\(\int^{x^4+4x^3y+6x^2y^2+4xy^3+y^4=6x^2y^2-215}_{xy\left(x^2+y^2\right)=-78}\)<==>\(\int^{x^4+y^4+4xy\left(x^2+y^2\right)=-215}_{xy\left(x^2+y^2\right)=-78}\)

                                                                 <==>\(\int^{x^4+y^4+4\cdot\left(-78\right)=-215}_{xy\left(x^2+y^2\right)=-78}\)<==>\(\int^{x^4+y^4=97}_{xy\left(x^2+y^2\right)=-78}\)

mk ra thế này rồi để mk nghĩ tiếp nhé

a/ \(\left\{{}\begin{matrix}xy+1+x+y=10\\\left(x+y\right)\left(xy+1\right)=1\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=a\\xy+1=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=10\\ab=1\end{matrix}\right.\)

Theo Viet đảo, a và b là nghiệm:

\(t^2-10t+1=0\) \(\Rightarrow\left[{}\begin{matrix}t=5+2\sqrt{6}\\t=5-2\sqrt{6}\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+y=5+2\sqrt{6}\\xy=4-2\sqrt{6}\end{matrix}\right.\)

Theo Viet đảo, x và y là nghiệm:

\(t^2-\left(5+2\sqrt{6}\right)t+4-2\sqrt{6}=0\) (bấm máy, số xấu quá)

TH2: \(\left\{{}\begin{matrix}x+y=5-2\sqrt{6}\\xy=4+2\sqrt{6}\end{matrix}\right.\)

Ta có \(\left(5-2\sqrt{6}\right)^2-4\left(4+2\sqrt{6}\right)=33-28\sqrt{6}< 0\) nên vô nghiệm

b/ \(\left\{{}\begin{matrix}x^4+y^4=97\\xy\left(x^2+y^2\right)=78\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y^2\right)^2-2x^2y^2=97\\xy\left(x^2+y^2\right)=78\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x^2+y^2=a>0\\xy=b\end{matrix}\right.\) với \(a\ge2b\) hệ trở thành:

\(\left\{{}\begin{matrix}a^2-2b^2=97\\ab=78\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^2-2b^2=97\\b=\frac{78}{a}\end{matrix}\right.\)

\(\Rightarrow a^2-2\left(\frac{78}{a}\right)^2=97\)

\(\Leftrightarrow a^4-97a^2-12168=0\Rightarrow\left[{}\begin{matrix}a^2=169\\a^2=-72\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a=13\Rightarrow b=6\\a=-13< 0\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^2+y^2=13\\xy=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2+y^2=13\\y=\frac{6}{x}\end{matrix}\right.\)

\(\Rightarrow x^2+\frac{36}{x^2}=13\Leftrightarrow x^4-13x^2+36=0\) \(\Rightarrow\left[{}\begin{matrix}x^2=9\\x^2=4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\Rightarrow y=2\\x=-3\Rightarrow y=-2\\x=2\Rightarrow y=3\\x=-2\Rightarrow y=-3\end{matrix}\right.\)

ko phải đối xứng loại 1 òi

Chả hỉu rì hết ó 

<=><=>(X+1)(Y+1)=6 và (x+1)^3+(y+1)^3=35đặt X+1;Y+1 biến đổi vế 2 giải ra đc(1;2);(2;1)

b,<=>\(\left[\sqrt{2}+1\right]^x+\left[\sqrt{2}-1\right]^x=6\)

<=>\(2\sqrt{2}^x+2=6\)

<=>x=2

Cộng vế với vế:

\(x^2+2xy+y^2+x+y=6\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x+y\right)-6=0\)

\(\Rightarrow\left[{}\begin{matrix}x+y=-3\\x+y=2\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+y=-3\\xy=5\end{matrix}\right.\)

Theo Viet đảo, x và y là nghiệm của:

\(t^2+3t+5=0\) (vô nghiệm)

TH2: \(\left\{{}\begin{matrix}x+y=2\\xy=0\end{matrix}\right.\)

Theo Viet đảo, x và y là nghiệm:

\(t^2-2t=0\Rightarrow\left[{}\begin{matrix}t=0\\t=2\end{matrix}\right.\)

\(\Rightarrow\left(x;y\right)=\left(2;0\right);\left(0;2\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}xy-3x+2y-6=xy+1\\2x+2y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2y-3x=7\\2x+2y=5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{5}\\y=\dfrac{29}{10}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\left(x+2\right)\left(y-3\right)=xy+1\\2\left(x+y\right)\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}\left(x+2\right)\left(y-3\right)=xy+1\\2x+2y=5\end{matrix}\right.\)

\(\left\{{}\begin{matrix}xy-3x+2y-6=xy+1\\y=\dfrac{5-2x}{2}\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}-3x+2y=7\\y=\dfrac{5-2x}{2}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}-3x+2.\dfrac{5-2x}{2}=7\\y=\dfrac{5-2x}{2}\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}x=-0,4\\y=2,9\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\)thì hệ trở thành

\(\left\{{}\begin{matrix}a^4=6b^2-215\\b\left(a^2-2b\right)=-78\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(\dfrac{-78}{b}+2b\right)^2=6b^2-215\left(1\right)\\a^2=\dfrac{-78}{b}+2b\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2b^4+97b^2-6084=0\)

\(\Leftrightarrow\left[{}\begin{matrix}b=6\\b=-6\end{matrix}\right.\)

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