a: ĐKXĐ: \(\begin{cases}4x^2+2x-4<>0\\ 4x^2-5x-4<>0\end{cases}\)
=>\(\begin{cases}x^2+\frac12x-1<>0\\ x^2-\frac54x-1<>0\end{cases}\)
=>\(\begin{cases}x^2+2\cdot x\cdot\frac14+\frac{1}{16}-\frac{17}{16}<>0\\ x^2-2\cdot x\cdot\frac58+\frac{25}{64}-\frac{89}{64}<>0\end{cases}\Rightarrow\begin{cases}\left(x+\frac14\right)^2<>\frac{17}{16}\\ \left(x-\frac58\right)^2<>\frac{89}{64}\end{cases}\)
=>\(\begin{cases}x+\frac14<>\pm\frac{\sqrt{17}}{4}\\ x-\frac58<>\pm\frac{\sqrt{89}}{8}\end{cases}\Rightarrow\begin{cases}x<>\pm\frac{\sqrt{17}}{4}-\frac14\\ x<>\pm\frac{\sqrt{89}}{8}+\frac58\end{cases}\)
Ta có: \(\frac{16x}{4x^2+2x-4}+\frac{3x}{4x^2-5x-4}=5\)
=>\(\frac{16x\left(4x^2-5x-4\right)+3x\left(4x^2+2x-4\right)}{\left(4x^2+2x-4\right)\left(4x^2-5x-4\right)}=5\)
=>\(5\left(4x^2+2x-4\right)\left(4x^2-5x-4\right)=64x^3-80x^2-64x+12x^3+6x^2-12x\)
=>\(5\left(4x^2+2x-4\right)\left(4x^2-5x-4\right)=76x^3-74x^2-76x\)
=>\(5\left\lbrack\left(4x^2-4\right)^2-3x\left(4x^2-4\right)-10x^2\right\rbrack=76x^3-74x^2-76x\)
=>\(5\left(4x^2-4\right)^2-15x\left(4x^2-4\right)-50x^2-76x^3+76x+74x^2=0\)
=>\(80\left(x^2-1\right)^2-60x\left(x^2-1\right)-76x\left(x^2-1\right)+24x^2=0\)
=>\(80\left(x^2-1\right)^2-136x\left(x^2-1\right)+24x^2=0\)
=>\(80\left(x^2-1\right)^2-120x\left(x^2-1\right)-16x\left(x^2-1\right)+24x^2=0\)
=>\(40\left(x^2-1\right)\left\lbrack2\left(x^2-1\right)-3x\right\rbrack-8x\left\lbrack2\left(x^2-1\right)-3x\right\rbrack=0\)
=>\(\left\lbrack2x^2-3x-2\right\rbrack\left(40x^2-40-8x\right)=0\)
=>\(\left(x-2\right)\left(2x+1\right)\cdot8\left(5x^2-x-5\right)=0\)
=>\(\left(x-2\right)\left(2x+1\right)\left(5x^2-x-5\right)=0\)
TH1: x-2=0
=>x=2(nhận)
TH2: 2x+1=0
=>2x=-1
=>\(x=-\frac12\) (loại)
TH3: \(5x^2-x-5=0\)
=>\(x^2-\frac15x-1=0\)
=>\(x^2-2\cdot x\cdot\frac{1}{10}+\frac{1}{100}-\frac{101}{100}=0\)
=>\(\left(x-\frac{1}{10}\right)^2=\frac{101}{100}\)
=>\(\left[\begin{array}{l}x-\frac{1}{10}=\frac{\sqrt{101}}{10}\\ x-\frac{1}{10}=-\frac{\sqrt{101}}{10}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\sqrt{101}+1}{10}\left(loại\right)\\ x=\frac{-\sqrt{101}+1}{10}\left(loại\right)\end{array}\right.\)
b: ĐKXĐ: x∈R
Ta có: \(\frac{6x}{x^2-x+2}+\frac{10x}{x^2+2x+2}=5\)
=>\(\frac{6x\left(x^2+2x+2\right)+10x\left(x^2-x+2\right)}{\left(x^2-x+2\right)\left(x^2+2x+2\right)}=5\)
=>\(5\left(x^2-x+2\right)\left(x^2+2x+2\right)=6x^3+12x^2+12x+10x^3-10x^2+20x\)
=>\(5\left\lbrack\left(x^2+2\right)^2+x\left(x^2+2\right)-2x^2\right\rbrack=16x^3+32x+2x^2\)
=>\(5\left(x^2+2\right)^2+5x\left(x^2+2\right)-10x^2-16x\left(x^2+2\right)-2x^2=0\)
=>\(5\left(x^2+2\right)^2-11x\left(x^2+2\right)-12x^2=0\)
=>\(5\left(x^2+2\right)^2-15x\left(x^2+2\right)+4x\left(x^2+2\right)-12x^2=0\)
=>\(5\left(x^2+2\right)\left(x^2+2-3x\right)+4x\left(x^2+2-3x\right)=0\)
=>\(\left(x^2-3x+2\right)\left(5x^2+4x+10\right)=0\)
TH1: \(5x^2+4x+10=0\)
\(\Delta=4^2-4\cdot5\cdot10=16-20\cdot10=16-200=-184<0\)
=>Phương trình vô nghiệm
TH2: \(x^2-3x+2=0\)
=>(x-1)(x-2)=0
=>\(\left[\begin{array}{l}x=1\left(nhận\right)\\ x=2\left(nhận\right)\end{array}\right.\)
