2)ĐK:x\(\ge\frac{1}{2}\)

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

Xét hàm số: f(t)=\(t^3\)+t

f'(t)=3\(t^2\)+1>0,\(\forall\)t

\(\Rightarrow\)hàm số liên tục và đồng biến trên R

\(\Rightarrow\)y+1=2x

Thay y=2x-1 vào pt(1) ta đc:

\(x^2\)-2x=2\(\sqrt{2x-1}\)

\(\Leftrightarrow\left(x^2-4x+2\right)\left(1+\frac{4}{2x-2+2\sqrt{2x-1}}\right)=0\)

\(\Leftrightarrow x^2\)-4x+2=0(do(...)>0)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2+\sqrt{2}\Rightarrow y=3+2\sqrt{2}\\x=2-\sqrt{2}\Rightarrow y=3-2\sqrt{2}\end{array}\right.\)

4)ĐK:\(y\ge\frac{2}{3}\)

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

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

Xét hàm số:\(f\left(t\right)=t\left(\sqrt{t^2+2}+1\right)\)

\(\Rightarrow\)hàm số liên tục và đồng biến trên R

\(\Rightarrow x=\sqrt{3y-2}\)

Thay vào pt(2) ta đc:\(\sqrt{3y-2}+y+\sqrt{y+3}=4\)

\(\Leftrightarrow\sqrt{3y-2}-1+\sqrt{y+3}-2+y-1=0\)

\(\Leftrightarrow\left(y-1\right)\left(\frac{3}{\sqrt{3y-2}+1}+\frac{1}{\sqrt{y+3}+2}+1\right)=0\)

\(\Leftrightarrow y=1\Rightarrow x=1\)(do...)>0)

KL:...

1/HPT\(\Leftrightarrow\hept{\begin{cases}x^2+y^2=6-\left(x+y\right)=3\\\left(x+y\right)^2=9\end{cases}}\Rightarrow2xy=\left(x+y\right)^2-\left(x^2+y^2\right)=9-3=6\Rightarrow xy=3\)

Kết hợp đề bài có được: \(\hept{\begin{cases}x+y=3\\xy=3\end{cases}}\). Dùng hệ thức Viet đảo là xong.

1) \(x^3-3x^2y-4x^2+4y^3+16xy=16y^2\Leftrightarrow x^3-3x^2y-4x^2+4y^3+16xy-16y^2=0\)

đưa về phương trình tích : \(\left(x-2y\right)^2\left(x+y-4\right)=0\) tới đây ok chưa

3)  ĐK : x \(\ge\)0 ; \(y\ge3\)\(\Rightarrow x+y>0\)

đặt \(\sqrt{x+y}=a;\sqrt{x+3}=b\)

\(\Rightarrow y-3=\left(x+y\right)-\left(x+3\right)=a^2-b^2\)

PT : \(\sqrt{x+y}+\sqrt{x+3}=\frac{1}{3}\left(y-3\right)\Leftrightarrow3\sqrt{x+y}+3\sqrt{x+3}=y-3\)

\(\Leftrightarrow3\left(a+b\right)=a^2-b^2\Leftrightarrow\left(a+b\right)\left(3-a+b\right)=0\Leftrightarrow\orbr{\begin{cases}a+b=0\\a-b=3\end{cases}}\)

Mà a + b = \(\sqrt{x+y}+\sqrt{x+3}>0\)nên loại

a - b  = 3 thì \(\sqrt{x+y}-\sqrt{x+3}=3\), ta có HPT : \(\hept{\begin{cases}\sqrt{x+y}-\sqrt{x+3}=3\\\sqrt{x+y}+\sqrt{x}=x+3\end{cases}}\)

\(\Rightarrow\)\(\sqrt{x}+\sqrt{x+3}=x\Leftrightarrow\sqrt{x+3}=x-\sqrt{x}\Leftrightarrow x^2-2x\sqrt{x}-3=0\Leftrightarrow x=\left(1+\sqrt[3]{2}\right)^2\)

từ đó tìm đc y

a: \(\left\{{}\begin{matrix}\left(\sqrt{3}+1\right)x+\left(\sqrt{3}-1\right)y=\sqrt{3}\\2\sqrt{3}x-2y=3\sqrt{3}+1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(\sqrt{3}+1\right)^2\cdot x+\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)y=\sqrt{3}\left(\sqrt{3}+1\right)\\2\sqrt{3}x-2y=3\sqrt{3}+1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\left(4+2\sqrt{3}\right)+2y=3+\sqrt{3}\\2\sqrt{3}\cdot x-2y=3\sqrt{3}+1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\left(4+2\sqrt{3}+2\sqrt{3}\right)=3+\sqrt{3}+3\sqrt{3}+1\\2\sqrt{3}\cdot x-2y=3\sqrt{3}+1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=1\\2y=2\sqrt{3}-3\sqrt{3}-1=-\sqrt{3}-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=1\\y=\dfrac{-\sqrt{3}-1}{2}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}x\sqrt{3}+y\sqrt{2}=1\\x\sqrt{2}+y\sqrt{3}=\sqrt{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\sqrt{6}+2y=\sqrt{2}\\x\sqrt{6}+3y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2y-3y=\sqrt{2}-3\\x\sqrt{3}+y\sqrt{2}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-y=\sqrt{2}-3\\x\sqrt{3}=1-y\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3-\sqrt{2}\\x\sqrt{3}=1-\sqrt{2}\left(3-\sqrt{2}\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=3-\sqrt{2}\\x\sqrt{3}=1-3\sqrt{2}+2=3-3\sqrt{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=3-\sqrt{2}\\x=\sqrt{3}-\sqrt{6}\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}xy-2y-y+2=xy-3x+y-3\\xy+4x-5y-20=xy+x-4y-4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-2x-y+2=-3x+y-3\\4x-5y-20=x-4y-4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-2x-y+3x-y=-3-2=-5\\4x-5y-x+4y=-4+20\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-2y=-5\\3x-y=16\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-6y=-15\\3x-y=16\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-5y=-15-16=-31\\x-2y=-5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=\dfrac{31}{5}\\x=-5+2y=-5+\dfrac{62}{5}=\dfrac{37}{5}\end{matrix}\right.\)