Đáp án:

\(\eqalign{ & a)\,\,m = \pm 2\sqrt {10} \cr & b)\,\,m = {9 \over 9};\,\,m = {7 \over 8} \cr} \)

Giải thích các bước giải:

\(\eqalign{ & a)\,\,d:\,\,y = - 2x + m,\,\,S = 10 \cr & d \cap Ox:\,\,Cho\,\,y = 0 \Rightarrow 2x = m \Leftrightarrow x = {m \over 2} \cr & \Rightarrow A\left( {{m \over 2};0} \right) \Rightarrow OA = \left| {{m \over 2}} \right| \cr & d \cap Oy:\,\,Cho\,\,x = 0 \Rightarrow y = m \cr & \Rightarrow B\left( {0;m} \right) \Rightarrow OB = \left| m \right| \cr & \Rightarrow {S_{OAB}} = {1 \over 2}OA.OB = {1 \over 2}\left| {{m \over 2}.m} \right| = 10 \cr & \Leftrightarrow {{{m^2}} \over 4} = 10 \Leftrightarrow {m^2} = 40 \Leftrightarrow m = \pm 2\sqrt {10} \cr & b)\,\,d:\,\,y = \left( {m - 1} \right)x + 2,\,\,S = 16. \cr & TH1:\,\,m = 1 \Rightarrow y = 2 \Rightarrow Khong\,\,cat\,\,hai\,\,truc\,\,toa\,\,do. \cr & TH2:\,\,d \cap Ox:\,\,y = 0 \Rightarrow x = - {2 \over {m - 1}} \cr & \Rightarrow A\left( { - {2 \over {m - 1}};0} \right) \Rightarrow OA = {2 \over {\left| {m - 1} \right|}} \cr & d \cap Oy:\,\,x = 0 \Rightarrow y = 2 \Rightarrow B\left( {0;2} \right) \Rightarrow OB = 2 \cr & \Rightarrow {S_{\Delta OAB}} = {1 \over 2}.{2 \over {\left| {m - 1} \right|}}.2 = 16 \cr & \Leftrightarrow {2 \over {\left| {m - 1} \right|}} = 16 \Leftrightarrow \left| {m - 1} \right| = {1 \over 8} \cr & \Leftrightarrow \left[ \matrix{ m - 1 = {1 \over 8} \hfill \cr m - 1 = - {1 \over 8} \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ m = {9 \over 9} \hfill \cr m = {7 \over 8} \hfill \cr} \right. \cr} \)