Giải phương trình:
1, \(\left(x+3\right)\left(3x^4+8x^2+12x+21\right)=5\left(x^2+1\right)^3\)
2, \(3\left(x^2+2x-1\right)^2-2\left(x^2+3x-1\right)^2+5x^2=0\)
3, \(\dfrac{x^2+x+1}{x+1}+\dfrac{x^2+2x+2}{x+2}-\dfrac{x^2+3x+3}{x+3}-\dfrac{x^2+4x+4}{x+4}=0\)
4, \(\left(\dfrac{x+6}{x-6}\right)\left(\dfrac{x+4}{x-4}\right)^2+\left(\dfrac{x-6}{x+6}\right)\left(\dfrac{x+9}{x-9}\right)^2=2.\dfrac{x^2+36}{x^2-36}\)
a/ \(\left(x+3\right)\left(3\left(x^2+1\right)^2+2\left(x+3\right)^2\right)=5\left(x^2+1\right)^3\)
\(\Leftrightarrow3\left(x+3\right)\left(x^2+1\right)^2+2\left(x+3\right)^3-5\left(x^2+1\right)^3=0\)
\(\Leftrightarrow3\left(x+3\right)\left(x^2+1\right)^2-3\left(x^2+1\right)^3+2\left(x+3\right)^3-2\left(x^2+1\right)^3=0\)
\(\Leftrightarrow3\left(x^2+1\right)^2\left(-x^2+x+2\right)+2\left(-x^2+x+2\right)\left(\left(x+3\right)^2+\left(x+3\right)\left(x^2+1\right)+\left(x^2+1\right)^2\right)=0\)
\(\Leftrightarrow\left(-x^2+x+2\right)\left[3\left(x^2+1\right)^2+2\left(x+3+\dfrac{x^2+1}{2}\right)^2+\dfrac{3\left(x^2+1\right)^2}{4}\right]=0\)
\(\Leftrightarrow-x^2+x+2=0\) (phần ngoặc phía sau luôn dương)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b/ \(3\left(x^2+2x-1\right)^2-2\left(x^2+3x-1\right)^2+5\left(x^2+3x-1-\left(x^2+2x-1\right)\right)^2=0\)
Đặt \(\left\{{}\begin{matrix}a=x^2+2x-1\\b=x^2+3x-1\end{matrix}\right.\)
\(3a^2-2b^2+5\left(b-a\right)^2=0\Leftrightarrow8a^2+3b^2-10ab=0\)
\(\Leftrightarrow\left(4a-3b\right)\left(2a-b\right)=0\Leftrightarrow\left[{}\begin{matrix}4a=3b\\2a=b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4\left(x^2+2x-1\right)=3\left(x^2+3x-1\right)\\2\left(x^2+2x-1\right)=x^2+3x-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\x^2+x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\\x=\dfrac{-1+\sqrt{5}}{2}\\x=\dfrac{-1-\sqrt{5}}{2}\end{matrix}\right.\)
c/ ĐKXĐ: \(x\ne-1;-2;-3;-4\)
\(\dfrac{x^2+x+1}{x+1}-1+\dfrac{x^2+2x+2}{x+2}-1+1-\dfrac{x^2+3x+3}{x+3}+1-\dfrac{x^2+4x+4}{x+4}=0\)
\(\Leftrightarrow\dfrac{x^2}{x+1}+\dfrac{x^2+x}{x+2}-\dfrac{x^2+2x}{x+3}-\dfrac{x^2+3x}{x+4}=0\)
\(\Leftrightarrow x\left(\dfrac{x}{x+1}+\dfrac{x+1}{x+2}-\dfrac{x+2}{x+3}-\dfrac{x+3}{x+4}\right)=0\)
\(\Leftrightarrow x\left(\dfrac{-1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+3}+\dfrac{1}{x+4}\right)=0\)
\(\Leftrightarrow x\left(\dfrac{-2}{\left(x+1\right)\left(x+3\right)}-\dfrac{2}{\left(x+4\right)\left(x+2\right)}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x+1\right)\left(x+3\right)+\left(x+2\right)\left(x+4\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x^2+10x+11=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-5-\sqrt{3}}{2}\\x=\dfrac{-5+\sqrt{3}}{2}\end{matrix}\right.\)
ĐK: \(x\ne4;9;\pm6\Rightarrow\dfrac{x+6}{x-6}\ne0\)
Nhân 2 vế với \(\dfrac{x+6}{x-6}\):
\(\left(\dfrac{\left(x+6\right)\left(x+4\right)}{\left(x-6\right)\left(x-4\right)}\right)^2+\left(\dfrac{x+9}{x-9}\right)^2=\dfrac{2\left(x^2+36\right)}{\left(x-6\right)^2}=2+\dfrac{24x}{\left(x-6\right)^2}\)
\(\Leftrightarrow\left(\dfrac{x^2+10x+24}{x^2-10x+24}\right)^2-1+\left(\dfrac{x+9}{x-9}\right)^2-1-\dfrac{24x}{\left(x-6\right)^2}=0\)
\(\Leftrightarrow\dfrac{40x\left(x^2+24\right)}{\left(x-6\right)^2\left(x-4\right)^2}+\dfrac{36x}{\left(x-9\right)^2}-\dfrac{24x}{\left(x-6\right)^2}=0\)
\(\Leftrightarrow4x\left(\dfrac{10x^2+240}{\left(x-6\right)^2\left(x-4\right)^2}+\dfrac{9}{\left(x-9\right)^2}-\dfrac{6}{\left(x-6\right)^2}\right)=0\)
\(\Leftrightarrow4x\left(\dfrac{2}{\left(x-6\right)^2}\left(\dfrac{5x^2+120}{\left(x-4\right)^2}-3\right)+\dfrac{9}{\left(x-9\right)^2}\right)=0\)
\(\Leftrightarrow4x\left(\dfrac{2}{\left(x-6\right)^2}\left(\dfrac{2x^2+24x+72}{\left(x-4\right)^2}\right)+\dfrac{9}{\left(x-9\right)^2}\right)=0\)
\(\Leftrightarrow4x\left(\dfrac{4\left(x+6\right)^2}{\left(x-6\right)^2\left(x-4\right)^2}+\dfrac{9}{\left(x-9\right)^2}\right)=0\)
\(\Leftrightarrow x=0\) (do \(\dfrac{4\left(x+6\right)^2}{\left(x-6\right)^2\left(x-4\right)^2}+\dfrac{9}{\left(x-9\right)^2}>0\))
Vậy pt có nghiệm duy nhất \(x=0\)
