b: \(\Leftrightarrow x^2+x+1+2x^2-5=4x-4\)
\(\Leftrightarrow3x^2-3x=0\)
hay x=0
Bài 1:
\(a,ĐK:x\ne1;x\ne-3\\ PT\Leftrightarrow\left(3x-1\right)\left(x+3\right)-x^2-2x+3=\left(2x+5\right)\left(x-1\right)-4\\ \Leftrightarrow2x^2+6x=2x^2+3x-9\\ \Leftrightarrow3x=9\Leftrightarrow x=3\left(tm\right)\\ b,ĐK:x\ne1\\ PT\Leftrightarrow x^2+x+1+2x^2-5=4x-4\\ \Leftrightarrow3x^2-3x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\)
Bài 2:
\(a,ĐK:x\ne2;x\ne3;x\ne4;x\ne5;x\ne6\\ PT\Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-3}+...+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-6}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{x-6-x+2}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\\ \Leftrightarrow x^2-8x+12=-32\\ \Leftrightarrow x^2-8x+44=0\\ \Leftrightarrow x\in\varnothing\)
\(b,ĐK:x\ne-1;x\ne-4;x\ne-6;x\ne3\\ PT\Leftrightarrow\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}=\dfrac{4}{3}+\dfrac{9}{\left(x-3\right)\left(x+6\right)}\\ \Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}=\dfrac{4}{3}+\dfrac{1}{x-3}-\dfrac{1}{x+6}\\ \Leftrightarrow\dfrac{1}{x+1}+\dfrac{1}{x-3}=\dfrac{4}{3}\\ \Leftrightarrow\dfrac{2x-2}{\left(x+1\right)\left(x-3\right)}=\dfrac{4}{3}\\ \Leftrightarrow6x-6=4x^2-8x-12\\ \Leftrightarrow4x^2-14x-6=0\\ \Leftrightarrow x=\dfrac{7\pm\sqrt{73}}{4}\)
