\(\left\{{}\begin{matrix}x^3+y^3=1+y-x+xy\left(1\right)\\7xy+y-x=7\left(2\right)\end{matrix}\right.\)
Từ(2)\(\Rightarrow x-y=7xy-7\)
\(\left(1\right)\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)=1+y-x+xy\)
\(\Leftrightarrow\left[\sqrt{\left(x-y\right)^2+4xy}\right]\left[\left(x-y\right)^2+xy\right]=1+7-7xy+xy\)
\(\Leftrightarrow7\left[\sqrt{\left(7xy-7\right)^2+4xy}\right]\left(7xy-7+xy\right)=-6xy+8\)
Đặt xy=a
\(\Rightarrow7\left[\sqrt{\left(7a-7\right)^2+4a}\right]\left(8a-7\right)=-6a+8\)
\(\Leftrightarrow49\left(\sqrt{\left(a-1\right)^2}\right)\left(8a-7\right)+6a-8=0\)
Với \(a-1\ge0\Leftrightarrow a\ge1\)
\(\Rightarrow49\left(8a^2-15a+7\right)+6a-8=0\)
\(\Leftrightarrow392a^2-729a+335=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{729+\sqrt{6161}}{784}\left(TM\right)\\a=\dfrac{729-\sqrt{6161}}{784}\left(KTM\right)\end{matrix}\right.\)\(\Rightarrow xy=\dfrac{729+\sqrt{6161}}{784}\)\(\Rightarrow y=\dfrac{\dfrac{729+\sqrt{6161}}{784}}{x}\)
Thay vào (2)\(\Rightarrow\)\(x\approx1,125;y\approx0,915\)
Với \(a-1< 0\Leftrightarrow a< 1\)
\(\Rightarrow49\left(-a+1\right)\left(8a-7\right)=-6a+8\)
\(\Leftrightarrow-49\left(8a^2-15a+7\right)+6a-8=0\)
\(\Leftrightarrow-392a^2+741a-351=0\)(vô nghiệm).
Vậy hpt có nghiệm (x;y)=(1,125;0,915).
\(\left\{{}\begin{matrix}x^3+y^3=1-x+y+xy\left(1\right)\\7xy+y-x=7\left(2\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^3+y^3=1-x+y+xy\\x-y=7xy-7\end{matrix}\right.\)
Từ pt (1) suy ra: \(x^3+y^3=1+xy-\left(x-y\right)\)
\(\Leftrightarrow x^3+y^3=1+xy-7xy+7\)
\(\Leftrightarrow x^3+y^3=-6xy+8\)
\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)=-6xy+8\)
\(\Leftrightarrow\left(x+y\right)^3-8=-6xy+3xy\left(x+y\right)\)
\(\Leftrightarrow\left(x+y-2\right)\left[\left(x+y\right)^2+2\left(x+y\right)+4\right]=3xy\left(x+y-2\right)\)
\(\Leftrightarrow\left(x+y-2\right)\left[\left(x+y\right)^2+2\left(x+y\right)+4-3xy\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y-2=0\\\left(x+y\right)^2+2\left(x+y\right)+4-3xy=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y=2\left(3\right)\\\left(x+y\right)^2+2\left(x+y\right)+4-3xy=0\left(4\right)\end{matrix}\right.\)
TH1: Từ (2) và (4) suy ra: \(\Leftrightarrow\left[{}\begin{matrix}x+y=2\\7xy+y-x=7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2-y\\7\left(2-y\right)y+y-2+y=7\end{matrix}\right.\)
Suy ra: 14y - 7y2 + y - 2 + y = 7
<=> 7y2 - 16y +9 = 0
\(\Leftrightarrow\left[{}\begin{matrix}y=1\rightarrow x=1\\y=\frac{9}{7}\rightarrow x=\frac{5}{9}\end{matrix}\right.\)
TH2:Thay vào tính cho kết quả ko thỏa mãn
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