a) \(\frac{5-x}{4x^2-8x}\) + \(\frac{7}{8x}\) = \(\frac{x-1}{2x\left(x-2\right)}\) +\(\frac{1}{8x-16}\)                               ĐKXĐ : x #0, x#2, x#-2

<=> \(\frac{5-x}{4x\left(x-2\right)}\) + \(\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}\) + \(\frac{1}{8\left(x-2\right)}\)

<=> \(\frac{2\left(5-x\right)}{8x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{4\left(x-1\right)}{8x\left(x-2\right)}+\frac{x}{8x\left(x-2\right)}\)

=> 10 - 2x + 7x - 14 = 4x - 4 + x

<=>-2x + 7x - 4x + x  = -4 - 10 + 14

<=>x=-14

\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)

<=> \(\left[x\left(x+1\right)\right]\left[\left(x-1\right)\left(x+2\right)\right]-24=0\)

<=> \(\left(x^2+x\right)\left(x^2+2x-x-2\right)-24=0\)

<=> \(\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)

Đặt t = x² + x 

<=> t(t - 2) - 24 = 0

<=> t² - 2t - 24 = 0

<=> t² - 6t + 4t - 24 = 0

<=> (t + 4)(t - 6) = 0

<=> \(\orbr{\begin{cases}x^2+x+4=0\\x^2+x-6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x^2+x+\frac{1}{4}\right)+\frac{15}{4}=0\\x^2+3x-2x-6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^2+\frac{15}{4}=0\left(ktm\right)\\\left(x-2\right)\left(x+3\right)=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)

Vậy S = {2; -3}

(lưu ý: thay "ktm" thành vô lý và giải thích thêm)

\(\left(x+3\right)^4+\left(x+5\right)^4=2\)

<=> (x + 4 - 1)⁴ + (x + 4 + 1)⁴ - 2 = 0

Đặt y = x + 4

<=> (y - 1)⁴ + (y + 1)⁴ - 2 = 0

<=> y⁴ - 4y³ + 6y² - 4y + 1 + y⁴ + 4y³ + 6y² + 4y + 1 - 2 = 0

<=> 2y⁴ + 12y² = 0

<=> 2y²(y² + 6) = 0

<=> \(\orbr{\begin{cases}y^2=0\\y^2+6=0\left(ktm\right)\end{cases}}\)

<=> y = 0

<=> x + 4 = 0

<=> x = -4

Vậy S = {-4}

\(\frac{x^2+x+4}{2}+\frac{x^2+x+7}{3}=\frac{x^2+x+13}{5}+\frac{x^2+x+16}{6}\)

<=> \(\frac{x^2+x+4}{2}-3+\frac{x^2+x+7}{3}-3=\frac{x^2+x+13}{5}-3+\frac{x^2+x+16}{6}-3\)

<=> \(\frac{x^2+x+4-6}{2}+\frac{x^2+x+7-9}{3}=\frac{x^2+x+13-15}{5}+\frac{x^2+x+16-18}{6}\)

<=> \(\frac{x^2+x-2}{2}+\frac{x^2+x-2}{3}=\frac{x^2+x-2}{5}+\frac{x^2+x-2}{6}\)

<=> \(\left(x^2+2x-x-2\right)\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{5}-\frac{1}{6}\right)=0\)

<=> (x + 2)(x - 1) = 0 (do \(\frac{1}{2}+\frac{1}{3}-\frac{1}{5}-\frac{1}{6}\ne0\))

<=> \(\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)

Vậy S = {-2; 1}

câu cuối: + 3 vào sau các phân số của pt như trên