\(\left(3x-2\right)\left(4x-5\right)-\left(2x-1\right)\left(6x+2\right)=0\)

<=> \(12x^2-15x-8x+10-\left(12x^2+4x-6x-2\right)=\) \(0\)

<=> \(12x^2-23x+10-12x^2-4x+6x+2=0\)

<=> \(12x^2-21x+12=0\)

(3x - 2)(4x - 5) - (2x - 1)(6x + 2) = 0

=> 12x² - 15x - 8x + 10 - (12x² - 4x - 6x - 2) = 0

=> 12x² - 15x - 8x + 10 - 12x² + 4x + 6x + 2 = 0

=> (12x² - 12x²) - (15x + 8x - 4x - 6x) + (10 + 2) = 0

=> 0 - 13x + 12 = 0

=> 13x = - 12

=> x = -12/13

ta có \(x\left(x+1\right)\left(x+2\right)=\left(x^2+x\right)\left(x+2\right)\)

                                              \(=x^3+2x^2+x^2+2x\)

                                               \(=x^3+3x^2+2x\)( đpcm )

học tốt

Biến đổi VT ta có :

\(x\left(x+1\right)\left(x+2\right)\)

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

\(=x^3+2x^2+x^2+2x\)

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

\(=x^3+3x^2+2x=VP\)

Vậy \(x\left(x+1\right)\left(x+2\right)=x^3+3x^2+2x\)

\(\left(x^2+x\right)^2-2x^2-2x-15\)

\(=\left(x^2+x\right)^2-\left(2x^2+2x+15\right)\)

\(=\left(x^2+x\right)^2-\left[\left(2x^2+2x\right)+15\right]\)

\(=\left(x^2+x\right)^2-\left[2.\left(x^2+x\right)+15\right]\)

\(=\left(x^2+x\right)^2-2\left(x^2+x\right)-15\) \(\left(1\right)\)

đặt \(x^2+x=t\)

\(\left(1\right)\)\(=\)  \(t^2-2t-15\)

            \(=\left(t-1\right)^2-16\)

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

           \(=\left(t-5\right)\left(t+3\right)\)

thay \(t=x^2+x\) ta có

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

các câu còn lại tương tự nha

học tốt 

\(\left(c^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)

\(=\left(a^2+b^2-2ab-9\right)\left(a^2+b^2+2ab-1\right)\)

\(=\left[\left(a-b\right)^2-9\right]\left[\left(a+b\right)^2-1\right]\)

\(=\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)

2)(2x^2-3x+1)/(x^2+x-2)

=(2x^2-2x-x+1)/(x^2-x-+2x-2)

=[2x(x-1)-(x-1)]/[x(x-1)+2(x-1)]

=[(x-1)(2x-1)]/[(x-1)(x+2)]= (2x-1)/(x+2)

câu 1 thì mik ko chắc

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

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

\(\Leftrightarrow\left(x-4\right)\left(3x+2\right)=0\)

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

\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

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

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

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

Áp dụng \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)

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

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

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

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

\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)

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

\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)