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Hãy luôn nhớ cảm ơn và vote 5*
nếu câu trả lời hữu ích nhé!
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\[\begin{array}{l} {d_1}:{\kern 1pt} {\kern 1pt} \left( {m - 1} \right)x + y = 5\\ {d_2}:{\kern 1pt} {\kern 1pt} 2x + my = 10.\\ a){\kern 1pt} {\kern 1pt} {\kern 1pt} Toa{\kern 1pt} {\kern 1pt} {\kern 1pt} do{\kern 1pt} {\kern 1pt} giao{\kern 1pt} {\kern 1pt} diem{\kern 1pt} {\kern 1pt} cua{\kern 1pt} {\kern 1pt} hai{\kern 1pt} {\kern 1pt} {\kern 1pt} duong{\kern 1pt} {\kern 1pt} thang{\kern 1pt} {\kern 1pt} {d_1},{\kern 1pt} {\kern 1pt} {d_2}{\kern 1pt} {\kern 1pt} {\kern 1pt} la{\kern 1pt} {\kern 1pt} {\kern 1pt} nghiem{\kern 1pt} {\kern 1pt} cua{\kern 1pt} {\kern 1pt} hpt:\\ \left\{ {\begin{array}{*{20}{l}} {\left( {m - 1} \right)x + y = 5}\\ {2x + my = 10} \end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}} {y = 5 - \left( {m - 1} \right)x{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left( 1 \right)}\\ {2x + my = 10{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left( 2 \right)} \end{array}} \right.\\ \Rightarrow \left( 2 \right) \Leftrightarrow 2x + m\left( {5 - \left( {m - 1} \right)x} \right) = 10\\ \Leftrightarrow 2x + 5m - m\left( {m - 1} \right)x = 10\\ \Leftrightarrow 2x + 5m - m\left( {m - 1} \right)x = 10\\ \Leftrightarrow \left( {{m^2} - m - 2} \right)x = 5m - 10{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \left( * \right)\\ \Rightarrow Hpt{\kern 1pt} {\kern 1pt} {\kern 1pt} co{\kern 1pt} {\kern 1pt} nghiem{\kern 1pt} {\kern 1pt} {\kern 1pt} duy{\kern 1pt} {\kern 1pt} nhat{\kern 1pt} {\kern 1pt} \Leftrightarrow \left( * \right){\kern 1pt} {\kern 1pt} {\kern 1pt} co{\kern 1pt} {\kern 1pt} nghiem{\kern 1pt} {\kern 1pt} duy{\kern 1pt} {\kern 1pt} nhat\\ \Leftrightarrow {m^2} - m - 2 \ne 0 \Leftrightarrow \left( {m + 1} \right)\left( {m - 2} \right) \ne 0 \Leftrightarrow \left\{ {\begin{array}{*{20}{l}} {m \ne - 1}\\ {m \ne 2} \end{array}} \right..\\ \Rightarrow \left( * \right) \Leftrightarrow x = \frac{{5m - 10}}{{{m^2} - m - 2}} = \frac{{5\left( {m - 2} \right)}}{{\left( {m + 1} \right)\left( {m - 2} \right)}} = \frac{5}{{m + 1}}\\ \Rightarrow y = 5 - \left( {m - 1} \right)x = 5 - \left( {m - 1} \right)\frac{5}{{m + 1}} = \frac{{5m + 5 - 5m + 5}}{{m + 1}} = \frac{{10}}{{m + 1}}.\\ \Rightarrow {d_1}{\kern 1pt} cat{\kern 1pt} {\kern 1pt} {d_2}{\kern 1pt} {\kern 1pt} tai{\kern 1pt} {\kern 1pt} {\kern 1pt} M\left( {\frac{5}{{m + 1}};\,\,\frac{{10}}{{m + 1}}} \right).\\ b){\kern 1pt} {\kern 1pt} {d_1}//{d_2} \Leftrightarrow \left( * \right){\kern 1pt} {\kern 1pt} {\kern 1pt} vo{\kern 1pt} {\kern 1pt} {\kern 1pt} nghiem \Leftrightarrow \left\{ {\begin{array}{*{20}{l}} {{m^2} - m - 2 \ne 0}\\ {5m - 10 = 0} \end{array}} \right. \\\Leftrightarrow \left\{ {\begin{array}{*{20}{l}} {m \ne - 1}\\ {m \ne 2}\\ {m = 2} \end{array}} \right. \Leftrightarrow m \in \emptyset .\\ c){\kern 1pt} {\kern 1pt} {d_1} \equiv {d_2} \Leftrightarrow \left( * \right){\kern 1pt} {\kern 1pt} {\kern 1pt} co{\kern 1pt} {\kern 1pt} {\kern 1pt} vo{\kern 1pt} {\kern 1pt} {\kern 1pt} so{\kern 1pt} {\kern 1pt} {\kern 1pt} nghiem \Leftrightarrow \left\{ {\begin{array}{*{20}{l}} {{m^2} - m - 2 = 0}\\ {5m - 10 = 0} \end{array}} \right. \Leftrightarrow \left\{ \begin{array}{l} \left[ \begin{array}{l} m = - 1\\ m = 2 \end{array} \right.\\ m = 2 \end{array} \right. \Leftrightarrow m = 2. \end{array}\]
Hãy giúp mọi người biết câu trả lời này thế nào?
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$$\eqalign{ & {d_1}:\,\,\left( {m - 1} \right)x + y = 5 \cr & {d_2}:\,\,2x + my = 1 \cr & a)\,\,{d_1}\,\,cat\,\,{d_2} \cr & m = 0 \Rightarrow {d_1}:\,\, - x + y = 5;\,\,{d_2}:\,\,2x = 1 \cr & \Rightarrow {d_1}\,\,cat\,\,{d_2} \cr & m \ne 0 \cr & {d_1}\,\,\,cat\,\,{d_2} \Leftrightarrow {{m - 1} \over 2} \ne {1 \over m} \cr & \Leftrightarrow m\left( {m - 1} \right) \ne 2 \Leftrightarrow {m^2} - m - 2 \ne 0 \cr & \Leftrightarrow m \ne - 1;\,\,m \ne 2 \cr & b)\,\,{d_1}//{d_2} \Leftrightarrow \left\{ \matrix{ m \ne 0 \hfill \cr {{m - 1} \over 2} = {1 \over m} \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ m \ne 0 \hfill \cr \left[ \matrix{ m = - 1 \hfill \cr m = 2 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ m = - 1 \hfill \cr m = 2 \hfill \cr} \right. \cr & c)\,\,{d_1} \equiv {d_2} \Leftrightarrow \left\{ \matrix{ m \ne 0 \hfill \cr {{m - 1} \over 2} = {1 \over m} = 5 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ m \ne 0 \hfill \cr \left[ \matrix{ m = - 1 \hfill \cr m = 2 \hfill \cr} \right. \hfill \cr m = {1 \over 5} \hfill \cr} \right. \Rightarrow Vo\,\,nghiem \cr & \Rightarrow Khong\,co\,\,m\,\,de\,\,{d_1} \equiv {d_2} \cr} $$
Hãy giúp mọi người biết câu trả lời này thế nào?
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còn tọa độ ??
Bảng tin
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Có 1 sự sai sai nhẹ ở câu . Cô đặt dấu (*)
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Rút x ra thì phải là m^2-m-2 chứ ??
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Ok e, cô đã sửa lại nhé!!!