(Bạn tự vẽ hình nhé)

a) Ta có: DE// AC (gt) hay DE//AF (do F\(\in\) AC)

               EF//AB (gt) hay EF // AD (do D \(\in\) AB)

\(\Rightarrow\) Tg ADEF là hbh

b) Vì AM là trung tuyến tam giác ABC \(\Rightarrow\) M là trung điểm BC

Xét tam giác ABC có:
      MG//AC (gt)

       M là TĐ BC 

\(\Rightarrow\) MG là đường trung bình tam giác ABC

\(\Rightarrow\) G là trung điểm AB

\(\Rightarrow\) GA=GB 

- Học tốt-

mình cần gấp giúp mình với

Lời giải:

a. $(x-2)^3+(x+2)^3-6x(x+2)(x-2)$

$=x^3-6x^2+12x-8+(x^3+6x^2+12x+8)-6x(x^2-4)$

$=2x^3+24x-6x^3+24x=-4x^3+48x$

b.

$(2x-y)^3+(2x+y)^3$

$=8x^3-12x^2y+6xy^2-y^3+8x^3+12x^2y+6xy^2+y^3$

$=16x^3+12xy^2$

c.

$(x-2)(x+2)-(x^2+2x+4)(x-2)$

$=(x^2-4)-(x^3-2^3)=x^2-4-x^3+8=x^2-x^3+4$

Lời giải:

a. $99^3+1+3(99^2+99)=99^3+3.99^2.1+3.99.1^2+1^3=(99+1)^3=100^3=1000000$

b. $11^3-1-3(11^2-11)=11^3-3.11^2.1+3.11.1^2-1^3=(11-1)^3=10^3=1000$

\(\dfrac{x^3+8}{x^2+2x+1}.\dfrac{x^2+3x+2}{1-x^2}\left(x\ne\pm1\right)\\ =\dfrac{x^3+2^3}{\left(x+1\right)^2}.\dfrac{\left(x^2+x\right)+\left(2x+2\right)}{1^2-x^2}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{x\left(x+1\right)+2\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{\left(x+2\right)\left(x+1\right)}{\left(1-x\right)\left(x+1\right)}\\ =\dfrac{\left(x+2\right)^2\left(x^2-2x+4\right)}{\left(1-x\right)\left(x+1\right)^2}\)

x(x + 2) + x(x - 5) - 5(x + 2)

= [x(x + 2) - 5(x + 2)] + x(x - 5)

= (x - 5)(x + 2) + x(x - 5)

= (x - 5)(x + 2 + x)

= (x - 5)(2x + 2)

= 2(x - 5)(x + 1)

a,A =  \(\dfrac{3x^2+6xy}{6x^2}\)  (đk  \(x\) ≠ 0)

  A = \(\dfrac{3x.\left(x+2y\right)}{6x^2}\)

 A = \(\dfrac{x+2y}{2x}\) 

b,B =  \(\dfrac{2x^2-x^3}{x^2-4}\) (đk \(x\)²  - 4 ≠ 0 ⇒ \(x\) ≠ \(\pm\) 2)

   B =  \(\dfrac{x^2\left(2-x\right)}{\left(x+2\right)\left(x-2\right)}\)

   B = \(\dfrac{-x^2.\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}\)

   B =  \(\dfrac{-x^2}{x+2}\)

a) \(\left(x^2-4x\right)+\left(12x-48\right)\)

\(=x\left(x-4\right)+12\left(x-4\right)\)

\(=\left(x-4\right)\left(x+12\right)\)

b) \(x(x+y)-3x-3y\)

\(=x\left(x+y\right)-\left(3x+3y\right)\)

\(=x\left(x+y\right)-3\left(x+y\right)\)

\(=\left(x+y\right)\left(x-3\right)\)

c) \(20xy^2-12x^2+12x^3y\)

\(=4x\left(5y^2-3x+3x^2y\right)\)

d) \(2x\left(x-y\right)+3y\left(y-x\right)\)

\(=2x\left(x-y\right)-3y\left(x-y\right)\)

\(=\left(x-y\right)\left(2x-3y\right)\)

\(9(x-y)^2-27(y-x)^3\\=9(x-y)^2+27(x-y)^3\\=(x-y)^2[9+27(x-y)]\\=(x-y)^2(9+27x-27y)\\=9(x-y)^2(1+3x-3y)\)