ĐKXĐ: \(x\ne-1,-\dfrac{1}{2}\)
\(\dfrac{x^2-4x+7}{x+1}-2=\dfrac{x^2-3x+7}{2x+1}\)
\(\Leftrightarrow\dfrac{x^2-4x+7-2\left(x+1\right)}{x+1}=\dfrac{x^2-3x+7}{2x+1}\)
\(\Leftrightarrow\dfrac{x^2-4x+7-2x-2}{x+1}=\dfrac{x^2-3x+7}{2x+1}\)
\(\Leftrightarrow\dfrac{x^2-6x+5}{x+1}=\dfrac{x^2-3x+7}{2x+1}\)
\(\Leftrightarrow\left(x^2-6x+5\right)\left(2x+1\right)=\left(x+1\right)\left(x^2-3x+7\right)\)
\(\Leftrightarrow2x^3+x^2-12x^2-6x+10x+5=x^3-3x^2+7x+x^2-3x+7\)
\(\Leftrightarrow x^3-9x^2-2=0\)
\(\Leftrightarrow x\approx9,025\) (TM)
\(Đk:\left\{{}\begin{matrix}x\ne-1\\x\ne-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\dfrac{x^2-4x+7}{x+1}-1=1-\dfrac{x^2-3x+7}{2x+1}\\ \Leftrightarrow\dfrac{x^2-5x+6}{x+1}=\dfrac{-x^2+5x-6}{2x+1}\\ \Leftrightarrow\left(x^2-5x+6\right).\left(\dfrac{1}{x+1}+\dfrac{1}{2x+1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x+6=0\\\dfrac{1}{x+1}+\dfrac{1}{2x+1}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\\x=\dfrac{-2}{3}\end{matrix}\right.\)
ĐK:\(x\ne-1,x\ne\dfrac{-1}{2}\)
=>\(\dfrac{x^2-4x+7-2\left(x+1\right)}{x+1}=\dfrac{x^2-3x+7}{2x+1}\)
=>\(\dfrac{x^2-4x+7-2x-2}{x+1}=\dfrac{x^2-3x+7}{2x+1}\)
=>\(\dfrac{x^2-6x+5}{x+1}-\dfrac{x^2-3x+7}{2x+1}=0\)
=>\(\dfrac{\left(x^2-6x+5\right)\left(2x+1\right)-\left(x^2-3x+7\right)\left(x+1\right)}{\left(x+1\right)\left(2x+1\right)}=0\)
=>\(2x^3+x^2-12x^2-6x+10x+5-\left(x^3+x^2-3x^2-3x+7x+7\right)=0\)
=>\(2x^3-11x^2+4x+5-x^3+2x^2+4x+7=0\)
=>\(x^3-9x^2+8x+12=0\)
=>\(\left(x^3-2x^2\right)-\left(7x^2-14x\right)-\left(6x-12\right)=0\)
=>\(x^2\left(x-2\right)-7x\left(x-2\right)-6\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x^2-7x-6\right)=0\)
=>\(\left(x-2\right)\left(x-\dfrac{7+\sqrt{73}}{2}\right)\left(x-\dfrac{7-\sqrt{73}}{2}\right)=0\)
=>\(\left[{}\begin{matrix}x=2\\x=\dfrac{7+\sqrt{73}}{2}\\x=\dfrac{7-\sqrt{73}}{2}\end{matrix}\right.\)
msc: \(\left(2x+1\right)\left(x+1\right)=2x^2+3x+1\)
\(pt\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-4x+7\right)}{\left(2x+1\right)\left(x+1\right)}-\dfrac{2.\left(x+1\right)\left(2x+1\right)}{\left(2x+1\right)\left(x+1\right)}=\dfrac{\left(x+1\right)\left(x^2-3x+7\right)}{\left(2x+1\right)\left(x+1\right)}\)
\(\Leftrightarrow\dfrac{2x^3-7x^2+10x+7}{2x^2+3x+1}-\dfrac{4x^2+6x+2}{2x^2+3x+1}-\dfrac{x^3-2x^2+4x+7}{2x^2+3x+1}=0\)
\(\Leftrightarrow2x^3-7x^2+10x+7-4x^2-6x-2-x^3+2x^2-4x-7=0\)
\(\Leftrightarrow x^3-9x^2-2=0\)
đề có nhầm j ko tr , muốn tìm nghiệm pt cần vẽ đồ thị á
x là giao điểm
\(x\approx9,02455716\)
