Cân thử nào!
\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
\(\Rightarrow\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}=\dfrac{1}{18}\)
\(\Rightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Rightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Rightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Rightarrow\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Rightarrow\dfrac{3}{x^2+11x+28}=\dfrac{1}{18}\)
\(\Rightarrow x^2+11x+28=54\)
\(\Rightarrow x^2+11x-26=0\)
\(\Rightarrow x^2-2x+13x-26=0\)
\(\Rightarrow x\left(x-2\right)+13\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+13=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)
Vậy................
Chúc bạn học tốt!!!
Xét mẫu 1: x2+9x+20=x2+4x+5x+20=x(x+4)+5(x+4)=(x+4)(x+5)
Xét mẫu 2: x2+11x+30=x2+5x+6x+30=x(x+5)+6(x+5)=(x+5)(x+6)
Xét mẫu3:x2+13x+42=x2+6x+7x+42=x(x+6)+7(x+6)=(x+6)(x+7)
Vậy .....=\(\dfrac{1}{\text{(x+4)(x+5)}}+\dfrac{1}{\text{(x+5)(x+6)}}+\dfrac{1}{\text{(x+6)(x+7)}}=\dfrac{1}{18}\)
<=>\(\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
<=>\(\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)=.....
