Môn Toán Lớp 7: Tìm x để 1.(x-2)^2 . (x+1) . (x-4) < 0 2. x^2.(x-3) / x-9 < 0

Giải đáp:

1)\({{\rm{\;}} – 1 < x < 4;x \ne 2}\)

\(\begin{array}{l}
2)3 < x < 9
\end{array}\)

Lời giải và giải thích chi tiết:

\(\begin{array}{*{20}{l}}
{1){{\left( {x – 2} \right)}^2}\left( {x + 1} \right)\left( {x – 4} \right) < 0}\\
{ \to \left( {x + 1} \right)\left( {x – 4} \right) < 0\left( {do:{{\left( {x – 2} \right)}^2} > 0\forall x \ne 2} \right)}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{x + 1 > 0}\\
{x – 4 < 0}
\end{array}} \right.}\\
{\left\{ {\begin{array}{*{20}{l}}
{x + 1 < 0}\\
{x – 4 > 0}
\end{array}} \right.}
\end{array}} \right.}\\
{{\rm{\;}} \to \left[ {\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{x >  – 1}\\
{x < 4}
\end{array}} \right.}\\
{\left\{ {\begin{array}{*{20}{l}}
{x <  – 1}\\
{x > 4}
\end{array}} \right.\left( l \right)}
\end{array}} \right.}\\
{ \to {\rm{ \;}} – 1 < x < 4;x \ne 2}\\
{2)DK:x \ne 9}\\
{\dfrac{{{x^2}\left( {x – 3} \right)}}{{x – 9}} < 0}\\
{ \to \dfrac{{x – 3}}{{x – 9}} < 0\left( {do:{x^2} \ge 0\forall x} \right)}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{x – 3 > 0}\\
{x – 9 < 0}
\end{array}} \right.}\\
{\left\{ {\begin{array}{*{20}{l}}
{x – 3 < 0}\\
{x – 9 > 0}
\end{array}} \right.}
\end{array}} \right.}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{\left\{ {\begin{array}{*{20}{l}}
{x > 3}\\
{x < 9}
\end{array}} \right.}\\
{\left\{ {\begin{array}{*{20}{l}}
{x < 3}\\
{x > 9}
\end{array}} \right.\left( l \right)}
\end{array}} \right.}\\
{ \to 3 < x < 9}
\end{array}\)