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

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

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

\(=30x+5\)

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

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

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

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

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

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

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

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

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

MHƯ VẬY ĐÚNG KHÔNG

Ta có (x^2 + y^2 )^3 + (z^2 – x^2 )^3 – (y^2 + z^2 )^3

= (x^2 + y^2 )^3 + (z^2 – x^2 )^3 + (-y^2 - z^2 )^3

Ta thấy x^2 + y^2 + z^2 – x^2 – y^2 – z^2 = 0

=> áp dụng nhận xét ta có: (x^2+y^2 )^3+ (z^2 -x^2 )^3 -y^2 -z^2 )^3

= 3(x^2 + y^2 ) (z^2 –x^2 ) (-y^2 – z^2 )

= 3(x^2+y^2 ) (x+z)(x-z)(y^2+z^2 )

\(=x^4-16x^2+100=x^4+20x^2+100-36x^2\)

\(=\left(x^2+10\right)^2-\left(6x\right)^2=\left(x^2-6x+10\right)\left(x^2+6x+10\right)\)

\(\left(x^2-8\right)^2+36\)

\(=x^4-16x^2+64+36\)

\(=x^4-16x^2+100\)

\(=x^4+20x^2+100-36x^2\)

\(=\left(x^2+10\right)^2-\left(6x\right)^2\)

\(=\left(x^2-6x+10\right)\left(x^2+6x+10\right)\)

(x -y)³  - 1 - 3(x -y)(x - y - 1)

= (x -y)³  - 3(x -y)(x - y - 1) - 1

Đặt x - y = t, khi đó ta có:

    t³  -  3t. (t  - 1) - 1

=   t³  -  3t²  + 3t - 1

=  (t  - 1)³

Thay t = x - y vào (t - 1)³ , ta có:  ( x - y - 1)³

Vậy (x -y)³  - 1 - 3(x -y)(x - y - 1) =  ( x - y - 1)³

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 :)) 

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 :

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

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

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

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

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

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

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

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

\(=\left(x-y\right)\left(x+5y\right)\)

( 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