Giải phương trình:
1. \(x^4-6x^2-12x-8=0\)
2. \(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)
3. \(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
4. \(2x^2.\sqrt{-4x^4+4x^2+3}=4x^4+1\)
5. \(x^2+4x+3=\sqrt{\dfrac{x}{8}+\dfrac{1}{2}}\)
6. \(\left\{{}\begin{matrix}4x^3+xy^2=3x-y\\4xy+y^2=2\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}\sqrt{x^2-3y}\left(2x+y+1\right)+2x+y-5=0\\5x^2+y^2+4xy-3y-5=0\end{matrix}\right.\)
8. \(\left\{{}\begin{matrix}\sqrt{2x^2+2}+\left(x^2+1\right)^2+2y-10=0\\\left(x^2+1\right)^2+x^2y\left(y-4\right)=0\end{matrix}\right.\)
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
2.
ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)
\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)
Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)
\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)
Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:
\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)
\(\Leftrightarrow10b+40=3\left(b+8\right)b\)
\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)
TH1: \(b=2\Leftrightarrow...\)
TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)
5.
ĐK: \(x\ge-4\)
Đặt \(\sqrt{\dfrac{x}{8}+\dfrac{1}{2}}=\dfrac{1}{2}y+1\Rightarrow x=2y^2+8y+4\)
Phương trình đã cho tương đương \(y=2x^2+8x+4\)
Ta có hệ \(\left\{{}\begin{matrix}x=2y^2+8y+4\left(1\right)\\y=2x^2+8x+4\end{matrix}\right.\)
Trừ vế theo vế ta được:
\(x-y=2y^2+8y-2x^2-8x\)
\(\Leftrightarrow\left(x-y\right)\left(2x+2y+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\\2x+2y+9=0\end{matrix}\right.\)
TH1: \(x=y\)
\(\left(1\right)\Leftrightarrow2x^2+7x+4=0\)
\(\Leftrightarrow x=\dfrac{-7\pm\sqrt{17}}{4}\left(tm\right)\)
Trường hợp còn lại tương tự
4.
ĐK: \(-\dfrac{\sqrt{6}}{2}\le x\le\dfrac{\sqrt{6}}{2}\)
Dễ thấy x=0 không phải nghiệm của phương trình, phương trình trở thành:
\(\sqrt{-4x^4+4x^2+3}=2x^2+\dfrac{1}{2x^2}\)
Ta có \(VT=\sqrt{-4x^4+4x^2+3}=\sqrt{-\left(2x^2-1\right)^2+4}\le2\)
\(VP=2x^2+\dfrac{1}{2x^2}\ge2\sqrt{2x^2.\dfrac{1}{2x^2}}=2\)
\(\Rightarrow\sqrt{-4x^4+4x^2+3}\le2x^2+\dfrac{1}{2x^2}\)
Đẳng thức xảy ra khi \(x=\dfrac{\pm\sqrt{2}}{2}\left(tm\right)\)
Vậy \(x=\dfrac{\pm\sqrt{2}}{2}\)
7.
Nếu \(x=0\Rightarrow y=5\Rightarrow\left(x;y\right)=\left(0;5\right)\) là nghiệm của hệ
Nếu \(x\ne0\)
\(\left\{{}\begin{matrix}\sqrt{2x^2+2}+\left(x^2+1\right)^2+2y-10=0\left(1\right)\\\left(x^2+1\right)^2+x^2y\left(y-4\right)=0\left(2\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow\dfrac{\left(x^2+1\right)^2}{x^2}+y\left(y-4\right)=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2+y\left(y-4\right)=0\)
Áp dụng BĐT Cosi:
\(\left(x+\dfrac{1}{x}\right)^2+y\left(y-4\right)\ge2^2+y^2-4y=\left(y-2\right)^2\ge0\)
Đẳng thức xảy ra khi \(\left(x;y\right)=\left(\pm1;2\right)\)
Thử lại vào phương trình \(\left(1\right)\) ta được \(\left(x;y\right)=\left(\pm1;2\right)\)
Vậy \(\left(x;y\right)\in\left\{\left(\pm1;2\right);\left(0;5\right)\right\}\)
6.
\(\left\{{}\begin{matrix}4x^3+xy^2=3x-y\\4xy+y^2=2\left(1\right)\end{matrix}\right.\)
Từ hệ phương trình suy ra: \(2\left(4x^3+xy^2\right)=\left(4xy+y^2\right)\left(3x-y\right)\)
\(\Leftrightarrow8x^3+2xy^2=12x^2y-xy^2-y^3\)
\(\Leftrightarrow8x^3-12x^2y+3xy^2+y^3=0\)
\(\Leftrightarrow\left(x-y\right)\left(8x^2-4xy-y^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y\\8x^2-4xy-y^2=0\end{matrix}\right.\)
TH1: \(x=y\)
\(\left(1\right)\Leftrightarrow5y^2=2\)
\(\Leftrightarrow y=\dfrac{\pm\sqrt{10}}{5}\)
\(\Rightarrow\left(x;y\right)\in\left\{\left(\dfrac{\sqrt{10}}{5};\dfrac{\sqrt{10}}{5}\right);\left(-\dfrac{\sqrt{10}}{5};-\dfrac{\sqrt{10}}{5}\right)\right\}\)
TH2: \(8x^2-4xy-y^2=0\left(2\right)\)
Cộng vế theo vế \(\left(1\right);\left(2\right)\) ta được:
\(8x^2=2\)
\(\Leftrightarrow x=\pm\dfrac{1}{2}\)
Nếu \(x=\dfrac{1}{2}\Rightarrow y=-1\pm\sqrt{3}\)
Nếu \(x=-\dfrac{1}{2}\Rightarrow y=1\pm\sqrt{3}\)
Thử lại rồi kết luận nghiệm
7.
ĐK: \(x^2-3y\ge0\)
\(\left\{{}\begin{matrix}\sqrt{x^2-3y}\left(2x+y+1\right)+2x+y-5=0\\5x^2+y^2+4xy-3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+y+1\right)\left(\sqrt{x^2-3y}+1\right)=6\\\left(2x+y\right)^2+\left(x^2-3y\right)=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+1\right)\left(b+1\right)=6\\a^2+b^2=5\end{matrix}\right.\) \(\left(a=2x+y;b=\sqrt{x^2-3y};b\ge0\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}2ab+2\left(a+b\right)=10\\\left(a+b\right)^2-2ab=5\end{matrix}\right.\)
Cộng vế theo vế hai phương trình ta được:
\(\left(a+b\right)^2+2\left(a+b\right)-15=0\)
\(\Leftrightarrow\left(a+b-3\right)\left(a+b+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=3\\a+b=-5\end{matrix}\right.\)
TH1: \(a+b=3\Rightarrow ab=2\)
\(\Leftrightarrow...\)
TH2: \(a+b=-5\Rightarrow ab=10\)
\(\Leftrightarrow...\)