\(a,3\left(x-1\right)\left(2x-1\right)=5\left(x+8\right)\left(x-1\right)\\ \Leftrightarrow\left(6x-3\right)\left(x-1\right)-\left(5x+40\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(6x-3-5x-40\right)=0\\\Leftrightarrow\left(x-1\right)\left(x-43\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=43\end{matrix}\right.\)
b, ĐKXĐ:\(x\ne\pm1\)
\(\dfrac{6}{x^2-1}+5=\dfrac{8x-1}{4x+4}-\dfrac{12x-1}{4-4x}\\ \Leftrightarrow\dfrac{6}{\left(x-1\right)\left(x+1\right)}+5-\dfrac{8x-1}{4\left(x+1\right)}-\dfrac{12x-1}{4\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{24}{4\left(x-1\right)\left(x+1\right)}+\dfrac{20\left(x-1\right)\left(x+1\right)}{4\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)\left(8x-1\right)}{4\left(x+1\right)\left(x-1\right)}-\dfrac{\left(12x-1\right)\left(x+1\right)}{4\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{24+20\left(x^2-1\right)-\left(8x^2-8x-x+1\right)-\left(12x^2-x+12x-1\right)}{4\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\dfrac{24+20x^2-20-8x^2+8x+x-1-12x^2+x-12x+1}{4\left(x-1\right)\left(x+1\right)}=0\)
\(\Rightarrow-2x+4=0\\ \Leftrightarrow x=2\left(tm\right)\)
a) \(3\left(x-1\right)\left(2x-1\right)=5\left(x+8\right)\left(x-1\right)\)
\(\Leftrightarrow3\left(x-1\right)\left(2x-1\right)-5\left(x+8\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[3\left(2x-1\right)-5\left(x+8\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x-3-5x-40\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-43\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-43=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=43\end{matrix}\right.\)
Vậy phương trình trên có 2 nghiệm là \(x_1=1;x_2=43\)
b) \(\dfrac{6}{x^2-1}+5=\dfrac{8x-1}{4x+4}-\dfrac{12x-1}{4-4x}\left(ĐKXĐ:x\ne\pm1\right)\)
\(\Leftrightarrow\dfrac{24}{4\left(x-1\right)\left(x+1\right)}+\dfrac{20\left(x-1\right)\left(x+1\right)}{4\left(x-1\right)\left(x+1\right)}=\dfrac{\left(8x-1\right)\left(x-1\right)}{4\left(x+1\right)\left(x-1\right)}+\dfrac{\left(12x-1\right)\left(x+1\right)}{4\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow24+20\left(x-1\right)\left(x+1\right)=\left(8x-1\right)\left(x-1\right)+\left(12x-1\right)\left(x+1\right)\)
\(\Leftrightarrow24+20\left(x^2-1\right)=8x^2-8x-x+1+12x^2+12x-x-1\)
\(\Leftrightarrow24+20x^2-20=20x^2+2x\)
\(\Leftrightarrow20x^2+4-20x^2-2x=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\) ( thỏa mãn)
Vậy phương trình đã cho có 1 nghiệm là x= 2
a, \(3\left(2x^2-3x+1\right)=5\left(x^2+7x-8\right)\Leftrightarrow6x^2-9x+3=5x^2+35x-40\)
\(\Leftrightarrow x^2-44x+43=0\Leftrightarrow\left(x-1\right)\left(x-43\right)=0\Leftrightarrow x=1;x=43\)
b, đk : x khác 1 ; -1
\(\Rightarrow24+20\left(x^2-1\right)=\left(8x-1\right)\left(x-1\right)+\left(12x-1\right)\left(x+1\right)\)
\(\Leftrightarrow20x^2+4=8x^2-9x+1+12x^2+11x-1\Leftrightarrow4=2x\Leftrightarrow x=2\)
