Question

[Đặng Gia Ân]

Giải phương trình: x+3/x-4 + x-1/x-2 = 2/6x-8-x^2

Phân tích đa thức thành nhân tử:(a+2)(a+3)(a^2+a+6)+4a^2

Answer 1

[Vũ Minh Tuấn]

a)

\(\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{6x-8-x^2}\left(ĐKXĐ:x\ne4;x\ne2\right)\)

\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{-x^2+6x-8}\)

\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{-x^2+4x+2x-8}\)

\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{\left(-x^2+4x\right)+\left(2x-8\right)}\)

\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{-x.\left(x-4\right)+2.\left(x-4\right)}\)

\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{\left(x-4\right).\left(2-x\right)}\)

\(\Leftrightarrow\frac{x+3}{x-4}-\frac{x-1}{2-x}=\frac{2}{\left(x-4\right).\left(2-x\right)}\)

\(\Leftrightarrow\frac{\left(x+3\right).\left(2-x\right)}{\left(x-4\right).\left(2-x\right)}-\frac{\left(x-1\right).\left(x-4\right)}{\left(x-4\right).\left(2-x\right)}=\frac{2}{\left(x-4\right).\left(2-x\right)}\)

\(\Rightarrow\left(x+3\right).\left(2-x\right)-\left(x-1\right).\left(x-4\right)=2\)

\(\Leftrightarrow2x-x^2+6-3x-\left(x^2-4x-x+4\right)=2\)

\(\Leftrightarrow2x-x^2+6-3x-x^2+4x+x-4=2\)

\(\Leftrightarrow4x-2x^2+2=2\)

\(\Leftrightarrow4x-2x^2+2-2=0\)

\(\Leftrightarrow4x-2x^2=0\)

\(\Leftrightarrow2x.\left(2-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0:2\\x=2-0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=2\left(KTM\right)\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{0\right\}.\)

Chúc bạn học tốt!