Xét ΔIDK vuông tại D và ΔIDC vuông tại D có 

ID chung

DK=DC

Do đó: ΔIDK=ΔIDC

Suy ra: \(\widehat{KID}=\widehat{DIC}\)

mà \(\widehat{KID}=\widehat{AIB}\)

nên \(\widehat{AIB}=\widehat{DIC}\)

MN là đường trung bình tam giác ABC \(\Rightarrow\overrightarrow{MN}=\dfrac{1}{2}\overrightarrow{BC}\)

NP là đường trung bình tam giác ABC \(\Rightarrow\overrightarrow{NP}=\dfrac{1}{2}\overrightarrow{AB}\Leftrightarrow\overrightarrow{AB}=2\overrightarrow{NP}\Leftrightarrow\overrightarrow{BA}=-2\overrightarrow{NP}\)

MP là đường trung bình \(\Rightarrow\overrightarrow{MP}=\dfrac{1}{2}\overrightarrow{AC}\)

Mà N là trung điểm AC \(\Rightarrow\overrightarrow{NA}=\dfrac{1}{2}\overrightarrow{CA}=-\dfrac{1}{2}\overrightarrow{AC}\)

\(\Rightarrow\overrightarrow{MP}=-\overrightarrow{NA}\)

3.

\(\left|\overrightarrow{AB}+\overrightarrow{BC}\right|=\left|\overrightarrow{BD}\right|=BD=\sqrt{AB^2+BC^2}=a\sqrt{2}\)

\(\left|\overrightarrow{u}\right|=\left|2\overrightarrow{AD}-\overrightarrow{BA}\right|\Rightarrow\left|\overrightarrow{u}\right|^2=\left(2\overrightarrow{AD}-\overrightarrow{BA}\right)^2=4AD^2+BA^2-4\overrightarrow{AD}.\overrightarrow{BA}\)

\(=4AD^2+BA^2=5a^2\Rightarrow\left|\overrightarrow{u}\right|=a\sqrt{5}\)

\(\left|\overrightarrow{CD}-2\overrightarrow{BO}\right|=\left|\overrightarrow{CD}-\overrightarrow{BD}\right|=\left|\overrightarrow{CD}+\overrightarrow{DB}\right|=\left|\overrightarrow{CB}\right|=a\)

\(\Leftrightarrow\dfrac{90}{x}-\dfrac{90}{x+5}=\dfrac{1}{5}\)

\(\Leftrightarrow x\left(x+5\right)=450x+2250-450x\)

\(\Leftrightarrow x^2+5x-2250=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=45\\x=-50\end{matrix}\right.\)

ĐKXĐ: \(x\ne0;x\ne-5\)

\(\dfrac{180}{x}-\dfrac{180}{x+5}=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{180.5.\left(x+5\right)}{5x\left(x+5\right)}-\dfrac{180.5.x}{5x\left(x+5\right)}=\dfrac{2.x.\left(x+5\right)}{5x\left(x+5\right)}\)

\(\Leftrightarrow900x+4500-900x=2x^2+10x\)

\(\Leftrightarrow2x^2+10x-4500=0\)

\(\Leftrightarrow x^2+5x-2250=0\)

\(\Leftrightarrow x^2+50x-45x-2250=0\)

\(\Leftrightarrow x\left(x+50\right)-45\left(x+50\right)=0\)

\(\Leftrightarrow\left(x+50\right)\left(x-45\right)=0\)

=>   x+50=0         hoặc                 x-45=0

         x=-50                                    x=45

Vậy....

\(y'=\left(m-1\right)x^2+2mx+3m-2\)

Với \(m=1\) ko thỏa mãn

\(y'>0\) ; \(\forall x\in R\) khi và chỉ khi:

\(\left\{{}\begin{matrix}m-1>0\\\Delta'=m^2-\left(3m-2\right)\left(m-1\right)< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m>1\\2m^2-5m+2>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m>1\\\left[{}\begin{matrix}m>2\\m< \dfrac{1}{2}\\\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow m>2\)