ĐKXĐ : \(\hept{\begin{cases}x\ne3\\x\ne-1\end{cases}}\)
<=> \(\frac{16x+16}{\left(x-3\right)\left(x+1\right)}-\frac{15x-45}{\left(x-3\right)\left(x+1\right)}=\frac{4\left(x-3\right)\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}\)
<=> \(\frac{x+61}{\left(x-3\right)\left(x+1\right)}=\frac{4x^2-8x-12}{\left(x-3\right)\left(x+1\right)}\)
=> 4x2 - 8x - 12 - x - 61 = 0
<=> 4x2 - 9x - 73 = 0
Δ = b2 - 4ac = (-9)2 - 4.4.(-73) = 1249
Δ > 0, áp dụng công thức nghiệm thu được \(\hept{\begin{cases}x_1=\frac{9+\sqrt{1249}}{8}\\x_2=\frac{9-\sqrt{1279}}{8}\end{cases}\left(tm\right)}\)
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