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`x^3+1=2y,y^3+1=2x`
`=>x^3-y^3=2y-2x`
`<=>(x-y)(x^2+xy+y^2)+2(x-y)=0`
`<=>(x-y)(x^2+xy+y^2+2)=0`
Vì `x^2+xy+y^2+2>=2>0`
`=>x-y=0<=>x=y` thay vào bthức
`=>x^3+1=2x`
`<=>x^3-2x+1=0`
`<=>x^3-x^2+x^2-2x+1=0`
`<=>x^2(x-1)+(x-1)^2=0`
`<=>(x-1)(x^2+x-1)=0`
`+)x=1=>x=y=1`
`+)x^2+x-1=0`
`\Delta=1+4=5`
`=>x_1=(-1-sqrt5)/2,x_2=(-1+sqrt5)/2`
`=>x=y=(-1-sqrt5)/2,x=y=z(-1+sqrt5)/2`
Vậy `(x,y)=(1,1),((-1-sqrt5)/2,(-1-sqrt5)/2),((-1+sqrt5)/2,(-1+sqrt5)/2)`
![](https://rs.olm.vn/images/avt/0.png?1311)
1: =>x^2+3x-4=0
=>(x+4)(x-1)=0
=>x=1 hoặc x=-4
2: =>2x-3y=1 và 3x=4y+2
=>2x-3y=1 và 3x-4y=2
=>x=2 và y=1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-6xy+2y^2=6\\x^2+2xy-2y^2=6\end{matrix}\right.\)
\(\Rightarrow3x^2-8xy+4y^2=0\)
\(\Rightarrow\left(3x-2y\right)\left(x-2y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}y=\dfrac{3}{2}x\\y=\dfrac{1}{2}x\end{matrix}\right.\)
Thế vào pt đầu...
\(\left\{{}\begin{matrix}2x^2-3xy+y^2=3\\x^2+2xy-2y^2=6\end{matrix}\right.\)\(\left(1\right)\)\(\Leftrightarrow\left\{{}\begin{matrix}4x^2-6xy+2y^2=6\\x^2+2xy-2y^2=6\end{matrix}\right.\)
\(\Leftrightarrow3x^2-8xy+4y^2=0\)
\(\Leftrightarrow3x\left(x-2y\right)-2y\left(x-2y\right)=0\)
\(\Leftrightarrow\left(x-2y\right)\left(3x-2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2y\\x=\dfrac{2y}{3}\end{matrix}\right.\)
Thay vào \(\left(1\right)\) ta được:
\(\Leftrightarrow\left[{}\begin{matrix}2.\left(2y\right)^2-3.2y.y+y^2=3\\2.\left(\dfrac{2y}{3}\right)^2-3.\dfrac{2y}{3}.y+y^2=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}y^2=1\\y^2=-27\left(VLý\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\end{matrix}\right.\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left\{{}\begin{matrix}18x^3+9y^3=90\\10x^2y-30xy^2+10x^3=-90\end{matrix}\right.\)
\(\Rightarrow28x^3+10x^2y-30xy^2+9y^3=0\)
\(\Leftrightarrow\left(2x-y\right)\left(14x^2+12xy-9y^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=2x\\y=\frac{2+2\sqrt{3}}{3}x\\y=\frac{2-2\sqrt{3}}{3}x\end{matrix}\right.\) thế vào pt đầu:
\(\Rightarrow\left[{}\begin{matrix}2x^3+\left(2x\right)^3=10\\2x^3+\left(\frac{2+2\sqrt{3}}{3}\right)^3x^3=10\\2x^3+\left(\frac{2-2\sqrt{3}}{3}\right)^3x^3=10\end{matrix}\right.\)
Bạn tự giải nốt
Khi \(x=0\), hệ đã cho trở thành \(\left\{{}\begin{matrix}y^3=6\\0=-9\end{matrix}\right.\Rightarrow\) vô nghiệm
Xét \(x\ne0\) từ hệ phương trình đã cho ta có:
\(9\left(2x^3+y^3\right)=-10\left(x^2y-3xy^2+x^3\right)\)
\(\Leftrightarrow18x^3+9y^3=-10x^2y+30xy^2-10x^3\)
\(\Leftrightarrow18+9\left(\frac{y}{x}\right)^3=-10\frac{y}{x}+30\left(\frac{y}{x}\right)^2-10\)
Đặt \(t=\frac{y}{x}\) khi đó:
\(9t^3-30t^2+10t+28=0\)
\(\Leftrightarrow\left(t-2\right)\left(9t^2-12t-14\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=\frac{2\pm3\sqrt{2}}{3}\end{matrix}\right.\)
Ta có phương trình thứ nhất tương đương
\(2x^3+t^3x^3=10\Leftrightarrow x^3=\frac{10}{2+t^3}\Leftrightarrow x=\sqrt[3]{\frac{10}{2+t^3}}\)
Nếu \(t=2\Rightarrow x=\sqrt[3]{\frac{10}{2+t^3}}=1\Rightarrow y=2\)
Nếu \(t=\frac{2\pm3\sqrt{2}}{3}\Rightarrow x=\sqrt[3]{\frac{10}{2+\left(\frac{2\pm3\sqrt{2}}{3}\right)^3}}\Rightarrow y=\sqrt[3]{\frac{10.\left(\frac{2\pm3\sqrt{2}}{3}\right)^3}{2+\left(\frac{2\pm3\sqrt{2}}{3}\right)^3}}\)
Vậy hệ pt đã cho có ba nghiệm \(\left(x;y\right)\in\left\{\left(1;2\right);\left(\sqrt[3]{\frac{10}{2+\left(\frac{2\pm3\sqrt{2}}{3}\right)^3}};\sqrt[3]{\frac{10.\left(\frac{2\pm3\sqrt{2}}{3}\right)^3}{2+\left(\frac{2\pm3\sqrt{2}}{3}\right)^3}}\right)\right\}\)
Bài j ghê vậy em, xỉu!!
Giải hệ phương trình: \(\left\{{}\begin{matrix}2x^3+y^3=10\\x^{2y}-3xy^2+x^3=-9\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Từ HPT ta có:
$9(2x^3+y^3)+10(x^2y-3xy^2+x^3)=0$
$\Leftrightarrow 28x^3+10x^2y-30xy^2+9y^3=0$
Dễ thấy $y=0$ không phải là nghiệm nên $y\neq 0$
Đặt $x=ty$. Khi đó PT trở thành:
$28(ty)^3+10(yt)^2y-30ty.y^2+9y^3=0$
$\Leftrightarrow y^3(28t^3+10t^2-30t+9)=0$
$\Leftrightarrow 28t^3+10t^2-30t+9=0$
$\Leftrightarrow (2t-1)(14t^2+12t-9)=0$
Nếu $2t-1=0\Rightarrow t=\frac{1}{2}\Rightarrow 2x=y$
Thay vào PT $(1)$ ta được $x=1; y=2$
Nếu $14t^2+12t-9=0\Rightarrow t=\frac{-6+9\sqrt{2}}{14}$
Thay vào PT$(1)$ ta được:
\((x,y)=(-3\sqrt[3]{\frac{-17-9\sqrt{2}}{127}}; (3\sqrt{2}-2)\sqrt[3]{\frac{-17-9\sqrt{2}}{127}})\) hoặc
\((x,y)=(3\sqrt[3]{\frac{17-9\sqrt{2}}{127}}; (3\sqrt{2}+2)\sqrt[3]{\frac{17-9\sqrt{2}}{127}})\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y^3+3x^2y-3xy^2-2x^3=0\)
\(\Leftrightarrow\left(y^3-xy^2+x^2y\right)-2\left(x^3-x^2y+xy^2\right)=0\)
\(\Leftrightarrow y\left(x^2-xy+y^2\right)-2x\left(x^2-xy+y^2\right)=0\)
\(\Leftrightarrow\left(y-2x\right)\left(x^2-xy+y^2\right)=0\)
\(\Rightarrow y=2x\)
Thế xuống dưới:
\(x^4-2x^3-x^2+2x+1=0\)
Nhận thấy \(x=0\) ko phải nghiệm, chia 2 vế cho \(x^2\)
\(x^2+\frac{1}{x^2}-2\left(x-\frac{1}{x}\right)-1=0\)
Đặt \(x-\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2+2\) pt trở thành:
\(t^2-2t+1=0\Leftrightarrow t=1\)
\(\Leftrightarrow x-\frac{1}{x}=1\Leftrightarrow x^2-x-1=0\Leftrightarrow...\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Leftrightarrow\left\{{}\begin{matrix}5x+15y=-10\\5x-4y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19y=-21\\5x-4y=11\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{21}{19}\\5x-4\left(-\dfrac{21}{19}\right)=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{25}{19}\\y=-\dfrac{21}{19}\end{matrix}\right.\)
\(c,\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\10x-5y=-40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\13x=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\\ d,\Leftrightarrow\left\{{}\begin{matrix}5x-10y=-30\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-3y=5\\-7y=-35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\\ e,\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\2\left(x+y\right)+4\left(x-y\right)=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=6\\2\left(x+y\right)+3\cdot6=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=6\\x+y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{13}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left\{{}\begin{matrix}x^2-xy-2y^2=0\\3x+y=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x\left(1-3x\right)-2\left(1-3x\right)^2=0\\y=1-3x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-14x^2+11x-2=0\\y=1-3x\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{7}\end{matrix}\right.\\y=1-3x\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{2}{7}\\y=\dfrac{1}{7}\end{matrix}\right.\end{matrix}\right.\)
Vậy...