1.

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

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x^2+y\right)+xy+xy\left(x^2+y\right)=-\dfrac{5}{4}\\\left(x^2+y\right)^2+xy=-\dfrac{5}{4}\end{matrix}\right.\left(1\right)\)

Đặt \(\left\{{}\begin{matrix}x^2+y=a\\xy=b\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}a+b+ab=-\dfrac{5}{4}\\a^2+b=-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-a^2-\dfrac{5}{4}-a\left(a^2+\dfrac{5}{4}\right)=-\dfrac{5}{4}\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a^2-a^3-\dfrac{1}{4}a=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-a\left(a^2-a+\dfrac{1}{4}\right)=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a\left(a-\dfrac{1}{2}\right)^2=0\\b=-a^2-\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=0\\b=-\dfrac{5}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=-\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}a=0\\b=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y=0\\xy=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{\sqrt[3]{10}}{2}\\y=-\dfrac{5}{2\sqrt[3]{10}}\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y=\dfrac{1}{2}\\xy=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-\dfrac{3}{2}\end{matrix}\right.\)

Kết luận: Phương trình đã cho có nghiệm \(\left(x;y\right)\in\left\{\left(\dfrac{\sqrt[3]{10}}{2};-\dfrac{5}{2\sqrt[3]{10}}\right);\left(1;-\dfrac{3}{2}\right)\right\}\)

2.

\(\left\{{}\begin{matrix}\left(x+1\right)^3-16\left(x+1\right)=\left(\dfrac{2}{y}\right)^3-4\left(\dfrac{2}{y}\right)\\1+\left(\dfrac{2}{y}\right)^2=5\left(x+1\right)^2+5\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+1=u\\\dfrac{2}{y}=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^3-16u=v^3-4v\\v^2=5u^2+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u^3-v^3=16u-4v\\4=v^2-5u^2\end{matrix}\right.\)

\(\Rightarrow4\left(u^3-v^3\right)=\left(16u-4v\right)\left(v^2-5u^2\right)\)

\(\Leftrightarrow21u^3-5u^2v-4uv^2=0\)

\(\Leftrightarrow u\left(7u-4v\right)\left(3u+v\right)=0\Rightarrow\left[{}\begin{matrix}u=0\Rightarrow v^2=4\\u=\dfrac{4v}{7}\Rightarrow4=v^2-5\left(\dfrac{4v}{7}\right)^2\\v=-3u\Rightarrow4=\left(-3u\right)^2-5u^2\end{matrix}\right.\) 

\(\Rightarrow...\)

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

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:

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

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

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

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

\(\Rightarrow\left[{}\begin{matrix}a=0\\7a=8b\\3a=-2b\end{matrix}\right.\) \(\Rightarrow...\)

e) Sửa đề: \(\left\{{}\begin{matrix}x\left(x^2-y^2\right)+x^2=2\sqrt{\left(x-y^2\right)^3}\\76x^2-20y^2+2=\sqrt[3]{4x\left(8x+1\right)}\end{matrix}\right.\)

PT(1) \(\Leftrightarrow x^3+x\left(x-y^2\right)=\sqrt{\left(x-y^2\right)^3}\)

Đặt \(\sqrt{x-y^2}=a.\text{Thay vào, ta có: }x^3+xa^2-2a^3=0\)

Làm tiếp như ở Câu hỏi của Nguyễn Mai - Toán lớp 9 - Học toán với OnlineMath

Băng Băng 2k6, Vũ Minh Tuấn, Nguyễn Việt Lâm, HISINOMA KINIMADO, Akai Haruma, Inosuke Hashibira, Nguyễn Thị Ngọc Thơ, Nguyễn Lê Phước Thịnh, Quân Tạ Minh, An Võ (leo), @tth_new

e nhiều bài quá giải k kịp mn giúp e vs ạ!cần gấp lắm ạ

thanks nhiều!

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