a.\(\Leftrightarrow2x^2+x-16x-8=0\)
\(\Leftrightarrow2x^2-15x-8=0\)
\(\Leftrightarrow\left(x-8\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{1}{2}\end{matrix}\right.\)
b.\(\Leftrightarrow\left(x+1\right)\left(7x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{2}{7}\end{matrix}\right.\)
c.\(ĐK:x\ne2;4\)
\(\Rightarrow\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\dfrac{\left(x-2\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2=-\left(x-2\right)\left(x-4\right)\)
\(\Leftrightarrow x^2-4x-3x+12+x^2-4x+4=-x^2+4x+2x-8\)
\(\Leftrightarrow3x^2-17x+24=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\)
d.\(ĐK:x\ne1\)
\(\Rightarrow\dfrac{1}{x-1}+\dfrac{2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{4}{x^2+x+1}\)
\(\Leftrightarrow\dfrac{\left(x^2+x+1\right)+2x^2-5}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{4\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow x^2+x+1+2x^2-5=4\left(x-1\right)\)
\(\Leftrightarrow3x^2+x-4=4x-4\)
\(\Leftrightarrow3x^2-3x=0\)
\(\Leftrightarrow3x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)
c: \(\Leftrightarrow\left(x-3\right)\left(x-4\right)+x^2-4x+4=-\left(x-2\right)\left(x-4\right)\)
\(\Leftrightarrow x^2-7x+12+x^2-4x+4=-\left(x^2-6x+8\right)\)
\(\Leftrightarrow2x^2-11x+16=-x^2+6x-8\)
\(\Leftrightarrow3x^2-17x+24=0\)
\(\text{Δ}=\left(-17\right)^2-4\cdot3\cdot24=1>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{17-1}{6}=\dfrac{16}{6}=\dfrac{8}{3}\\x_2=\dfrac{17+1}{6}=\dfrac{18}{6}=3\end{matrix}\right.\)
d: \(\Leftrightarrow x^2+x+1+2x^2-5=4x-4\)
\(\Leftrightarrow3x^2+x-4-4x+4=0\)
\(\Leftrightarrow3x^2-3x=0\)
=>3x(x-1)=0
=>x=0
