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\(\frac{1}{x}+\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+9\right)\left(x+10\right)}\)
\(=\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+9}-\frac{1}{x+10}\)
\(=\frac{2}{x}-\frac{1}{x+10}=\frac{2\left(x+10\right)}{x\left(x+10\right)}-\frac{x}{x\left(x+10\right)}=\frac{2x+20-x}{x\left(x+10\right)}=\frac{x+20}{x^2+10x}\)
ĐKXĐ: 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 ...
ĐK x khác +- 1 :
\(pt\Leftrightarrow\left(\frac{x}{x-1}\right)^2+\left(\frac{x}{x+1}\right)^2-\frac{2x^2}{\left(x-1\right)\left(x+1\right)}+\frac{2x^2}{\left(x-1\right)\left(x+1\right)}=\frac{10}{9}\)
<=> \(\left(\frac{x}{x-1}+\frac{x}{x+1}\right)^2-\frac{2x^2}{\left(x-1\right)\left(x+1\right)}=\frac{10}{9}\)
<=> \(\left[\frac{2x^2}{\left(x-1\right)\left(x+1\right)}\right]^2-\frac{2x^2}{\left(x-1\right)\left(x+1\right)}=\frac{10}{9}\)
Đặt \(\frac{2x^2}{\left(x-1\right)\left(x+1\right)}=t\)
pt <=> \(t^2-t=\frac{10}{9}\Leftrightarrow9t^2-9t-10=0\)
Đến đây tự làm tiếp nha
\(\left(\frac{x}{x-1}\right)^2+\left(\frac{x}{x+1}\right)^2=\frac{10}{9}\Leftrightarrow\frac{x^2}{\left(x-1\right)^2}+\frac{x^2}{\left(x+1\right)^2}=\frac{10}{9}\)
\(\Leftrightarrow\frac{x^2\left(x+1\right)^2+x^2\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)^2}=\frac{10}{9}\Leftrightarrow\frac{x^2\left[\left(x+1\right)^2-\left(x-1\right)^2\right]}{\left[\left(x-1\right)\left(x+1\right)\right]^2}=\frac{10}{9}\)
\(\Leftrightarrow\frac{x^2\left(x+1-x+1\right)\left(x+1+x-1\right)}{\left(x^2-1\right)^2}=\frac{10}{9}\Leftrightarrow\frac{x^2.2.2x}{x^4-2x^2+1}=\frac{10}{9}\)
\(\Leftrightarrow36x^3=10x^4-20x^2+10\Leftrightarrow18x^3=5x^4-10x^2+5\Leftrightarrow5x^4-18x^3-10x^2\)+5=0
đến đây tự giải tiếp
ĐK:\(x\ne1;x\ne-1\)
\(pt\Leftrightarrow\frac{x^2}{\left(x-1\right)^2}+\frac{x^2}{\left(x+1\right)^2}=\frac{10}{9}\)
\(\Leftrightarrow\frac{9x^2\left(x+1\right)^2+9x^2\left(x-1\right)^2-10\left(x-1\right)^2\left(x+1\right)^2}{9\left(x-1\right)^2\left(x+1\right)^2}=0\)
\(\Leftrightarrow9x^2\left(x+1\right)^2+9x^2\left(x-1\right)^2-10\left(x-1\right)^2\left(x+1\right)^2=0\)
\(\Leftrightarrow9x^4+18x^3+9x^2+9x^4-18x^3+9x^2-10x^4+20x^2-10=0\)
\(\Leftrightarrow8x^4+38x^2-10=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=\frac{1}{4}\\x^2=5\left(l\right)\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
