\(16y^2-4x^2-12x-9=16y^2-\left(2x-3\right)^2\)

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

( 4y + 2x - 3 )

\(\frac{\left(x-3\right)^3}{3x^2}:\frac{x^2-6x+9}{6x}\)

\(=\frac{\left(x-3\right)^3}{3x^2}.\frac{6x}{\left(x-3\right)^2}\)

\(=\frac{2\left(x-3\right)}{x}\)

\(=\frac{2x-6}{x}\)

#H

Trả lời:

\(\frac{\left(x-3\right)^3}{3x^2}:\frac{x^2-6x+9}{6x}\)

\(=\frac{\left(x-3\right)^3}{3x^2}.\frac{6x}{x^2-6x+9}\)

\(=\frac{\left(x-3\right)^3}{3x^2}.\frac{6x}{\left(x-3\right)^2}\)

\(=\frac{\left(x-3\right)^3.6x}{3x^2.\left(x-3\right)^2}\)

\(=\frac{2\left(x-3\right)}{x}\)

a, Vì M là trung điểm AB

N là trung điểm AC

=> MN là đường trung bình tam giác ABC 

=> MN // BC mà ABC cân tại A => ^ABC = ^ACB 

 => BMNC là hình thang cân 

b, Vì MN là đường trung bình tam giác ABC ( cma ) 

=> MN // BC ; MN = 1/2 BC

( 4 - x ) mũ 2 - 16 = 0

<=>(4-x)^2=4^2

<=>4-x=4

=>x=0

\(\left(4-x\right)^2-16=0\)

\(\Leftrightarrow\left(4-x\right)^2=16\)

\(\Leftrightarrow\orbr{\begin{cases}4-x=4\\4-x=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=8\end{cases}}}\)

#H

100-46=i06

a, \(\left(4x-1\right)^2-\left(4x-5\right)\left(4x+5\right)=3\)

\(\Leftrightarrow16x^2-8x+1-16x^2+25=3\Leftrightarrow-8x+23=0\Leftrightarrow x=\frac{23}{8}\)

b, \(\left(x+2\right)^3-\left(x^3+8\right)=0\Leftrightarrow\left(x+2\right)^3-\left(x+2\right)\left(x^2-2x+4\right)=0\)

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

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

\(\Leftrightarrow6x\left(x+2\right)=0\Leftrightarrow x=-2;x=0\)

a/ (4x-1)²-(4x-5)(4x+5)=3

     x = 23/8

b/(x+2)³-(x³+8)=0

   x = -2 

  x = 0