I'll let you draw the figure.

a) AM is a median of the triangle ABC. Therefore, M is the midpoint of BC.

E is symmetric to A through M, so M is the midpoint of AE.

Consider the quadrilateral ABEC, it has M is the midpoint of both AE and BC. Thus, ABEC is a parallelogram.

b) Consider the triangle ADE, M is the midpoint of AE, H is the midpoint of AD. Therefore, HM is the average line of this triangle. This means \(HM//DE\) or \(DE//BC\), which means BCED is a trapezoid.

Triangle ABD has the height BH, which is also a median. Thus, ABD must be an isosceles triangle, which means the height BH is also a bisector, or \(\widehat{ABH}=\widehat{DBH}\) or \(\widehat{ABC}=\widehat{DBC}\)

On the other hand, ABEC is a parallelogram. So, \(AB//CE\) and this leads to \(\widehat{ABC}=\widehat{ECB}\) (2 staggered angles of 2 parallel lines)

From these, we have \(\widehat{DBC}=\widehat{ECB}\left(=\widehat{ABC}\right)\)

The trapezoid BCED (DE//BC) has \(\widehat{DBC}=\widehat{ECB}\). Therefore, BCED must be an isosceles trapezoid.

   5. (2 + x) (x+2) - \(\dfrac{1}{2}\) . (6-8x)² + 17

= 5 ( x² + 4x + 4) - \(\dfrac{1}{2}\). 4 ( 3 - 4x) ²+ 17

=  5x² + 20 x + 20 - 2 ( 9 - 24x + 16x²) + 17

= 5x² + 20x + 20 - 18 + 58x - 32x² + 17

= -27x² + 78 x  + 19

Ab mới đúng ghi lộn an

\(v_{tb}=\dfrac{S}{\dfrac{1}{v}+\dfrac{1}{v'}}=\dfrac{120}{\dfrac{1}{60}+\dfrac{1}{37}}\approx2746\left(\dfrac{km}{h}\right)\)

CTHH: Fe^a(CO₃)^II

Theo quy tắc hóa trị: a.1 = II.1

=> a = II

Fe hoá trị 2.

Vì gốc = CO₃ có hoá trị 2

A = x² + 5x - 5 

A = x² + 2. \(\dfrac{5}{2}\) x + ( \(\dfrac{5}{2}\))² - \(\dfrac{45}{4}\)

A = (x + \(\dfrac{5}{2}\))² - \(\dfrac{45}{4}\)

(x + \(\dfrac{5}{2}\))² ≥ 0 ⇔ A = (x + \(\dfrac{5}{2}\))² - \(\dfrac{45}{4}\) ≥ \(\dfrac{-45}{4}\)

A(min) = -45/4   = xảy ra ⇔ x + 5/2 = 0 ⇔ x = -5/2