b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)

\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)

\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)

\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)

\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)

\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)

caau a) binh phuong len ra no x=y tuong tu

\(\left\{{}\begin{matrix}x^3+xy^2+3\left(x-2y\right)=0\\x^2+xy=3\end{matrix}\right.\)\(\Rightarrow x^3+xy^2+\left(x^2+xy\right)\left(x-2y\right)=0\)\(\Leftrightarrow x^3+xy^2+x^3-x^2y-2xy^2=0\Leftrightarrow2x^3-x^2y-xy^2=0\)\(\Leftrightarrow x\left(2x+y\right)\left(x-y\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\y=-2x\\x=y\end{matrix}\right.\)

+) \(x=0\Rightarrow0y=3\)(vô nghiệm)

+) y=-2x \(\Rightarrow x^2-2x^2=3\Leftrightarrow-x^2=3\)(vô nghiệm)

+) x=y\(\Rightarrow2x^2=3\Leftrightarrow x^2=\dfrac{3}{2}\Leftrightarrow\left[{}\begin{matrix}x=y=\sqrt{\dfrac{3}{2}}\\x=y=-\sqrt{\dfrac{3}{2}}\end{matrix}\right.\)

a, ĐKXĐ : \(\left[{}\begin{matrix}x\le-3\\x\ge0\end{matrix}\right.\)

TH1 : \(x\le-3\) ( LĐ )

TH2 : \(x\ge0\)

BPT \(\Leftrightarrow x^2+2x+x^2+3x+2\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge4x^2\)

\(\Leftrightarrow\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge x^2-\dfrac{5}{2}x\)

\(\Leftrightarrow2\sqrt{\left(x+2\right)\left(x+3\right)}\ge2x-5\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\4x^2+20x+24\ge4x^2-20x+25\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}0\le x< \dfrac{5}{2}\\x\ge\dfrac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow x\ge0\)

Vậy \(S=R/\left(-3;0\right)\)

\(x^3-7x^2y+16xy^2-12y^3=0\)

\(\Leftrightarrow\left(x-3y\right)\left(x-2y\right)^2=0\)

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

Thế xuống pt dưới giải đơn giản

1/ \(\left\{{}\begin{matrix}x^3+y^3=1\left(1\right)\\x^2y+2xy^2+y^3=2\left(2\right)\end{matrix}\right.\)

Lấy (1). 2 - (2) ta được:

\(2x^3+y^3-x^2y-2xy^2=0\)

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

Đến đây dễ rồi nhé ^^

2/ Ta viết lại pt thứ 2 của hệ:

\(y^2-4\left(x+2\right)y+16+16x-5x^2=0\)

\(\Leftrightarrow y^2-4\left(x+2\right)y+4\left(x+2\right)^2-9x^2=0\)

\(\Leftrightarrow\left[y-2\left(x+2\right)\right]^2-\left(3x\right)^2=0\)

\(\Leftrightarrow\left(x+y-4\right)\left(y-5x-4\right)=0\)

Bạn làm tiếp nhé!

3/ Ta viết lại pt thứ nhất của hệ

\(x^2-x\left(2y-3\right)+y^2-3y-4=0\)

\(\Leftrightarrow x^2-x\left(2y-3\right)+\dfrac{4y^2-12y+9}{4}-\dfrac{25}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{2y+3}{2}\right)^2-\left(\dfrac{5}{2}\right)^2=0\)

\(\Leftrightarrow\left(x-y-4\right)\left(x-y+1\right)=0\)

Bạn làm tiếp được chứ?

4/ Viết lại pt thứ 2 của hệ

\(\left(y+\sqrt{x}\right)^2-\left(y\sqrt{x}\right)^2=0\)

\(\Leftrightarrow\left(y-\sqrt{x}-y\sqrt{x}\right)\left(y-\sqrt{x}+y\sqrt{x}\right)=0\)

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

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

Đặt \(\left\{{}\begin{matrix}x^2-2x=u\\y^2-4y=v\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2u-v=1\\u^2+2=uv\end{matrix}\right.\) \(\Rightarrow u^2+2=u\left(2u-1\right)\)

\(\Leftrightarrow u^2-u-2=0\Leftrightarrow...\)