Đáp án:

$a)m=\dfrac{57}{7}\\ b)\left[\begin{array}{l} m=1\\ m=-2\end{array} \right.\\ c)\left[\begin{array}{l} m=3\\ m=\dfrac{1}{3}\end{array} \right.\\ d)\left[\begin{array}{l} m=2\\ m=\dfrac{4}{3}\end{array} \right.$

Giải thích các bước giải:

$a)(m+1)x^2-2(m+1)x+m-3=0(m \ne -1)\\ \Delta'=(m+1)^2-(m+1)(m-3)=4m+4\\ \Delta' >0 \Leftrightarrow m>-1\\ Vi-et:x_1+x_2=2\\ x_1x_2=\dfrac{m-3}{m+1}\\ \circledast (4x_1+1)(4x_2+1)=18\\ \Leftrightarrow 16x_1x_2+4(x_1+x_2)-17=0\\ \Leftrightarrow \dfrac{16(m-3)}{m+1}-9=0\\ \Leftrightarrow \dfrac{16(m-3)-9(m+1)}{m+1}=0\\ \Leftrightarrow \dfrac{7m−57}{m+1}=0\\ \Leftrightarrow m=\dfrac{57}{7}(TM)\\ b)mx^2-(m-4)x+2m=0(m \ne 0)\\ \Delta=(m-4)^2-4m.2m=−7m^2−8m+16\\=−7m^2-2.\sqrt{7}m.\dfrac{4}{\sqrt{7}}-\dfrac{16}{7} +\dfrac{128}{7}\\$$=\sqrt{\dfrac{128}{7}}^2−\left(\sqrt{7}m+\dfrac{4}{\sqrt{7}}\right)^2 $$\\= \sqrt{\dfrac{8\sqrt{14}}{7}}^2−\left(\sqrt{7}m+\dfrac{4}{\sqrt{7}}\right)^2$ $\\=\left(\dfrac{8\sqrt{14}}{7}-\sqrt{7}m-\dfrac{4}{\sqrt{7}}\right)\left(\dfrac{8\sqrt{14}}{7}+\sqrt{7}m+\dfrac{4}{\sqrt{7}}\right)=\left(\dfrac{8\sqrt{14}-4\sqrt{7}}{7}-\sqrt{7}m\right)\left(\dfrac{8\sqrt{14}+4\sqrt{7}}{7}+\sqrt{7}m\right)=7\left(\dfrac{8\sqrt{2}-4}{\sqrt{7}}-m\right)\left(\dfrac{8\sqrt{2}+4}{\sqrt{7}}+m\right)\\\Delta>0\Leftrightarrow \dfrac{-4-8\sqrt{2}}{7}<m<\dfrac{-4+8\sqrt{2}}{7}\\ Vi-et:x_1+x_2=\dfrac{m-4}{m}\\ x_1x_2=2\\ \circledast 2(x_1^2+x_2^2)=5x_1x_2\\ \Leftrightarrow 2(x_1^2+2x_1x_2+x_2^2-2x_1x_2)-5x_1x_2=0\\ \Leftrightarrow 2(x_1+x_2)^2-4x_1x_2-5x_1x_2=0\\ \Leftrightarrow 2(x_1+x_2)^2-9x_1x_2=0\\ \Leftrightarrow \dfrac{2(m-4)^2}{m^2}-18=0\\ \Leftrightarrow \dfrac{2(m-4)^2-18m^2}{m^2}=0\\ \Leftrightarrow \dfrac{−16m^2−16m+32}{m^2}=0\\ \Leftrightarrow \left[\begin{array}{l} m=1\\ m=-2\end{array} \right.\\ c)(m-1)x^2-2mx+m+1=0(m \ne 1)\\ \Delta'=m^2-(m-1)(m+1)=1 >0 \ \forall \ m\\ Vi-et:x_1+x_2=\dfrac{2m}{m-1}\\ x_1x_2=\dfrac{m+1}{m-1}\\ \circledast 4(x_1^2+x_2^2)=5x_1^2x_2^2\\ \Leftrightarrow 4(x_1^2+2x_1x_2+x_2^2-2x_1x_2)-5x_1^2x_2^2=0\\ \Leftrightarrow 4(x_1+x_2)^2-8x_1x_2-5x_1^2x_2^2=0\\ \Leftrightarrow 4.\dfrac{4m^2}{(m-1)^2}-8.\dfrac{m+1}{m-1}-5.\dfrac{(m+1)^2}{(m-1)^2}=0\\ \Leftrightarrow \dfrac{16m^2-8(m+1)(m-1)-5(m+1)^2}{(m-1)^2}=0\\ \Leftrightarrow \dfrac{3m^2−10m+3}{(m-1)^2}=0\\ \Leftrightarrow \left[\begin{array}{l} m=3\\ m=\dfrac{1}{3}\end{array} \right.\\ d)x^2-(2m+1)x+m^2+2=0\\ \Delta=(2m+1)^2-4(m^2+2)=4m−7\\ \Delta >0 \Leftrightarrow m> \dfrac{7}{4}\\ Vi-et:x_1+x_2=2m+1\\ x_1x_2=m^2+2\\ \circledast 3x_1x_2-5(x_1+x_2)+7=0\\ \Leftrightarrow 3(m^2+2)-5(2m+1)+7=0\\ \Leftrightarrow 3m^2−10m+8=0\\ \Leftrightarrow \left[\begin{array}{l} m=2\\ m=\dfrac{4}{3}\end{array} \right.$