(x² - 1)(x+6) = 0

x² -1 = 0 hoặc x + 6 = 0

⇔ x = +-1; x = - 6

vậy x \(\in\){ -6; -1; 1}

`(x^2 - 1).(x+6)=0`

`=>`  \(\left[{}\begin{matrix}x^2-1=0\\x+6=0\end{matrix}\right.\)

`=>` \(\left[{}\begin{matrix}x^2=0\\x=-6\end{matrix}\right.\)

`=>` \(\left[{}\begin{matrix}x=\pm1\\x=-6\end{matrix}\right.\)

Vậy \(S=\left\{\pm1;-6\right\}\)

Em nên gõ công thức trực quan để được hỗ trợ tốt nhất nhé

Nhân cả 2 vế với 1/2

\(\Leftrightarrow\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{11}{80}\)

\(\Leftrightarrow\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+\dfrac{8-7}{7.8}+\dfrac{9-8}{8.9}+...+\dfrac{\left(x+1\right)-x}{x\left(x+1\right)}=\dfrac{11}{80}\)

\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{11}{80}\)

\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{x+1}=\dfrac{11}{80}\)

\(\Leftrightarrow16\left(x+1\right)-80=11\left(x+1\right)\Rightarrow x=15\)

    \(\dfrac{12}{25}\) . \(\dfrac{23}{7}\) - \(\dfrac{12}{7}\) . \(\dfrac{13}{25}\)

=  \(\dfrac{12}{25}\). ( \(\dfrac{23}{7}\) - \(\dfrac{13}{7}\))

= \(\dfrac{12}{25}\) . \(\dfrac{10}{7}\)

= \(\dfrac{24}{35}\)

không thấy hình em ơi