\(log_{\sqrt{3}}\left(2x+y\right)-log_{\sqrt{3}}\left(4x^2+y^2+2xy+2\right)=\left(4x^2+y^2+2xy+2\right)-3\left(2x+y\right)-2\)

\(\Leftrightarrow log_{\sqrt{3}}\left(2x+y\right)+2+3\left(2x+y\right)=log_{\sqrt{3}}\left(4x^2+y^2+2xy+2\right)+\left(4x^2+y^2+2xy+2\right)\)

\(\Leftrightarrow log_{\sqrt{3}}\left(6x+3y\right)+\left(6x+3y\right)=log_{\sqrt{3}}\left(4x^2+y^2+2xy+2\right)+\left(4x^2+y^2+2xy+2\right)\)

Xét hàm \(f\left(t\right)=log_{\sqrt{3}}t+t\) với \(t>0\)

\(f'\left(t\right)=\dfrac{1}{t.ln\sqrt{3}}+1>0\Rightarrow f\left(t\right)\) đồng biến

\(\Rightarrow6x+3y=4x^2+y^2+2xy+2\)

\(\Leftrightarrow4x+y=\left(x+y-1\right)^2+1+3\left(x^2+1\right)-3\ge2\left(x+y-1\right)+6x-3\)

\(\Leftrightarrow4x+y\ge2\left(4x+y\right)-5\)

\(\Leftrightarrow4x+y\le5\)

\(\Rightarrow P=\dfrac{2x+y+6+\left(4x+y-5\right)}{2x+y+6}=1+\dfrac{4x+y-5}{2x+y+6}\le1\)

\(P_{max}=1\) khi \(x=y=1\)

ĐKXĐ: \(x\le\dfrac{1}{2}\)

\(4x^2+y^2+2x+y=2-4xy\)

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

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

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

Thế vào pt dưới:

\(\Rightarrow\left[{}\begin{matrix}8\sqrt{y}+y^2-9=0\\8\sqrt{y+3}+y^2-9=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(x^2+2xy+y^2+x^2-6x+9+\left|y+3\right|=0\)

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

\(\Rightarrow\left\{{}\begin{matrix}x+y=0\\x-3=0\\y+3=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\)

Điều kiện: \(y\ge0\)

pt thứ nhất của hệ \(\Leftrightarrow\left(y-x+3\right)^2=0\) \(\Leftrightarrow y-x+3=0\) \(\Leftrightarrow y=x-3\)

Thay vào pt thứ hai của hệ, ta được  \(2x^2+3x+x-3-\left(3x+1\right)\sqrt{x-3}-2=0\)

\(\Leftrightarrow2x^2+4x-5=\left(3x+1\right)\sqrt{x-3}\)         \(\left(x\ge3\right)\)

\(\Rightarrow\left(2x^2+4x-5\right)^2=\left[\left(3x+1\right)\sqrt{x-3}\right]^2\)

\(\Leftrightarrow4x^4+16x^2+25+16x^3-20x^2-40x=\left(3x+1\right)^2\left(x-3\right)\)

\(\Leftrightarrow4x^4+16x^3-4x^2-40x+25=9x^3-21x^2-17x-3\)

\(\Leftrightarrow4x^4+7x^3+17x^2-23x+28=0\)

Đặt \(f\left(x\right)=4x^4+7x^3+17x^2-23x+28\)

\(f\left(x\right)=4x^4+7x^3+17x^2+4+4+...+4-23x+4\) (có 6 số 4 ở giữa)

\(f\left(x\right)\ge9\sqrt[9]{4x^4.7x^3.17x^2.4^6}-23x+4\) \(=\left(9\sqrt[9]{1949696}-23\right)x+4\)

Hiển nhiên \(9\sqrt[9]{1949696}>23\). Lại có \(x\ge3\) nên \(f\left(x\right)>0\), Như vậy pt \(f\left(x\right)=0\) vô nghiệm. Điều đó có nghĩa là phương trình đã cho vô nghiệm.

a.

\(\left\{{}\begin{matrix}x^3-y^3=16x-4y\\-4=5x^2-y^2\end{matrix}\right.\)

Nhân vế:

\(-4\left(x^3-y^3\right)=\left(16x-4y\right)\left(5x^2-y^2\right)\)

\(\Leftrightarrow21x^3-5x^2y-4xy^2=0\)

\(\Leftrightarrow x\left(7x-4y\right)\left(3x+y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4y}{7}\\y=-3x\end{matrix}\right.\)

Thế vào \(y^2=5x^2+4...\)

b. Đề bài không hợp lý ở \(4x^2\)

c.

\(\Leftrightarrow\left\{{}\begin{matrix}x^3-y^3=9\\3x^2+6y^2=3x-12y\end{matrix}\right.\)

Trừ vế:

\(x^3-y^3-3x^2-6y^2=9-3x+12y\)

\(\Leftrightarrow x^3-3x^2+3x-1=y^3+6y^2+12y+8\)

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

\(\Leftrightarrow x-1=y+2\)

\(\Leftrightarrow y=x-3\)

Thế vào \(x^2=2y^2=x-4y\) ...

b.

\(\Leftrightarrow\left\{{}\begin{matrix}4x^2+y^4-4xy^3=1\\4x^2+2y^2-4xy=2\end{matrix}\right.\)

\(\Rightarrow y^4-2y^2-4xy^3+4xy=-1\)

\(\Leftrightarrow\left(y^2-1\right)^2-4xy\left(y^2-1\right)=0\)

\(\Leftrightarrow\left(y^2-1\right)\left(y^2-1-4xy\right)=0\)

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

Thế vào \(2x^2+y^2-2xy=1\) ...

Với \(x=\dfrac{y^2-1}{4y}\) ta được:

\(2\left(\dfrac{y^2-1}{4y}\right)^2+y^2-2\left(\dfrac{y^2-1}{4y}\right)y=1\)

\(\Leftrightarrow5y^4-6y^2+1=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=y^2-1\\\sqrt{y^2+3}+y^2+3-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y^2-1\\\left(\sqrt{y^2+3}-2\right)\left(\sqrt{y^2+3}+3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y^2-1=0\\y^2=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\pm1\end{matrix}\right.\)

\(HPT\Leftrightarrow\left\{{}\begin{matrix}2x^2+4xy-6y^2=18\left(1\right)\\18x^2+18xy+9y^2=18\left(2\right)\end{matrix}\right.\)

Lấy \(PT\left(2\right)\) trừ đi \(PT\left(1\right)\) ta có : \(16x^2+14xy+15y^2=0\)

\(\Leftrightarrow\left(4x\right)^2+2.4x.\dfrac{7}{4}y+\dfrac{49}{16}y^2+\dfrac{191}{16}y^2=0\Leftrightarrow\left(4x+\dfrac{7}{4}y\right)^2+\dfrac{191}{16}y^2=0\)

\(\Rightarrow\left(x;y\right)=\left(0;0\right)\left(loại\right)\)

Vậy hệ PT vô nghiệm