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

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

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

b) \(25-\left(3-x\right)^2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}3-x=5\\3-x=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}}\)

c)\(3x^2-6x+3-27=0\)

\(\Leftrightarrow3x^2-6x-24=0\)

\(\Leftrightarrow\left(3x+6\right)\left(x-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x+6=0\\x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=4\end{cases}}}\)

#H

1.(4-x)²-16 =0
<=> 16 -8x+x² -16 =0
<=> -x(8-x) =0
<=> TH1: x=0
.      TH2: 8-x=0
.              => x= -8

2. 25 - (3-x)² = 0
<=> 25 - (9-6x+x²) = 0
<=> 25 - 9+6x-x² = 0
<=> -x²+6x+16 = 0
<=> -(x-8)(x+2) = 0 (bước này bạn nhập phương trình trên mtinh là nó sẽ ra nghiệm nhe)
<=> TH1:x-8=0
.             x= 8
.      TH2:x+2=0
.             x=-2

3.(bạn tự làm nhé, giải bth thui)

\(1,x^2-y^2+4x-4y\)

\(\left(x-y\right)\left(x+y\right)+4\left(x-y\right)\)

\(\left(x-y\right)\left(x+y+4\right)\)

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

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

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

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

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

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

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

\(x^4-6x^3+54x-81\)

\(x^4+3x^3-9x^3+27x^2-27x^2+81x-27x-81\)

\(\left(x^4+3x^3\right)-\left(9x^3+27x^2\right)+\left(27x^2+81x\right)-\left(27x+81\right)\)

\(x^3\left(x+3\right)-9x^2\left(x+3\right)+27x\left(x+3\right)-27\left(x+3\right)\)

\(\left(x+3\right)\left(x^3-9x^2+27x-27\right)\)

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

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

7, \(10ax-5ay-2x+y=5a\left(2x-y\right)-\left(2x-y\right)=\left(5a-1\right)\left(2x-y\right)\)

8, xem lại đề bạn nhé

9, \(4x^2-y^2+8y-16=4x^2-\left(y^2-8y+16\right)=4x^2-\left(y-4\right)^2\)

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

Trả lời:

6, x²y + xy² - 4x - 4y = ( x²y + xy² ) - ( 4x + 4y ) = xy ( x + y ) - 4 ( x + y ) = ( x + y )( xy - 4 )

7, 10ax - 5ay - 2x + y = ( 10ax - 5ay ) - ( 2x - y ) = 5a ( 2x - y ) - ( 2x - y ) = ( 2x - y )( 5a - 1 ) 

8, Sửa đề: x³ - 2x² + 2x - 4 = ( x³ - 2x² ) + ( 2x - 4 ) = x² ( x - 2 ) + 2 ( x - 2 ) = ( x - 2 )( x² + 2 )

9, 4x² - y² + 8y - 16 = 4x² - ( y² - 8y + 16 ) = 4x² - ( y - 4 )² = ( 2x - y + 4 )( 2x + y - 4 )

1, \(x^2\left(x-3\right)-4x+12=x^2\left(x-3\right)-4\left(x-3\right)\)

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

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

3, \(2x-4+5x^2-10x=2\left(x-2\right)+5x\left(x-2\right)=\left(2+5x\right)\left(x-2\right)\)

4, sửa đề : 

 \(6x^2-12x-7x+14=6x\left(x-2\right)-7\left(x-2\right)=\left(6x-7\right)\left(x-2\right)\)

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

a) x²(x-3)-4x+12

=x²(x-3)-4(x-3)

=(x-3)(x²-4)

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

b) 2a(x+y)-x-y

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

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

c) 2x-4+5x²-10x

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

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

d) 5x²-12x-7x+14

=5x²-19x+14

e) xy-y²-3x+3y

=y(x-y)-3(x-y)

=(x-y)(y-3)

#H

D = x² + 4xy + 4y² - z² + 2xt - t² 

 = (x + 2y)² - (z - t)² 

= (x + 2y - z + t)(x + 2y + z - t) 

Thay x = 10 ; y = 40 ; z = 30 ; t = 20 vào D 

\(\Rightarrow D=\left(10+40.2-30+20\right)\left(10+40.2+30-20\right)=80.100=8000\)

D = x\(^2\) + 4xy + 4y \(^2\) - z \(^2\) + 2zt - t \(^2\)

D = (x + 2y)\(^2\) - z\(^2\)+ z\(^2\) + 2zt + t\(^2\) - t\(^2\) 

D = (10 + 80)\(^2\) - 30\(^2\) + (z + t)\(^2\) - 20\(^2\) 

D = 90\(^2\) - 900 - 900 + (30 + 20)\(^2\) - 400

D = 8100 - 900 + 2500 - 400 

D =8600

HT

\(A=x\left(x+2\right)\left(x+4\right)\left(x+6\right)+8\)

\(=\left(x^2+6x\right)\left(x^2+6x+8\right)+8\)

\(=\left(x^2+6x+4\right)^2-4^2+8\)

\(=\left(x^2+6x+4\right)^2-8\ge-8\)

Dấu \(=\)khi \(x^2+6x+4=0\Leftrightarrow x=-3\pm\sqrt{5}\).

\(B=5+\left(1-x\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(=5-\left[\left(x-1\right)\left(x+6\right)\right].\left[\left(x+2\right)\left(x+3\right)\right]\)

\(=5-\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=5-\left(x^2+5x\right)^2+6^2\)

\(=41-\left(x^2+5x\right)^2\le41\)

Dấu \(=\)khi \(x^2+5x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

\(C=\left(x+3\right)^4+\left(x-7\right)^4=\left[\left(x-2\right)+5\right]^4+\left[\left(x-2\right)-5\right]^4\)

\(=2\left(x-2\right)^4+300\left(x-2\right)^2+1250\ge1250\)

Dấu \(=\)khi \(x-2=0\Leftrightarrow x=2\).