1/ \(\Leftrightarrow x^2+2=y\left(x+1\right)\)
Với \(x=-1\) ko phải nghiệm
Với \(x\ne-1\Rightarrow y=\frac{x^2+2}{x+1}=x-1+\frac{3}{x+1}\)
\(\Rightarrow3⋮x+1\Rightarrow x+1=\left\{-3;-1;1;3\right\}\)
\(\Rightarrow x=....\Rightarrow y=...\)
2/ ĐKXĐ: ...
\(\Leftrightarrow x^2-2+\frac{1}{x^2}+16y^2-8+\frac{1}{y^2}=0\)
\(\Leftrightarrow\left(x-\frac{1}{x}\right)^2+\left(4y-\frac{1}{y}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-\frac{1}{x}=0\\4y-\frac{1}{y}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2=1\\4y^2=1\end{matrix}\right.\)
3/ Giả sử a; b làm biểu thức xác định
GTLN xảy ra khi \(7-\left(a+b\right)\) dương, do đó:
\(M=\frac{b}{7-\left(a+b\right)}\le\frac{a+b}{7-\left(a+b\right)}=\frac{a+b}{7-\left(a+b\right)}+1-1=\frac{7}{7-\left(a+b\right)}-1\le\frac{7}{7-6}-1=6\)
\(\Rightarrow M_{max}=6\) khi \(\left\{{}\begin{matrix}a=0\\b=6\end{matrix}\right.\)