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 :)) 

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

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

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

\(=30x+5\)

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

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

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

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

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

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

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

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

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

MHƯ VẬY ĐÚNG KHÔNG

(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

\(\left(a+b+c\right)^2+\left(a+b-c\right)^2-4c^2\)

\(=\left[\left(a+b+c\right)^2-\left(2c\right)^2\right]+\left(a+b-c\right)^2\)

\(=\left(a+b+3c\right)\left(a+b-c\right)+\left(a+b-c\right)^2\)

\(=\left(a+b-c\right)\left(a+b+3c+a+b-c\right)\)

\(=\left(a+b-c\right)\left(2a+2b+2c\right)\)

\(=2\left(a+b-c\right)\left(a+b+c\right)\)

\(=\left(a+b\right)^2+2\left(a+b\right)c+c^2+\left(a+b\right)^2-2\left(a+b\right)c+c^2-4c^2\)

\(=2\left(a+b\right)^2-2c^2=2\left[\left(a+b\right)^2-c^2\right]=2\left(a+b+c\right)\left(a+b-c\right)\)

Mình đã làm bài này rồi.

Link: https://hoc24.vn/hoi-dap/question/824554.html