Toán giải và biện luận phương trình (3m^2+m-4)x=2m^2+m-3 10/09/2021 By Raelynn giải và biện luận phương trình (3m^2+m-4)x=2m^2+m-3
Giải thích các bước giải: $\begin{array}{l} + xet\,pt\,3{m^2} + m – 4 = 0 \Leftrightarrow \left[ \begin{array}{l} m = 1\\ m = – \frac{4}{3} \end{array} \right.\\ + 2{m^2} + m – 3 = 0 \Leftrightarrow \left[ \begin{array}{l} m = 1\\ m = – \frac{3}{2} \end{array} \right.\\ – TH1:\left\{ \begin{array}{l} 3{m^2} + m – 4 = 0\\ 2{m^2} + m – 3 = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m = 1\,hoac\,m = \frac{{ – 4}}{3}\\ m = 1\,hoac\,m = \frac{{ – 3}}{2} \end{array} \right. \Leftrightarrow m = 1\\ thi\,pt \Leftrightarrow 0.x = 0\,dung\,\forall x\\ vay\,m = 1\,thi\,pt\,co\,tap\,nghiem\,x = R\\ – TH2:\,\left\{ \begin{array}{l} 3{m^2} + m – 4 \ne 0\\ 2{m^2} + m – 3 = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m \ne 1va\,m \ne – \frac{4}{3}\\ m = 1\,hoac\,m = – \frac{3}{2} \end{array} \right. \Leftrightarrow m = – \frac{3}{2}\\ thi\,pt \Leftrightarrow \left( {3{m^2} + m – 4} \right)x = 0 \Leftrightarrow x = 0\\ vay\,m = – \frac{3}{2}thi\,x = 0\\ – TH3:\left\{ \begin{array}{l} 3{m^2} + m – 4 \ne 0\\ 2{m^2} + m – 3 \ne 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m \ne 1;\,m \ne – \frac{4}{3}\\ m \ne 1\,;m \ne – \frac{3}{2} \end{array} \right. \Leftrightarrow \left\{ {m \ne 1;m \ne – \frac{4}{3};m \ne – \frac{3}{2}} \right.\\ thi\,pt\,co\,nghiem\,x = \frac{{3{m^2} + m – 4}}{{2{m^2} + m – 3}} = \frac{{3m + 4}}{{2m + 3}} \end{array}$ Trả lời
Giải thích các bước giải:
$\begin{array}{l}
+ xet\,pt\,3{m^2} + m – 4 = 0 \Leftrightarrow \left[ \begin{array}{l}
m = 1\\
m = – \frac{4}{3}
\end{array} \right.\\
+ 2{m^2} + m – 3 = 0 \Leftrightarrow \left[ \begin{array}{l}
m = 1\\
m = – \frac{3}{2}
\end{array} \right.\\
– TH1:\left\{ \begin{array}{l}
3{m^2} + m – 4 = 0\\
2{m^2} + m – 3 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m = 1\,hoac\,m = \frac{{ – 4}}{3}\\
m = 1\,hoac\,m = \frac{{ – 3}}{2}
\end{array} \right. \Leftrightarrow m = 1\\
thi\,pt \Leftrightarrow 0.x = 0\,dung\,\forall x\\
vay\,m = 1\,thi\,pt\,co\,tap\,nghiem\,x = R\\
– TH2:\,\left\{ \begin{array}{l}
3{m^2} + m – 4 \ne 0\\
2{m^2} + m – 3 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m \ne 1va\,m \ne – \frac{4}{3}\\
m = 1\,hoac\,m = – \frac{3}{2}
\end{array} \right. \Leftrightarrow m = – \frac{3}{2}\\
thi\,pt \Leftrightarrow \left( {3{m^2} + m – 4} \right)x = 0 \Leftrightarrow x = 0\\
vay\,m = – \frac{3}{2}thi\,x = 0\\
– TH3:\left\{ \begin{array}{l}
3{m^2} + m – 4 \ne 0\\
2{m^2} + m – 3 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m \ne 1;\,m \ne – \frac{4}{3}\\
m \ne 1\,;m \ne – \frac{3}{2}
\end{array} \right. \Leftrightarrow \left\{ {m \ne 1;m \ne – \frac{4}{3};m \ne – \frac{3}{2}} \right.\\
thi\,pt\,co\,nghiem\,x = \frac{{3{m^2} + m – 4}}{{2{m^2} + m – 3}} = \frac{{3m + 4}}{{2m + 3}}
\end{array}$