Ta có: 8x.(x - 3) - 8.(x - 1)(x + 1) = 20

=> 8x² - 24x - 8.(x² - 1) = 20

=> 8x² - 24x - 8x² + 8 - 20 = 0

=> -24x - 12 = 0

=> -24x = 12

=> x = -1/2

1/2 là kết quả đúng nhất

\(8x\left(x-3\right)-8\left(x-1\right)\left(x+1\right)=20\)

Áp dụng hằng đẳng thức : \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)

\(pt\Leftrightarrow8x^2-24x-8\left(x^2-1\right)=20\)

\(\Leftrightarrow8x^2-24x-8x^2+8=20\)

\(\Leftrightarrow-24x+8=20\Leftrightarrow-24x=12\Leftrightarrow x=\frac{12}{-24}=-\frac{1}{2}\)

Vậy x=-1/2

khó hiểu quá

éo tìm được x đâu 

x=-0,5 nhé

\(\frac{37-x}{x+13}=\frac{5}{3}\)

\(\Rightarrow3\left(37-x\right)=5\left(x+13\right)\)

\(111-3x=5x+65\)

\(8x=46\)

\(x=\frac{23}{4}\)

Vậy..................

Đây nhé Yến ^^!: 

\(\frac{x-2}{x-6}\)nhận giá trị dương 

\(\Leftrightarrow\orbr{\begin{cases}..\\,,\end{cases}}\)

Ghép vào chỗ \(..\): \(\hept{\begin{cases}x-2>0\\x-6>0\end{cases}\Rightarrow\hept{\begin{cases}x>2\\x>6\end{cases}\Rightarrow}x>6}\)

Ghép vào chỗ \(,,\): \(\hept{\begin{cases}x-2< 0\\x-6< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 2\\x< 6\end{cases}\Rightarrow}x< 2}\)

Vậy x > 6 hoặc x < 2 .

Sửa đề:

 \(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)

ĐKXĐ: \(x\notin\left\{1;3;8;20\right\}\)

PT=>\(-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-8}-\dfrac{1}{x-8}+\dfrac{1}{x-20}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)

=>\(-\dfrac{1}{x-4}=-\dfrac{3}{4}\)

=>\(x-1=\dfrac{4}{3}\)

=>\(x=\dfrac{4}{3}+1=\dfrac{7}{3}\)(nhận)

ngu như con bò tót, ko biết 1+1=2.

ngu như con bò tót, ko biết 1+1=2.

ngu như con bò tót, ko biết 1+1=2.

M viết j đó

\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)

\(\frac{1}{x-1}-\frac{1}{x-3}+\frac{1}{x-3}-\frac{1}{x-8}+\frac{1}{x-8}-\frac{1}{x-20}-\frac{1}{x-20}=-\frac{3}{4}\)

\(\frac{1}{x-1}-\frac{2}{x-20}=-\frac{3}{4}\)

\(\frac{x-20-2x+2}{x^2-21x+20}=\frac{-3}{4}\)

\(-4x-72=-3x^2+63x-60\)

\(-3x^2+63x-60+4x+72=0\)

\(-3x^2+67x+12=0\)

Ko có x t/m

\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}\)

\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-1}{4}\)

\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-1}{4}\)

\(\Leftrightarrow\dfrac{1}{x-1}=\dfrac{-1}{4}\)

\(\Leftrightarrow\left(x-1\right)\left(-1\right)=4\)

\(\Leftrightarrow x-1=-4\)

\(\Leftrightarrow x=-3\left(tm\right)\)

Vậy ..............