1. Giải phương trình:
\(\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+12x+35}=\frac{1}{9}\)
2.Tìm x:
\(\frac{14}{20-6x-2x^2}+\frac{x^2+4x}{x^2+5x}-\frac{x+3}{2-x}+3=0\)
3. Rút gọn:
\(\frac{2}{a}-\left(\frac{a^2}{a^2+ab}-\frac{a^2-b^2}{ab}-\frac{b^2}{ab+b^2}\right)\times\frac{a+b}{a^2+ab+b^2}\)
1. ĐKXĐ : \(x\ne-1;-3;-5;-7\)
\(\frac{1}{x^2+x+3x+3}+\frac{1}{x^2+3x+5x+15}+\frac{1}{x^2+7x+5x+35}=\frac{1}{9}\)=1/9
\(\frac{1}{x\left(x+1\right)+3\left(x+1\right)}+\frac{1}{x\left(x+3\right)+5\left(x+3\right)}+\frac{1}{x\left(x+7\right)+5\left(x+7\right)}=\frac{1}{9}\)
\(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}=\frac{1}{9}\)
nhân cả 2 vế với 2 ta được
\(\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{2}{\left(x+3\right)\left(x+5\right)}+\frac{2}{\left(x+5\right)\left(x+7\right)}=\frac{2}{9}\)
\(< =>\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+7}=\frac{2}{9}\)
\(< =>\frac{1}{x+1}-\frac{1}{x+7}=\frac{2}{9}\)
\(< =>\frac{\left(x+7\right)-\left(x+1\right)}{\left(x+1\right)\left(x+7\right)}=\frac{2}{9}\)
\(< =>\frac{6}{x^2+8x+7}=\frac{2}{9}\)
\(=>6.9=2x^2+16x+14\)
\(< =>2x^2+16x+14-54=0\)
\(< =>2\left(x^2+8x-20\right)=0\)
\(< =>x^2+8x-20=0\)
\(< =>x^2+10x-2x-20=0\)
\(< =>x\left(x+10\right)-2\left(x+10\right)=0\)
\(< =>\left(x-2\right)\left(x+10\right)=0\)
\(=>\hept{\begin{cases}x-2=0\\x+10=0\end{cases}< =>\hept{\begin{cases}x=2\\x=-10\end{cases}}}\)(thỏa mãn ĐKXĐ)
