\(a,ĐKXĐ:x\ne\pm\frac{1}{2}\)
Ta có: \(\frac{2}{2x+1}-\frac{3}{2x-1}=\frac{4}{4x^2-1}\)
\(\Leftrightarrow2\left(2x-1\right)-3\left(2x+1\right)=4\)
\(\Leftrightarrow4x-2-6x-3=4\)
\(\Leftrightarrow-2x=9\)
\(\Leftrightarrow x=-\frac{9}{2}\)(Tm ĐKXĐ)
Vậy pt có nghiệm duy nhất \(x=-\frac{9}{2}\)
\(b,ĐKXĐ:x\ne\pm1;-3\)
Ta có: \(\frac{2x}{x+1}+\frac{18}{x^2+2x-3}=\frac{2x-5}{x+3}\)
\(\Leftrightarrow\frac{2x}{x+1}+\frac{18}{\left(x-1\right)\left(x+3\right)}=\frac{2x-5}{x+3}\)
\(\Leftrightarrow2x\left(x-1\right)\left(x+3\right)+18\left(x+1\right)=\left(2x-5\right)\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow2x\left(x^2+2x-3\right)+18x+18=\left(2x-5\right)\left(x^2-1\right)\)
\(\Leftrightarrow2x^3+4x^2-6x+18x+18=2x^3-2x-5x^2+5\)
\(\Leftrightarrow9x^2+14x+13=0\)
\(\Leftrightarrow\left(9x^2+14x+\frac{49}{9}\right)+\frac{68}{9}=0\)
\(\Leftrightarrow\left(3x+\frac{7}{3}\right)^2+\frac{68}{9}=0\)
Pt vô nghiệm
\(c,ĐKXĐ:x\ne1\)
Ta có: \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
\(\Leftrightarrow x^2+x+1+2x^2-5=x-1\)
\(\Leftrightarrow3x^2=3\)
\(\Leftrightarrow x^2=1\)
\(\Leftrightarrow x=\pm1\)
Kết hợp vs ĐKXĐ được x = -1
Vậy pt có nghiệm duy nhất x = -1
làm lần lượt nha(bài nào k bt bỏ qua)
\(a,\frac{2}{2x+1}-\frac{3}{2x-1}=\frac{4}{4x^2-1}\)
\(\Rightarrow\frac{2\left(2x-1\right)-3\left(2x+1\right)}{4x^2-1}=\frac{4}{4x^2-1}\)
\(\Rightarrow-2x-5=4\)
\(\Rightarrow-2x=9\)
\(\Rightarrow x=\frac{9}{-2}\)
