(4x² +  8xy    - 3xy² )(-\(\dfrac{3}{4}\)x²y)

= -3x²y  - 6x³y² +\(\dfrac{9}{4}\)x³y³

4x³(2x² - x + 5) 5x

= 20x⁴(2x² - x + 5)

= 40x⁶ - 20x⁵ +100x⁴

\(\left(x-1\right)^2-9\left(x+1\right)^2=0\)

\(\Leftrightarrow x^2-2x+1-9\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow x^2-2x+1-9x^2-2x-9=0\)

\(\Leftrightarrow-8x^2-4x-8=0\)

\(\Leftrightarrow-4\left(2x^2+x+2\right)=0\)

\(\Leftrightarrow2x^2+x+2=0\)

Xét \(\Delta=1^2-4\cdot2\cdot2=-15< 0\)

⇒ Phương trình vô nghiệm

Vậy \(S=\varnothing\).

(x-1)² - 9 (x-1)² = 0

(x-1)².(1-9) =0

x-1 = 0

x = 1 

Ta có: \(\left(x-1\right)^2-9\left(x+1\right)^2=0\)

\(\Leftrightarrow\left[\left(x-1\right)-3\left(x+1\right)\right]\left[x-1+3\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(-2x-4\right)\left(4x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\4x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-\dfrac{1}{2}\end{matrix}\right.\)

...

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

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