x}+4}=\frac{x+5}{x+8}\left(1\right)\end{matrix}\right.\)

Giải tiếp pt 1 và kết hợp vs đk t tìm được nghiệm