\(A=\frac{1-x^2}{x}.\left(\frac{x^2}{x+3}-1\right)+\frac{3x^2-14x+3}{x^2+3x}\)
\(A=\frac{\left(x^2-x-3\right)\left(-x^2+1\right)}{x\left(x+3\right)}+\frac{3x^2-14x+3}{x^2+3x}\)
\(A=\frac{\left(x^2-x-3\right)\left(1-x^3\right)}{\left(x+3\right)x}+\frac{3x^2-14x+3}{x\left(x+3\right)}\)
\(A=\frac{\left(x^2-x-3\right)\left(1-x^2\right)+3x^2-14x+3}{\left(x+3\right)x}\)
\(A=\frac{-x^4+x^3+7x^2-15x}{x\left(x+3\right)}\)
\(A=\frac{x\left(-x^3+x^2+7x-15\right)}{x\left(x+3\right)}\)
\(A=\frac{-x^3+x^2+7x-15}{x+3}\)
\(A=-\frac{\left(x+3\right)\left(x^2-4x+5\right)}{x+3}\)
\(A=-\left(x^2-4x+5\right)\)
\(A=-x^2+4x-5\)
Trình độ hơi thấp, có gì sai sót xin bỏ qua cho :)
Phân thức A xác định \(\Leftrightarrow\hept{\begin{cases}x\ne0\\x+3\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-3\end{cases}}}\)
\(A=\frac{1-x^2}{x}\left(\frac{x^2}{x+3}-1\right)+\frac{3x^2-14x+3}{x^2+3x}\)
\(A=\frac{\left(1-x\right)\left(1+x\right)}{x}\cdot\frac{x^2-x-3}{x+3}+\frac{3x^2-14x+3}{x\left(x+3\right)}\)
\(A=\frac{\left(1-x\right)\left(1+x\right)\left(x^2-x-3\right)}{x\left(x+3\right)}+\frac{3x^2-14x+3}{x\left(x+3\right)}\)
\(A=\frac{-x^4+x^3+4x^2-x-3+3x^2-14x+3}{x\left(x+3\right)}\)
\(A=\frac{-x^4+x^3+7x^2-15x}{x\left(x+3\right)}\)
\(A=\frac{x\left(-x^3+x^2+7x-15\right)}{x\left(x+3\right)}\)
\(A=\frac{-x^3+x^2+7x-15}{x+3}\)
Phân tích tử thức :
\(-x^3+x^2+7x-15\)
\(=-x^3-3x^2+4x^2+12x-5x-15\)
\(=-x^2\left(x+3\right)+4x\left(x+3\right)-5\left(x+3\right)\)
\(=\left(x+3\right)\left(-x^2+4x-5\right)\)
\(\Rightarrow A=\frac{\left(x+3\right)\left(-x^2+4x-5\right)}{x+3}=-x^2+4x-5\)
\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x+3\Leftrightarrow0\\x^2+3x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-3\\x\left(x+3\right)\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-3\end{cases}}\)
Rút gọn : \(A=\frac{1-x^2}{x}.\left(\frac{x^2}{x+3}-1\right)+\frac{3x^2-14x+3}{x^2+3x}\)
\(=\frac{1-x^2}{x}.\frac{x^2-x-3}{x+3}+\frac{3x^2-14x+3}{x\left(x+3\right)}\)
\(=\frac{\left(1-x^2\right)\left(x^2-x-3\right)}{x\left(x+3\right)}+\frac{3x^2-14x+3}{x\left(x+3\right)}\)
\(=\frac{x^2-x-3-x^4+x^3+3x^2+3x^2-14x+3}{x\left(x+3\right)}\)
\(=\frac{-x^4+x^3+7x^2-15x}{x\left(x+3\right)}=\frac{x\left(-x^3+x^2+7x-15\right)}{x\left(x+3\right)}=\frac{-x^3+x^2+7x-15}{x+3}\)
bạn ơi cho mk hỏi phân tích \(-x^3+x^2+7x\)\(-15\)ra thành nhân tử là ntn thế
@Quyết Tâm Chiến Thắng mình làm rồi mà ?
