Câu hỏi của Phương Thùy Lê

Question

Giải pt sau bằng cách đặt ẩn phụ

1, $\left\{{}\begin{matrix}x^2+y+x^3y+xy^2+xy=-\frac{5}{4}\x^4+y^2+xy\left(1+2x\right)=-\frac{5}{4}\end{matrix}\right.$

2, \(\left\{{}\begin{matrix}x^3+3x^2-13x-1...

Nguyễn Việt Lâm · Accepted Answer

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: $\left\{{}\begin{matrix}x^2+y=-\frac{1}{2}\xy+1=-\frac{1}{2}\end{matrix}\right.$ $\Rightarrow\left\{{}\begin{matrix}y=-\frac{1}{2}-x^2\x\left(-\frac{1}{2}-x^2\right)=-\frac{5}{4}\end{matrix}\right.$ $\Rightarrow...$Câu trả lời: 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: $\left\{{}\begin{matrix}x^2+y=-\frac{1}{2}\xy+1=-\frac{1}{2}\end{matrix}\right.$ $\Rightarrow\left\{{}\begin{matrix}y=-\frac{1}{2}-x^2\x\left(-\frac{1}{2}-x^2\right)=-\frac{5}{4}\end{matrix}\right.$ $\Rightarrow...$

Nguyễn Việt Lâm · Answer

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:

$4\left(a^3-8b^3\right)=4\left(4a-2b\right)\left(-5a^2+4b^2\right)$

$\Leftrightarrow21a^3-10a^2b-16ab^2=0$

$\Leftrightarrow a\left(21a^2-10ab-16b^2\right)=0$

$\Leftrightarrow a\left(7a-8b\right)\left(3a+2b\right)=0$

$\Rightarrow\left[{}\begin{matrix}a=0\7a=8b\3a=-2b\end{matrix}\right.$ $\Rightarrow...$Câu trả lời: b/ ĐKXĐ; ...