\(\dfrac{2x-1}{3}=\dfrac{27}{\left(2x-1\right)}\left(ĐKXĐ:x\ne\dfrac{1}{2}\right)\)

\(\Rightarrow\left(2x-1\right)^2=81\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=9\\2x-1=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\left(n\right)\\x=-4\left(n\right)\end{matrix}\right.\)

Hình bạn tự vẽ

Ta có : \(\left\{{}\begin{matrix}NN'//EF\subset\left(CDFE\right)\\MM'//CD\subset\left(CDFE\right)\end{matrix}\right.\) ; \(\left\{{}\begin{matrix}NN'\subset\left(MNN'M'\right)\\MM'\subset\left(MNN'M'\right)\end{matrix}\right.\) ; \(NN'//MM'\)

\(\Rightarrow\left(MNN'M'\right)//\left(CDFE\right)\Rightarrow MN//\left(CDFE\right)\)

\(x^2-mx+1=0\)

phương trình có nghiệm \(\Leftrightarrow\Delta=m^2-4\ge0\Leftrightarrow\left[{}\begin{matrix}m\le-2\\m\ge2\end{matrix}\right.\)

Theo Viéte : \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=1\end{matrix}\right.\)

Ta có : \(\dfrac{1}{\sqrt{x_1^2+1}+x_1}=2\sqrt{2}-x_1-\sqrt{x_2^2+1}\)

\(\Leftrightarrow\dfrac{\sqrt{x_1^2+1}-x_1}{x_1^2+1-x_1^2}=2\sqrt{2}-x_1-\sqrt{x_2^2+1}\)

\(\Leftrightarrow\sqrt{x_1^2+1}-x_1=2\sqrt{2}-x_1-\sqrt{x_2^2+1}\)

\(\Leftrightarrow\sqrt{x_1^2+1}+\sqrt{x_2^2+1}=2\sqrt{2}\) 

Mà \(x^2-mx+1=0\left(gt\right)\Rightarrow mx-1=x^2\)

\(\Rightarrow\sqrt{mx_1-1+1}+\sqrt{mx_2-1+1}=2\sqrt{2}\)

\(\Leftrightarrow\left(\sqrt{mx_1}+\sqrt{mx_2}\right)^2=8\)

\(\Leftrightarrow mx_1+2\sqrt{m^2x_1x_2}+mx_2=8\)

\(\Leftrightarrow m\left(x_1+x_2\right)+2\sqrt{m^2}=8\)

\(m^2+2m=8\Leftrightarrow\left[{}\begin{matrix}m=2\left(n\right)\\m=-4\left(n\right)\end{matrix}\right.\)

Vậy m = 2 hoặc m = -4 thì thỏa ycbt

Xét \(\left(ABCD\right)\) có AB//CD

Theo Thales có \(\dfrac{IA}{IC}=\dfrac{AB}{CD}=\dfrac{a}{2a}=\dfrac{1}{2}\)

\(\Rightarrow IC=2IA\) \(\Rightarrow\dfrac{IC}{AC}=\dfrac{2}{3}\)

Vì IM//CD//AB nên theo Thales ta có :

\(\dfrac{CM}{BM}=\dfrac{IC}{CA}=\dfrac{2}{3}\)

Mà \(\dfrac{CN}{SC}=\dfrac{2}{3}\) do \(CN=2NS\left(gt\right)\)

\(\Rightarrow\dfrac{CN}{SC}=\dfrac{CM}{BM}=\dfrac{2}{3}\) (mặt phẳng SBC)

\(\Rightarrow NM//SB\left(1\right)\)

Lại có \(IM//CD//AB\left(2\right)\)

mà \(\left\{{}\begin{matrix}MN\subset\left(IMN\right)\\\left(AB\right)\subset\left(SAB\right)\end{matrix}\right.\) (3)

\(\left(1\right)\left(2\right)\left(3\right)\Rightarrow\left(IMN\right)//\left(SAB\right)\) => A đúng

hình bạn tự vẽ

\(A.\left(MNK\right)//\left(SAD\right)\)

Nhận thấy có AD cắt MN trên (ABCD)

Mà \(\left\{{}\begin{matrix}AD\subset\left(SAD\right)\\MN\subset\left(MNK\right)\end{matrix}\right.\)\(\Rightarrow\left(MNK\right)\cap\left(SAD\right)\)

=> A sai

\(B.\left(MNK\right)//\left(SCD\right)\)

Tương tự ý A do MN cắt CD trên ( ABCD) 

và \(\left\{{}\begin{matrix}DC\subset\left(SCD\right)\\MN\subset\left(MNK\right)\end{matrix}\right.\)\(\Rightarrow\left(MNK\right)\cap\left(SCD\right)\)

=> B sai

\(C.\left(MNK\right)//\left(SAC\right)\)

Ta có : \(\left\{{}\begin{matrix}KN//SC\left(đtb\right)\\MN//AC\left(đtb\right)\end{matrix}\right.\) và \(\left\{{}\begin{matrix}MN\subset\left(MNK\right)\\AC\subset\left(SAC\right)\end{matrix}\right.\)\(\Rightarrow\left(MNK\right)//\left(SAC\right)\)

=> C đúng

\(D.\left(MNK\right)//\left(SAB\right)\)

Có \(\left(MNK\right)\cap\left(SAB\right)=\left\{MK\right\}\) nên hai mặt phẳng này không thể song song

=> D sai