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b/ ĐKXĐ; ...
\(\Leftrightarrow\left\{{}\begin{matrix}x^3+3x^2+3x+1-16x-16=\frac{8}{y^3}-\frac{8}{y}\\5\left(x^2+2x+2\right)=1+\frac{4}{y^2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^3-16\left(x+1\right)=\frac{8}{y^3}-\frac{8}{y}\\5\left(x+1\right)^2=\frac{4}{y^2}-4\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+1=a\\\frac{1}{y}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^3-16a=8b^3-8b\\5a^2=4b^2-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^3-8b^3=16a-8b\\4=-5a^2+4b^2\end{matrix}\right.\)
Nhân vế với vế:
\(4\left(a^3-8b^3\right)=4\left(4a-2b\right)\left(-5a^2+4b^2\right)\)
\(\Leftrightarrow21a^3-10a^2b-16ab^2=0\)
\(\Leftrightarrow a\left(21a^2-10ab-16b^2\right)=0\)
\(\Leftrightarrow a\left(7a-8b\right)\left(3a+2b\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=0\\7a=8b\\3a=-2b\end{matrix}\right.\) \(\Rightarrow...\)
a/ \(\left\{{}\begin{matrix}x^2+y+xy\left(x^2+y\right)+xy+1=-\frac{1}{4}\\x^4+y^2+2x^2y+xy+1=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y+1\right)\left(xy+1\right)=-\frac{1}{4}\\\left(x^2+y\right)^2+xy+1=-\frac{1}{4}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x^2+y=a\\xy+1=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a+1\right)b=-\frac{1}{4}\\a^2+b=-\frac{1}{4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(a+1\right)b=-\frac{1}{4}\\b=-\frac{1}{4}-a^2\end{matrix}\right.\)
\(\Rightarrow\left(a+1\right)\left(-\frac{1}{4}-a^2\right)=-\frac{1}{4}\)
\(\Leftrightarrow4a^3+4a^2+a=0\Leftrightarrow a\left(2a+1\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=0\Rightarrow b=-\frac{1}{4}\\a=-\frac{1}{2}\Rightarrow b=-\frac{1}{2}\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x^2+y=0\\xy+1=-\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=-x^2\\-x^3=-\frac{5}{4}\end{matrix}\right.\) \(\Rightarrow...\)
TH2: \(\left\{{}\begin{matrix}x^2+y=-\frac{1}{2}\\xy+1=-\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=-\frac{1}{2}-x^2\\x\left(-\frac{1}{2}-x^2\right)=-\frac{5}{4}\end{matrix}\right.\) \(\Rightarrow...\)
\(\left\{{}\begin{matrix}\frac{4x-5}{7}< x+3\\\frac{3x+8}{4}>2x-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-5< 7x+21\\3x+8>8x-20\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x>-26\\5x< 28\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\frac{-26}{3}\\x< \frac{28}{5}\end{matrix}\right.\Leftrightarrow\frac{-26}{3}< x< \frac{28}{5}\)