\(đặt:\sqrt{x+y+1}=a\ge0;\sqrt{3x+3y}=b\ge0\)\(\left(đk:x+y+1\ge0;3x+3y\ge0\right)\)
\(\Rightarrow b^2-a^2=3\left(x+y\right)-\left(x+y+1\right)=2\left(x+y\right)-1\Leftrightarrow\left(x+y\right)=\dfrac{b^2-a^2+1}{2}\)
\(\Rightarrow a+1=4\left(\dfrac{b^2-a^2+1}{2}\right)^2+b=\left(b^2-a^2+1\right)^2+b\)
\(\Leftrightarrow-\left(a-b\right)\left(a^3+a^2b-ab^2-2a-b^3-2b-1\right)=0\Rightarrow a=b\)
\(\Leftrightarrow\sqrt{x+y+1}=\sqrt{3x+3y}\Leftrightarrow x+y+1=3x+3y\Leftrightarrow2\left(x+y\right)=1\Leftrightarrow y=\dfrac{1-2x}{2}\)
\(với:y=\dfrac{1-2x}{2}\Rightarrow pt\) \(dưới\)
\(\Leftrightarrow6x\left(-10x^2+x+3\right)=-1-3\left(2x-1\right)^2\left(3+5x\right)\)
\(rút\) \(gọn\Leftrightarrow-\left(3x-2\right)\left(6x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\Rightarrow y=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\Rightarrow y=\dfrac{4}{3}\end{matrix}\right.\)(thỏa)
bài này chắc có cánh làm đạo hàm f(t') để ra cái a=b nhưng mình chưa học nên phân tích thủ công tí
\(b;\left\{{}\begin{matrix}\sqrt{x}+\sqrt[4]{32-x}-y^2+3=0\left(1\right)\\\sqrt[4]{x}+\sqrt{32-x}+6y-24=0\left(2\right)\end{matrix}\right.\)
\(đặt:\left\{{}\begin{matrix}\sqrt[4]{32-x}=a\ge0\\\sqrt[4]{x}=b\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{32-x}=a^2\\\sqrt{x}=b^2\end{matrix}\right.\)
\(\Rightarrow hpt\Leftrightarrow\left\{{}\begin{matrix}b^2+a-y^2+3=0\left(1\right)\\b+a^2+6y-24=0\left(2\right)\end{matrix}\right.\)
\(\left(1\right)+\left(2\right)\Leftrightarrow a^2+b^2+a+b-y^2+6y-21=0\Leftrightarrow a^2+b^2+a+b=y^2-6y+21\)
\(có:a^2+b^2\le\sqrt{2\left(a^4+b^4\right)}=\sqrt{2.32}=8\left(bunhia\right)\)
\(có:a+b\le\sqrt{2\left(a^2+b^2\right)}=\sqrt{2.8}=4\)
\(\Rightarrow a^2+b^2+a+b\le12\)
\(mà:y^2-6y+21=\left(y-3\right)^2+12\ge12\)
\(\Rightarrow dấu"="xảy\) \(ra\Leftrightarrow\left\{{}\begin{matrix}a=b\\y=3\end{matrix}\right.\)
\(\Rightarrow a^4=b^4\Leftrightarrow x=32-x\Leftrightarrow x=16\)
\(x=16;y=3\) \(thử\) \(lại\) \(thấy\) \(thỏa\)
