Đề sai nhé .Sửu lại

\(x^2-4x^2y^2+4+4x\)

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

\(=\left(x+2\right)^2-\left(2xy\right)^2\)

\(=\left(x+2+2xy\right)\left(x+2-2xy\right)\)

Lời giải:

1. 
$x^3+3x^2-16x-48=(x^3+3x^2)-(16x+48)=x^2(x+3)-16(x+3)$

$=(x+3)(x^2-16)=(x+3)(x-4)(x+4)$

2.

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

3.

$x^3+2x^2-2x-1=(x^3-x^2)+(3x^2-3x)+(x-1)=x^2(x-1)+3x(x-1)+(x-1)$

$=(x-1)(x^2+3x+1)$

Ta có: 4x² - y² + 4x + 4y - 3

= (4x² - 4x + 1) - (y² - 4y + 4)

= (2x - 1)² - (y - 2)²

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

= (2x - y + 1)(2x + y - 3)

\(4x^2-y^2+4x+4y-3\)

\(=\left(4x^2+4x+1\right)-\left(y^2-4y+4\right)\)

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

\(=\left(2x+1+y-2\right)\left(2x+1-y+2\right)\)

\(=\left(2x+y-1\right)\left(2x-y+3\right)\)

Ta có : 4x² - y² + 4x + 4y - 3

         = 4x² - y² + 4x + 4y + 1 - 4 

        = ( 4x² + 4x + 1 ) - ( y² - 4y + 4 )

       = ( 2x + 1 )² - ( y - 2 )²

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

     = ( 2x - y + 3 )( 2x + y - 1 )

\(\left(1+x^2\right)^2-4x\left(1-x^2\right)\)

\(\Leftrightarrow\left(1+x^2\right)^2+4x\left(1+x^2\right)\)

\(\Leftrightarrow\left(1+x^2\right)\times\left[\left(1+x^2\right)+4\right]\)

( 1+x² )² -4x( 1- x² )

=x⁴+2x²+1-4x+4x³

=x³+2x²-x+2x³+4x²-2x-x²-2x+1

=x(x²+2x-1)+2x(x²+2x-1)-(x²+2x-1)

=(x²+2x-1)(x²+2x-1)

=(x²+2x-1)²

\(\left(1+x\right)^2-4x\left(1-x^2\right)\)

\(=\left(1+x\right)^2-4x\left(1-x\right)\left(1+x\right)\)

\(=\left(1+x\right)\left(1+x-4\left(1-x\right)\right)\)

\(=\left(1+x\right)\left(1+x-4+4x\right)\)

\(=\left(1+x\right)\left(5x-3\right)\)

-b giải sai đầu bài rồi .

sai đề rồi kìa, đề mà cho như bạn thì tớ đã biết làm rồi

\(\left(x^2+2x\right)^2-2x^2-4x-3=0\Leftrightarrow x^4+4x^3+4x^2-2x^2-4x-3=0\Leftrightarrow x^4+4x^3+2x^2-4x-3=0\Leftrightarrow\left(x-1\right)\left(x+1\right)^2\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=3\end{matrix}\right.\)

Ta có: \(\left(x^2+2x\right)^2-2x^2-4x-3=0\)

\(\Leftrightarrow\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3=0\)

\(\Leftrightarrow\left(x^2+2x-3\right)\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x+3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\\x=1\end{matrix}\right.\)

a) (2x - 1)² - (x + 3)²

= (2x - 1 - x - 3).(2x - 1 + x + 3)

= (x - 4).(3x + 2)

b) x².(x - 3) + 12 - 4x

= x².(x - 3) - 4x + 12

= x².(x - 3) - 4.(x - 3)

= (x - 3).(x² - 4)

= (x - 3).(x - 2).(x + 2)

Áp dụng HĐT:

    a² - b² = (a - b)(a + b)

\(\left(2x-1\right)^2-\left(x+3\right)^2\)

\(=\left(2x-1-x-3\right)\left(2x-1+x+3\right)\)

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

\(-4x^2+12xy-9y^2+25=25-\left(2x-3y\right)^2=\left(5-2x+3y\right)\left(5+2x-3y\right)\)

\(-4x^2+12xy-9y^2+25\)

\(=25-\left(4x^2-12xy+9y^2\right)\)

\(=25-\left(2x-3\right)^2\)

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

\(-4x^2+12xy-9y^2+25\)
\(=\left(-4x^2+12xy-9y^2\right)+25\)
\(=-\left(4x^2-12xy+9y^2\right)+25\)
\(=-\left(2x-3y\right)^2+25\)
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