\(a, 3x^2-6x+3=(x-1)(x+4)\)

\(⇔(3x^2-6x+3)-(x-1)(x+4)=0\)

\(⇔3x^2-6x+3-x^2-3x+4=0\)

\(⇔2x^2-9x+7=0\)

\(⇔2x^2-2x-7x+7=0\)

\(⇔2x(x-1)-7(x-1)=0\)

\(⇔(x-1)(2x-7)=0\)

\(⇔\left[\begin{array}{} x-1=0\\ 2x-7=0 \end{array}\right.\)

\(⇔\left[\begin{array}{} x=1\\ x=\dfrac{7}{2} \end{array}\right.\)

Vậy pt có tập nghiệm là S={\(1; \dfrac{7}{2}\)}

\(b, 18x^2(x+4)-12(x^2+4x)=0\)

\(⇔18x^2(x+4)-12x(x+4)=0\)

\(⇔6x(x+4)(3x-2)=0\)

\(⇔\left[\begin{array}{} 6x=0\\\ x+4=0\\ 3x-2=0 \end{array}\right. \)

\(⇔\left[\begin{array}{} x=0\\\ x=-4\\ x=\dfrac{2}{3} \end{array}\right.\)

vậy pt có tập nghiệm là S={\(0;-4;\dfrac{2}{3}\)}

\(c,25x^2-10x+1=2(5x-1)(3x-4)\)

\(⇔(5x-1)^2=(5x-1)(6x-8)\)

\(⇔(5x-1)^2-(5x-1)(6x-8)=0\)

\(⇔(5x-1)[(5x-1)-(6x-8)]=0\)

\(⇔(5x-1)(7-x)=0\)

\(⇔\left[\begin{array}{} 5x-1=0\\ 7-x=0 \end{array}\right.\)

\(⇔\left[\begin{array}{} x=\dfrac{1}{5}\\ x=7 \end{array}\right.\)

Vậy pt có tập nghiệm là S={\(\dfrac{1}{5};7\)}

\(3x^2-6x+3=\left(x-1\right)\left(x+4\right)\\ \Leftrightarrow3x^2-6x+3=x\left(x+4\right)-\left(x+4\right)\\ \Leftrightarrow3x^2-6x+3=x^2+4x-x-4\\ \Leftrightarrow3x^2-6x+3=x^2+3x-4\\ \Leftrightarrow3x^2-6x+3-x^2-3x+4=0\\ \Leftrightarrow2x^2-9x+7=0\\ \Leftrightarrow x\left(2x-9\right)=-7\)

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

\(b,x^3+6x^2+12x+8=\left(x+2\right)^3\)

[9x³(x² - 1) - 6x²(x² - 1) + 12x(x² - 1)] : 3x(x² - 1)

= [9x³(x² - 1) : 3x(x² - 1)] - [6x²(x² - 1) : 3x(x² - 1) + [12x(x² - 1) : 3x(x² - 1)]

= 3x² - 2x + 4

Gúp mik vs

a: \(\dfrac{3x+2}{4}-\dfrac{3x+1}{3}=\dfrac{5}{6}\)

=>3(3x+2)-4(3x+1)=10

=>9x+6-12x-4=10

=>-3x+2=10

=>-3x=8

=>x=-8/3

b: \(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{9x-10}{4-x^2}\)

=>(x-1)(x-2)-x(x+2)=-9x+10

=>x^2-3x+2-x^2-2x=-9x+10

=>-5x+2=-9x+10

=>x=2(loại)

a)\(x^3y+3x^2y\)

b)\(-24x^2+4xy\)

[9x3(x2-1)-6x2(x2-1)2+12x(x2-1)]:3x(x2-1)

Gọi x2-1 là y tá đc

(9x3y-6x2y2+12xy):3xy

=3x2-2xy+4

=3x2-2x(x2-1)+4

h) \(=3x\left(2y-3z\right)\left[x^2-5\left(2y-3z\right)\right]=3x\left(2y-3z\right)\left(x^2-10y+15z\right)\)

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

l) \(=\left(18^2+3\right)\left(x+3\right)=327\left(x+3\right)\)

m) \(=7xy\left(2x-3y+4xy\right)\)

n) \(=2\left(x-y\right)\left(5x-4y\right)\)