ĐKXĐ: x<>0
Ta có: \(\frac{4}{4x^2-8x+7}+\frac{3}{4x^2-10x+7}=\frac{1}{x}\)
=>\(\frac{4\left(4x^2-10x+7\right)+3\left(4x^2-8x+7\right)}{\left(4x^2-8x+7\right)\left(4x^2-10x+7\right)}=\frac{1}{x}\)
=>\(x\left(16x^2-40x+28+12x^2-24x+21\right)=\left(4x^2-8x+7\right)\left(4x^2-10x+7\right)\)
=>\(\left(4x^2-8x+7\right)\left(4x^2-10x+7\right)=x\left(28x^2-64x+49\right)\)
=>\(\left(4x^2-8x+7\right)^2-2x\left(4x^2-8x+7\right)-x\left(28x^2-64x+49\right)=0\)
=>\(\left(4x^2-8x+7\right)^2+x\left(-8x^2+16x-14-28x^2+64x-49\right)=0\)
=>\(\left(4x^2-8x+7\right)^2+x\left(-36x^2+80x-63\right)=0\)
=>\(\left(4x^2-8x+7\right)^2+x\left(-36x^2+72x-63+8x\right)=0\)
=>\(\left(4x^2-8x+7\right)^2-9x\left(4x^2-8x+7\right)+8x^2=0\)
=>\(\left(4x^2-8x+7-x\right)\left(4x^2-8x+7-8x\right)=0\)
=>\(\left(4x^2-9x+7\right)\left(4x^2-16x+7\right)=0\)
TH1: \(4x^2-9x+7=0\)
\(\Delta=\left(-9\right)^2-4\cdot4\cdot7=81-16\cdot7=81-112=-31<0\)
=>Phương trình vô nghiệm
TH2: \(4x^2-16x+7=0\)
=>\(4x^2-14x-2x+7=0\)
=>2x(2x-7)-(2x-7)=0
=>(2x-7)(2x-1)=0
=>\(\left[\begin{array}{l}2x-7=0\\ 2x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac72\left(nhận\right)\\ x=\frac12\left(nhận\right)\end{array}\right.\)
