Question

[mai dao]

giải phương trình

a) \(\frac{x+1}{x-1}\)-\(\frac{x-1}{x+1}\)=\(\frac{4}{x^2-1}\)

b) \(\frac{8x^2}{3\left(1-4x^2\right)}\)=\(\frac{2x}{6x-3}\)-\(\frac{1+8x}{4+8x}\)

c) \(\frac{3}{4\left(x-5\right)}\)+\(\frac{15}{50-2x^2}\)= - \(\frac{7}{6\left(x+5\right)}\)

d) \(\frac{13}{\left(x-3\right)\left(2x+7\right)}\)+\(\frac{1}{2x+7}\)=\(\frac{6}{x^2-9}\)

Answer 1

[An Võ (leo)]

a) Đề ( \(x\ne\pm1\))

>\(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}=\frac{4}{\left(x+1\right)\left(x-1\right)}\\ \Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=4\\ \Leftrightarrow\left(x+1-x+1\right)\left(x+1+x-1\right)=4\\ \Leftrightarrow2.2x=4\Leftrightarrow x=1\left(kothỏa\right)\)

Vậy \(S=\varnothing\)

Answer 2

[An Võ (leo)]

b) đề \(\left(x\ne-\frac{1}{2},\frac{1}{2}\right)\)

\(\frac{32x^2}{12\left(1-2x\right)\left(1+2x\right)}=\frac{-8x\left(1+2x\right)}{12\left(1-2x\right)\left(1+2x\right)}-\frac{3\left(1+8x\right)\left(1-2x\right)}{12\left(1-2x\right)\left(1+2x\right)}\\ \Leftrightarrow32x^2=-8x-16x^2-3-12x+48x^2\\ \Leftrightarrow20x+3=0\Leftrightarrow x=\frac{20}{3}\left(thỏadk\right)\)

Vậy \(S=\left\{\frac{20}{3}\right\}\)