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ĐKXĐ: ...
\(\left\{{}\begin{matrix}x+\frac{1}{x}+y+\frac{1}{y}=5\\x^2+\frac{1}{x^2}+y^2+\frac{1}{y^2}=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{x}+y+\frac{1}{y}=5\\\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2=13\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+\frac{1}{x}=u\\y+\frac{1}{y}=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u+v=5\\u^2+v^2=13\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u+v=5\\\left(u+v\right)^2-2uv=13\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}u+v=5\\uv=6\end{matrix}\right.\)
Theo Viet đảo, u và v là nghiệm của: \(t^2-5t+6=0\Rightarrow\left[{}\begin{matrix}t=2\\t=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+\frac{1}{x}=2\\y+\frac{1}{y}=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+\frac{1}{x}=3\\y+\frac{1}{y}=2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow...\)
Đặt \(x+\frac{1}{x}=a\)và\(y+\frac{1}{y}=b\)
ta cm được\(a+b=\left(x+y\right)\left(1+\frac{1}{xy}\right)\)
\(a^2+b^2+4=\left(x^2+y^2\right)\left(1+\frac{1}{x^2y^2}\right)\)
vậy hệ pt trở thành\(\hept{\begin{cases}a+b=5\\a^2+b^2+4=9\end{cases}}\)
từ đó tìm đc a và b rồi x và y
\(\frac{1}{x}+\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+9\right)\left(x+10\right)}\)
\(=\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+9}-\frac{1}{x+10}\)
\(=\frac{2}{x}-\frac{1}{x+10}=\frac{2\left(x+10\right)}{x\left(x+10\right)}-\frac{x}{x\left(x+10\right)}=\frac{2x+20-x}{x\left(x+10\right)}=\frac{x+20}{x^2+10x}\)
ĐKXĐ: ...
\(\Leftrightarrow\frac{2\left(x^2+1\right)}{\left(1-x^2\right)^2}+\frac{1}{4x^2}=\frac{\left(3x^2+1\right)^2}{144}\)
Đặt \(\left\{{}\begin{matrix}1-x^2=a\\4x^2=b\end{matrix}\right.\)
\(\Rightarrow\frac{2a+b}{a^2}+\frac{1}{b}=\frac{\left(a+b\right)^2}{144}\)
\(\Leftrightarrow\frac{\left(a+b\right)^2}{a^2b}=\frac{\left(a+b\right)^2}{144}\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=0\left(vn\right)\\a^2b=144\end{matrix}\right.\)
\(\Leftrightarrow\left(1-x^2\right)^2.4x^2=144\)
\(\Leftrightarrow\left(2x-2x^3\right)^2=12^2\)
\(\Leftrightarrow...\)
Đặt \(\hept{\begin{cases}x-\frac{9}{2}=a\\x-\frac{11}{4}=b\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a^4+b^4=1\\a=b-\frac{7}{4}\end{cases}}\)
\(\Rightarrow\left(b-\frac{7}{4}\right)^4+b^4-1=0\)
Giờ chứng minh nó vô nghiệm thôi
\(\left(\frac{x}{x-1}\right)^2+\left(\frac{x}{x+1}\right)^2=\frac{10}{9}\Leftrightarrow\frac{x^2}{\left(x-1\right)^2}+\frac{x^2}{\left(x+1\right)^2}=\frac{10}{9}\)
\(\Leftrightarrow\frac{x^2\left(x+1\right)^2+x^2\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}=\frac{10}{9}\Leftrightarrow\frac{x^2\left[\left(x+1\right)^2-\left(x-1\right)^2\right]}{\left[\left(x-1\right)\left(x+1\right)\right]^2}=\frac{10}{9}\)
\(\Leftrightarrow\frac{x^2\left(x+1-x+1\right)\left(x+1+x-1\right)}{\left(x^2-1\right)^2}=\frac{10}{9}\Leftrightarrow\frac{x^2.2.2x}{x^4-2x^2+1}=\frac{10}{9}\)
\(\Leftrightarrow36x^3=10x^4-20x^2+10\Leftrightarrow18x^3=5x^4-10x^2+5\Leftrightarrow5x^4-18x^3-10x^2\)+5=0
đến đây tự giải tiếp
ĐK:\(x\ne1;x\ne-1\)
\(pt\Leftrightarrow\frac{x^2}{\left(x-1\right)^2}+\frac{x^2}{\left(x+1\right)^2}=\frac{10}{9}\)
\(\Leftrightarrow\frac{9x^2\left(x+1\right)^2+9x^2\left(x-1\right)^2-10\left(x-1\right)^2\left(x+1\right)^2}{9\left(x-1\right)^2\left(x+1\right)^2}=0\)
\(\Leftrightarrow9x^2\left(x+1\right)^2+9x^2\left(x-1\right)^2-10\left(x-1\right)^2\left(x+1\right)^2=0\)
\(\Leftrightarrow9x^4+18x^3+9x^2+9x^4-18x^3+9x^2-10x^4+20x^2-10=0\)
\(\Leftrightarrow8x^4+38x^2-10=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=\frac{1}{4}\\x^2=5\left(l\right)\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
ĐKXĐ: x ≠ \(\pm\) 1
Từ phương trình ban đầu suy ra:
\(x^2\left(x+1\right)^2+x^2\left(x-1\right)^2=\frac{10}{9}.\left(x^2-1\right)^2\)
⇒ \(x^4+2x^3+x^2+x^4-2x^3+x^2=\frac{10}{9}\left(x^4-2x^2+1\right)\)
⇒ \(18\left(x^4+x^2\right)=10\left(x^4-2x^2+1\right)\)
⇒ \(4x^4+19x^2-5=0\Leftrightarrow\left(x^2+5\right)\left(4x^2-1\right)=0\)
⇔ \(x^2=\frac{1}{4}\Leftrightarrow x=\pm\frac{1}{2}\)( thỏa mãn ĐKXĐ)
Vậy ...