Cộng vế:

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

\(\Leftrightarrow\left(x+1\right)^3-\left(y-1\right)^3+x-y+2=0\)

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

\(\Leftrightarrow\left(x-y+2\right)\left[\left(x+\dfrac{y}{2}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\left(y-1\right)^2+1\right]=0\)

\(\Leftrightarrow y=x+2\)

2:

a: y1+y2=-(x1+x2)=-5

y1*y2=(-x1)(-x2)=x1x2=6

Phương trình cần tìm có dạng là;

x^2+5x+6=0

b: y1+y2=1/x1+1/x2=(x1+x2)/x1x2=5/6

y1*y2=1/x1*1/x2=1/x1x2=1/6

Phương trình cần tìm là:

a^2-5/6a+1/6=0

\(5x^4+y^2-4x^2y-85=0\)

\(\Leftrightarrow x^4=4x^2-4x^2y+y^2-85=0\)

\(\Leftrightarrow x^4+\left(2x^2-y\right)^2=85\)

\(\Leftrightarrow x^4\in\left\{3^4;2^4;1^4;0^4\right\}\)

tiếp tục xét lần lượt các trường hợp:

+) nếu \(x^4=0^4\Rightarrow x=0\Rightarrow y^2=85\Rightarrow y\in\varnothing\)

+) nếu \(x^4=1^4\Rightarrow x=\pm1\Rightarrow\left(y-2\right)^2=84\Rightarrow y\in\varnothing\)

+) nếu \(x^4=2^4\Rightarrow x=\pm2\Rightarrow\left(y-8\right)^2=69\Rightarrow x\in\varnothing\)

+) nếu \(x^4=3^4\Rightarrow x=\pm3\Rightarrow\left(y-18\right)^2=2^2\)

\(\Leftrightarrow\orbr{\begin{cases}y-18=2\\y-18=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}y=20\\y=16\end{cases}}}\)( nhận ) 

P/s nhận cả hai nhé

a/ ĐKXĐ: \(x\ge4\)

Đặt \(\sqrt{x+4}+\sqrt{x-4}=a>0\)

\(\Rightarrow a^2=2x+2\sqrt{x^2-16}\)

Phương trình trở thành:

\(a=a^2-12\Leftrightarrow a^2-a-12=0\Rightarrow\left[{}\begin{matrix}a=4\\a=-3\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x+4}+\sqrt{x-4}=4\)

\(\Leftrightarrow2x+2\sqrt{x^2-16}=16\)

\(\Leftrightarrow\sqrt{x^2-16}=8-x\left(x\le8\right)\)

\(\Leftrightarrow x^2-16=x^2-16x+64\)

\(\Rightarrow x=5\)

b/ \(x\ge-\frac{1}{2}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{2x+1}=a\\\sqrt{4x^2-2x+1}=b\end{matrix}\right.\) ta được:

\(a+3b=3+ab\)

\(\Leftrightarrow ab-a-\left(3b-3\right)=0\)

\(\Leftrightarrow a\left(b-1\right)-3\left(b-1\right)=0\)

\(\Leftrightarrow\left(a-3\right)\left(b-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=3\\b=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+1}=3\\\sqrt{4x^2-2x+1}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x+1=9\\4x^2-2x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\\x=\frac{1}{2}\end{matrix}\right.\)

Bài 2:

a/ \(\left\{{}\begin{matrix}\left(x+2y\right)^2-4xy-5=0\\4xy\left(x+2y\right)+5\left(x+2y\right)-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+2y\right)^2-\left(4xy+5\right)=0\\\left(4xy+5\right)\left(x+2y\right)-1=0\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+2y=a\\4xy+5=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2-b=0\\ab=1\end{matrix}\right.\) \(\Rightarrow a^2-\frac{1}{a}=0\Rightarrow a^3-1=0\)

\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x+2y=1\\4xy+5=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1-2y\\4y\left(1-2y\right)+4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1-2y\\-8y^2+4y+4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=1\Rightarrow x=-1\\y=-\frac{1}{2}\Rightarrow x=2\end{matrix}\right.\)

b/Cộng vế với vế:

\(17x^2-2\left(4y^2+1\right)x+y^4+1=0\)

\(\Delta'=\left(4y^2+1\right)^2-17\left(y^4+1\right)=-y^4+8y^2-16\)

\(\Delta'=-\left(y^2-4\right)^2\ge0\Rightarrow y^2-4=0\Rightarrow\left[{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\)

- Với \(y=2\) \(\Rightarrow x^2-2x+1=0\Rightarrow x=1\)

\(\)- Với \(y=-2\Rightarrow x^2-2x-7=0\Rightarrow x=1\pm2\sqrt{2}\)