08:43 :vvvv

Vì \(\widehat{MIA}=90^0\left(\text{góc nt chắn nửa đường tròn}\right)\) nên \(MI\perp IA\)

Xét \(\Delta MBP\) có \(\left\{{}\begin{matrix}PK\perp MB\left(PK\perp MN\right)\\MI\perp PB\left(MI\perp IA\right)\\\left\{H\right\}=PK\cap MI\end{matrix}\right.\) nên H là trực tâm 

Do đó \(HB\perp PM\)

Mà \(AM\perp PM\Rightarrow HB\text{//}AM\)

Vì \(HB\text{//}OA\Rightarrow\dfrac{PB}{PA}=\dfrac{HB}{OA}\)

Ta có \(\sin MPB=\sin MPA=\dfrac{MA}{PA}=\dfrac{2OA}{PA}\)

\(\Rightarrow\dfrac{1}{2}BP\cdot\sin MPB=\dfrac{PB\cdot\dfrac{2OA}{PA}}{2}=\dfrac{PB\cdot2OA}{2PA}=\dfrac{PB}{PA}\cdot OA=\dfrac{HB}{OA}\cdot OA=HB\left(đpcm\right)\)

c: Ta có: AM//BC

AE⊥BC

Do đó:AM⊥AE

Suy ra: \(\widehat{AME}+\widehat{AEM}=90^0\)

hay \(\widehat{AME}+\widehat{BAD}=90^0\)

\(b,N=\left(2x-1\right)^2-4\ge-4\\ N_{min}=-4\Leftrightarrow x=\dfrac{1}{2}\\ c,P=\left(2x-5\right)^2+6\left(2x-5\right)+9-4\\ P=\left(2x-5+3\right)^2-4=\left(2x-2\right)^2-4\ge-4\\ P_{min}=-4\Leftrightarrow x=1\\ d,Q=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\\ Q=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1\\ Q_{min}=1\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

6a.

$M=x^2-x+1=(x^2-x+\frac{1}{4})+\frac{3}{4}$

$=(x-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}$

Vậy $M_{\min}=\frac{3}{4}$ khi $x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}$

a: CH=6cm

\(a,m=0\Leftrightarrow y=3x+2\)

Vì \(3>0\) nên hàm đồng biến

\(b,\text{Thay }x=-1;y=3\\ \Leftrightarrow-m-3+2=3\Leftrightarrow m=-4\\ c,\text{PT giao Ox: }y=0\Leftrightarrow x=-\dfrac{2}{m+3}\Leftrightarrow A\left(-\dfrac{2}{m+3};0\right)\Leftrightarrow OA=\dfrac{2}{\left|m+3\right|}\\ \text{PT giao Oy: }x=0\Leftrightarrow y=2\Leftrightarrow B\left(0;2\right)\Leftrightarrow OB=2\\ \text{Ta có }S_{OAB}=4\\ \Leftrightarrow\dfrac{1}{2}OA\cdot OB=4\Leftrightarrow\dfrac{2}{\left|m+3\right|}\cdot2=8\\ \Leftrightarrow\dfrac{4}{\left|m+3\right|}=8\\ \Leftrightarrow\left|m+3\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}m=-\dfrac{5}{2}\\m=-\dfrac{7}{2}\end{matrix}\right.\)