Câu hỏi của Vinh Lê Thành

Question

giải phương trình $\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}$Giúp mik nha ai nhanh nhất mik tik

Pham Van Hung · Accepted Answer

$\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}$$\Leftrightarrow\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}+\frac{1}{x\left(x+6\right)+7\left(x+6\right)}=\frac{1}{18}$(điều kiện: $x\ne\left\{-4;-5;-6;-7\right\}$ )$\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}$$\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}$$\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}$$\Rightarrow54=\left(x+4\right)\left(x+7\right)$$\Leftrightarrow x^2+11x-26=0$$\Leftrightarrow x\left(x+13\right)-2\left(x+13\right)=0\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x=-13\x=2\end{cases}}$(thỏa mãn ĐKXĐ)Vậy tập nghiệm của pt là: $S=\left\{-13;2\right\}$Câu trả lời: $\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}$$\Leftrightarrow\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}+\frac{1}{x\left(x+6\right)+7\left(x+6\right)}=\frac{1}{18}$(điều kiện: $x\ne\left\{-4;-5;-6;-7\right\}$ )$\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}$$\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}$$\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}$$\Rightarrow54=\left(x+4\right)\left(x+7\right)$$\Leftrightarrow x^2+11x-26=0$$\Leftrightarrow x\left(x+13\right)-2\left(x+13\right)=0\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x=-13\x=2\end{cases}}$(thỏa mãn ĐKXĐ)Vậy tập nghiệm của pt là: $S=\left\{-13;2\right\}$

Kaori Miyazono · Answer

Lâu lắm không làm nhể$\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}$$\Rightarrow\frac{1}{x^2+4x+5x+20}+\frac{1}{x^2+5x+6x+30}+\frac{1}{x^2+6x+7x+42}=\frac{1}{18}$$\Rightarrow\frac{1}{x.\left(x+4\right)+5.\left(x+4\right)}+\frac{1}{x.\left(x+5\right)+6.\left(x+5\right)}+\frac{1}{x.\left(x+6\right)+7.\left(x+6\right)}=\frac{1}{18}$$\Rightarrow\frac{1}{\left(x+4\right).\left(x+5\right)}+\frac{1}{\left(x+5\right).\left(x+6\right)}+\frac{1}{\left(x+6\right).\left(x+7\right)}=\frac{1}{18}$Dùng công thứ $\frac{1}{x.\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}$Khi đó $\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}$$\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}$$\Rightarrow\frac{x+7}{\left(x+4\right).\left(x+7\right)}-\frac{\left(x+4\right)}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}$$\Rightarrow\frac{3}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}\Rightarrow\left(x+4\right).\left(x+7\right)=54$$\Rightarrow\hept{\begin{cases}x+4=6\x+7=9\end{cases}}$hoặc $\hept{\begin{cases}x+4=-6\x+7=-9\end{cases}}$Suy ra $x=3$hoặc $x=-3$Câu trả lời: Lâu lắm không làm nhể$\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}$$\Rightarrow\frac{1}{x^2+4x+5x+20}+\frac{1}{x^2+5x+6x+30}+\frac{1}{x^2+6x+7x+42}=\frac{1}{18}$$\Rightarrow\frac{1}{x.\left(x+4\right)+5.\left(x+4\right)}+\frac{1}{x.\left(x+5\right)+6.\left(x+5\right)}+\frac{1}{x.\left(x+6\right)+7.\left(x+6\right)}=\frac{1}{18}$$\Rightarrow\frac{1}{\left(x+4\right).\left(x+5\right)}+\frac{1}{\left(x+5\right).\left(x+6\right)}+\frac{1}{\left(x+6\right).\left(x+7\right)}=\frac{1}{18}$Dùng công thứ $\frac{1}{x.\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}$Khi đó $\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}$$\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}$$\Rightarrow\frac{x+7}{\left(x+4\right).\left(x+7\right)}-\frac{\left(x+4\right)}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}$$\Rightarrow\frac{3}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}\Rightarrow\left(x+4\right).\left(x+7\right)=54$$\Rightarrow\hept{\begin{cases}x+4=6\x+7=9\end{cases}}$hoặc $\hept{\begin{cases}x+4=-6\x+7=-9\end{cases}}$Suy ra $x=3$hoặc $x=-3$

kudo shinichi · Answer

Thêm đk: $x\in Z$$\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}$$\frac{1}{\left(x^2+4x\right)+\left(4x+20\right)}+\frac{1}{\left(x^2+5x\right)+\left(6x+30\right)}+\frac{1}{\left(x^2+6x\right)+\left(7x+42\right)}=\frac{1}{18}$$\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}+\frac{1}{x\left(x+6\right)+7\left(x+6\right)}=\frac{1}{18}$$\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}$$\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}$$\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}$$\frac{\left(x+7\right)-\left(x+4\right)}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}$$\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}$$\Rightarrow\left(x+4\right)\left(x+7\right)=54$đến đây b tự lập bảng nhé~Câu trả lời: Thêm đk: $x\in Z$$\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}$$\frac{1}{\left(x^2+4x\right)+\left(4x+20\right)}+\frac{1}{\left(x^2+5x\right)+\left(6x+30\right)}+\frac{1}{\left(x^2+6x\right)+\left(7x+42\right)}=\frac{1}{18}$$\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}+\frac{1}{x\left(x+6\right)+7\left(x+6\right)}=\frac{1}{18}$$\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}$$\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}$$\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}$$\frac{\left(x+7\right)-\left(x+4\right)}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}$$\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}$$\Rightarrow\left(x+4\right)\left(x+7\right)=54$đến đây b tự lập bảng nhé~

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