( 3x - 1 )² - 16

= ( 3x - 1 )² - 4²

= ( 3x - 1 - 4 )( 3x - 1 + 4 )

= ( 3x - 5 )( 3x + 3 )

= 3( 3x - 5 )( x + 1 )

( x + 2 ) ( x + 3 ) ( x + 4 ) ( x + 5 ) - 24

= ( x² + 7x + 10 ) ( x² + 7x + 12 ) - 24

Đặt x² + 7x + 10 = y

Ta có : 

y² + 2y - 24 = ( y - 4 ) ( y + 6 ) = ( x² + 7x + 6 ) ( x² + 7x + 16 )

                    = ( x + 1 ) ( x + 6 ) ( x² + 7x + 16 )

Đặt x²+7x+10=t

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=t\left(t+2\right)-24=t^2+2t-24\)

\(=\left(t^2+2t+1\right)-25=\left(t+1\right)^2-5^2=\left(t-4\right)\left(t+6\right)\)=(x²+7x+6)(x²+7x+16)

=(x²+x+6x+6)(x²+7x+16)=[x(x+1)+6(x+1)](x²+7x+16)=(x+1)(x+6)(x²+7x+16)

\(B=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt:  \(x^2+7x+10=t\)Khi đó B trở thành:

\(B=t\left(t+2\right)-24\)

\(=t^2+2t-24=\left(t-4\right)\left(t+6\right)\)

đến đây bạn thay trở lại

Ta có

a,    x²-x-y²-y

=x²-y²-(x+y)

=(x-y)(x+y) - (x+y)

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

b,   x²-2xy+y²-z²

=(x-y)²-z²

=(x-y-z)(x-y+z)

con bai 32, 33 neu ban tra loi duoc minh h them

nhanh nha

a)7²  -27² =(7-27)(7+27)

=-20.34

=-680

b)37² -13²=(37-13)(37+13)=24.50

=1200

c)2002²-2²=(2002-2)(2002+2)

=2000.2004

=4008000

a) 16x²-(x²+4)²= (4x)²-(x²+4)²

                      = (4x-x²-4)(4x+x²+4)

\(\text{b) 27x^3-54x^2+36x-8=[(3x)^3-3.(3x)^2.2+3.3x.2^2-2^3}]\)

                                                      = (3x-2)³
\(\text{c) (x+y)^3 - (x-y)^3= (x+y-x+y)[(x+y)^2+(x+y)(x-y)+(x-y)^2]}\)

                                   =2y(x²+2xy+y²+x²-y²+x²-2xy+y²)

                                   = 2y(3x²+y²)

\(9\left(x+1\right)^2-\left(3x-2\right)^2\)

\(=9\left(x^2+2x+1\right)-\left(9x^2-12x+4\right)\)

\(=9x^2+18x+9-9x^2+12x-4\)

\(=30x+5\)

\(=5\left(6x+1\right)\)

\(9\left(x+1\right)^2-\left(3x-2\right)^2\)

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

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

\(=5\left(6x+1\right)\)

\(9\left(x+1\right)^2-\left(3x-2\right)^2\)

=\(\left(9x^2+18x+9\right)-\left(9x^2-12x+4\right)\)

=\(9x^2+18x+9-9x^2+12x-4\)

=\(5\left(6x+1\right)\)

MHƯ VẬY ĐÚNG KHÔNG

a) \(-y^2+\dfrac{1}{9}\)

\(=-\left(y^2-\left(\dfrac{1}{3}\right)^2\right)\)

\(=-\left(y+\dfrac{1}{3}\right)\left(y-\dfrac{1}{3}\right)\)

b) \(4^4-256\)

\(=4^4-4^4\)

\(=0\)

c) \(9\left(x-3\right)^2-4\left(x+1\right)^2\)

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

\(=\left(3x-9+2x+2\right)\left(3x-9-2x-2\right)\)

\(=\left(5x-7\right)\left(x-11\right)\)

\(a,=\left(\dfrac{1}{3}-y\right)\left(\dfrac{1}{3}+y\right)\\ b,=\left(x^2-16\right)\left(x^2+16\right)\\ =\left(x-4\right)\left(x+4\right)\left(x^2+16\right)\\ c,=\left[3\left(x-3\right)-2\left(x+1\right)\right]\left[3\left(x-3\right)+2\left(x+1\right)\right]\\ =\left(3x-9-2x-2\right)\left(3x-9+2x+2\right)\\ =\left(x-11\right)\left(5x-7\right)\\ d,=\left(5x-\dfrac{1}{9}xy\right)\left(5x+\dfrac{1}{9}xy\right)=x^2\left(5-\dfrac{1}{9}y\right)\left(5+\dfrac{1}{9}y\right)\)

Câu a :

\(\left(x-5\right)^2+\left(x-5\right)\left(x+5\right)-\left(5-x\right)\left(2x+1\right)\)

\(=x^2-10x+25+x^2-25-10x-5+2x^2+x\)

\(=4x^2-19x-5\)

Câu b :

\(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)

\(=12x^2-9x-8x+6-2x+2+3x^2-3x-6x^2-6x+4x+4\)

\(=9x^2-24x+2\)

PTĐTTNT ??? :)) bn phân tích rồi đấy, đề là tìm x thôi 

Giải ( suỵt :), đừng ai nhìn thấy ... :v 

\(\left(2x-10\right)\left(x+10\right)\left(x+\sqrt{3}\right)=0\)

TH1 : \(2x-10=0\Leftrightarrow x=5\)

TH2 : \(x+10=0\Leftrightarrow x=-10\)

TH3 : \(x+\sqrt{3}=0\Leftrightarrow x=-\sqrt{3}\)( vô lí )

Vậy x = {5;-10}

sao lại "vô lí" vậy bạn 

lp 8 chưa học số vô tỉ babe nhá :))