\(\hept{\begin{cases}x\sqrt{12-y}+\sqrt{y\left(12-x^2\right)}=12\left(1\right)\\x^3-8x-1=2\sqrt{y-2}\left(2\right)\end{cases}}\)
\(\Rightarrow\left(1\right)\Leftrightarrow\sqrt{y\left(12-x^2\right)}=12-x\sqrt{12-y}\)
\(\Leftrightarrow\left(\sqrt{y\left(12-x^2\right)}\right)^2=\left(12-x\sqrt{12-y}\right)^2\)
\(\Leftrightarrow x^2-2x\sqrt{12-y}+\left(12-y\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{12-y}\right)^2=0\)
\(\Leftrightarrow3-y=x^2-9\left(3\right)\)
Ta lại có:
\(\left(2\right)\Leftrightarrow\left(x^3-8x-3\right)=2\left(\sqrt{y-2}-1\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+1\right)=\frac{2\left(y-3\right)}{\sqrt{y-2}+1}\left(4\right)\)
Thay (3) vào (4) ta được:
\(\left(x-3\right)\left(x^2+3x+1\right)+\frac{2\left(x^2-9\right)}{\sqrt{y-2}+1}=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+1+\frac{2\left(x+3\right)}{\sqrt{y-2}+1}\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=3\\y=3\end{cases}}\)
Đặt \(\sqrt{x^2-x+1}=a"ĐK:a>0"\)
\(pt\Leftrightarrow\frac{"6^2+3x^4a""4-a^2"}{4"2+a"a^2}=a"2-a"\)
\(\Leftrightarrow"x^6+3x^4a""4-a^2"=4a^3"4-a^2"\)
\(\Leftrightarrow"4-a^2""x^6+3x^4a-4a^3"=0\)
TH1: \(4-a^2=0\Leftrightarrow\orbr{\begin{cases}a=-2\\a=2\end{cases}}\)
Với \(a=2,\sqrt{x^2-x+1}=2\Rightarrow x^2-x-3=0\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{3}+1}{2}\\x=\frac{-\sqrt{13}+1}{2}\end{cases}}\)
TH2: \(x^6+3x^4a-4a^3=0\Rightarrow x^6-4x^4a-4x^2a^2+4x^2a^2-4a^3=0\)
\(\Leftrightarrow"x^2-a""x^4+4x^2a+4a^2"=0\Leftrightarrow"x^2-a""x^2+2a"^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=a\\x^2=-2a\end{cases}}\)
Với \(x^2=a\Rightarrow x^2=\sqrt{x^2-x+1}\)
P/s: Tham khảo thôi đừng có chép nguyên vào
Thay dấu ngoặc kép thành ngoặc đơn nha
\(\hept{\begin{cases}x\sqrt{12-y}+\sqrt{y\left(12-x^2\right)}=12\left(1\right)\\x^3-8x-1=2\sqrt{y-2}\left(2\right)\end{cases}}\)
\(ĐK:-2\sqrt{3}\le x\le2\sqrt{3},2\le y\le12\)
Ta có: \(\left(a-b\right)^2\ge0\forall a,b\inℝ\Leftrightarrow\frac{a^2+b^2}{2}\ge ab\forall a,b\inℝ\)
Do đó: \(\hept{\begin{cases}x\sqrt{12-y}\le\left|x\right|\sqrt{12-y}\le\frac{x^2+12-y}{2}\\\sqrt{y\left(12-x^2\right)}\le\frac{y+12-x^2}{2}\end{cases}}\)
\(\Rightarrow x\sqrt{12-y}+\sqrt{y\left(12-x^2\right)}\le\frac{\left(x^2+12-y\right)+\left(y+12-x^2\right)}{2}=12\)
Suy ra \(\left(1\right)\Leftrightarrow\hept{\begin{cases}x\ge0\\y=12-x^2\end{cases}}\)
Thay \(y=12-x^2\)vào (2), ta được: \(x^3-8x-1=2\sqrt{10-x^2}\)
\(\Leftrightarrow x^3-8x-3+2\left(1-\sqrt{10-x^2}\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+1\right)+\frac{2\left(x^2-9\right)}{1+\sqrt{10-x^2}}=0\)
\(\Leftrightarrow\left(x-3\right)\left[x^2+3x+1+\frac{2\left(x+3\right)}{1+\sqrt{10-x^2}}\right]=0\)
Mà ta dễ thấy được: \(x^2+3x+1+\frac{2\left(x+3\right)}{1+\sqrt{10-x^2}}>0\forall x\ge0\)nên \(x-3=0\Leftrightarrow x=3\left(tm\right)\Rightarrow y=12-3^2=3\)
Vậy hệ có 1 nghiệm duy nhất \(\left(x,y\right)=\left(3,3\right)\)
