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bạn tham khảo thêm cách này nha Shonogeki No Soma

ĐK: \(\hept{\begin{cases}x\ne0\\x\ne1\\x\ne-1\end{cases}}\)

Đặt  \(a=\left(x-1\right)^3;b=x^3;c=\left(x+1\right)^3\)

pt đã cho đc viết lại thành

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}a=-b\\b=-c\\c=-a\end{cases}}\)  (kí hiệu [..] mới đúng nha)

- TH1: a = -b hay  \(\left(x-1\right)^3=-x^3\)  \(\Leftrightarrow2x^3-3x^2+3x-1=0\)  \(\Leftrightarrow x=\frac{1}{2}\)  (Nhận)

- TH2: b = -c hay  \(\left(x+1\right)^3=-x^3\)  \(\Leftrightarrow2x^3+3x^2+3x+1=0\)  \(\Leftrightarrow x=-\frac{1}{2}\)  (Nhận)

- TH3: c = -a hay  \(\left(x+1\right)^3=-\left(x-1\right)^3\)  \(\Leftrightarrow x=0\)  (Loại)

KL:  \(S=\left\{\frac{1}{2};-\frac{1}{2}\right\}\)

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

\(\Leftrightarrow4x^8+15x^6+12x^4+8x^2-6=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)\left(x^2+3\right)\left(x^2-x+1\right)\left(x^2+x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{1}{2}\end{cases}}\)

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

\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+3}=\frac{3}{10}\)

\(\Leftrightarrow\frac{\left(x+3\right)-x}{x\left(x+3\right)}=\frac{3}{10}\)

\(\Leftrightarrow\frac{3}{x\left(x+3\right)}=\frac{3}{10}\)

\(\Rightarrow x\left(x+3\right)=10=2.\left(2+3\right)\)

\(\Rightarrow x=2\)

pt <=> \(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}=\frac{3}{10}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+3}=\frac{3}{10}\)

\(\Leftrightarrow\frac{3}{x\left(x+3\right)}=\frac{3}{10}\)

\(\Leftrightarrow x^2+3x-10=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)

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

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

\(\Leftrightarrow\frac{x}{\left(x^2-1\right)^3}+\frac{1}{3x^3\left(x^2+2\right)}=0\)

\(\Leftrightarrow4x^6+3x^4+3x^2-1=0\)

Đặt \(x^2=a\)

\(\Rightarrow4a^3+3a^2+3a-1=0\)

\(\Leftrightarrow\left(4a-1\right)\left(a^2+a+1\right)=0\)

\(\Leftrightarrow4a=1\)

\(\Rightarrow4x^2=1\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)

Bài lớp mấy mà khó vậy!Mình ko hiểu!