Câu hỏi của Nguyễn Lê Việt ANh

Question

Giải phương trình?:

a) A=$\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}$

b) B=$\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}$

Mysterious Person · Accepted Answer

a) điều kiện xác định : $x\ne0$

ta có :  $A=\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}$

$\Leftrightarrow A=\dfrac{\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2-x+1\right)}=\dfrac{3}{x\left(x^4+x^2+1\right)}$

$\Leftrightarrow A=\dfrac{x^3-x^2+x+x^2-x+1-\left(x^3+x^2+x-x^2-x-1\right)}{x^4-x^3+x^2+x^3-x^2+x+x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}$

$\Leftrightarrow A=\dfrac{x^3-x^2+x+x^2-x+1-x^3-x^2-x+x^2+x+1}{x^4+x^2+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}$

$\Leftrightarrow A=\dfrac{2}{x^4+x^2+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\Leftrightarrow\left(x^4+x^2+1\right)A=2=\dfrac{3}{x}$

$\Leftrightarrow2x=3\Leftrightarrow x=\dfrac{3}{2}\left(tmđk\right)$ vậy $x=\dfrac{3}{2}$Câu trả lời: a) điều kiện xác định : $x\ne0$

ta có :  $A=\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}$

$\Leftrightarrow A=\dfrac{\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x^2-x+1\right)}=\dfrac{3}{x\left(x^4+x^2+1\right)}$

$\Leftrightarrow A=\dfrac{x^3-x^2+x+x^2-x+1-\left(x^3+x^2+x-x^2-x-1\right)}{x^4-x^3+x^2+x^3-x^2+x+x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}$

$\Leftrightarrow A=\dfrac{x^3-x^2+x+x^2-x+1-x^3-x^2-x+x^2+x+1}{x^4+x^2+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}$

$\Leftrightarrow A=\dfrac{2}{x^4+x^2+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\Leftrightarrow\left(x^4+x^2+1\right)A=2=\dfrac{3}{x}$

$\Leftrightarrow2x=3\Leftrightarrow x=\dfrac{3}{2}\left(tmđk\right)$ vậy $x=\dfrac{3}{2}$

Mysterious Person · Answer

b) điều kiện : $x\notin\left\{-4;-5;-6;-7\right\}$

$B=\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}$

$\Leftrightarrow B=\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}$

$\Leftrightarrow B=\dfrac{\left(x+6\right)\left(x+7\right)+\left(x+4\right)\left(x+7\right)+\left(x+4\right)\left(x+5\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}$

$\Leftrightarrow B=\dfrac{x^2+13x+42+x^2+11x+28+x^2+9x+20}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}$

$\Leftrightarrow B=\dfrac{3x^2+33x+90}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}$

$\Leftrightarrow B=\dfrac{3\left(x+5\right)\left(x+6\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}$

$\Leftrightarrow B=\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}$ $\Leftrightarrow54=\left(x+4\right)\left(x+7\right)$

$\Leftrightarrow54=x^2+11x+28\Leftrightarrow x^2+11x+28-54=0$

$\Leftrightarrow x^2+11x-26=0\Leftrightarrow\left(x-2\right)\left(x+13\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x-2=0\x+13=0\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=2\x=-13\end{matrix}\right.\left(tmđk\right)$

vậy $x=2;x=-13$Câu trả lời: b) điều kiện : $x\notin\left\{-4;-5;-6;-7\right\}$