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

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

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

2,\(=\left(x^2+2xy+y^2\right)-\left(z^2-2zt+t^2\right)\)

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

     \(=\left(x+y+z-t\right)\left(x+y-z+t\right)\)

3,\(=9x\left(x-y\right)-7\left(x-y\right)\)

    \(=\left(x-y\right)\left(9x-7\right)\)

4\(=3\left(x-y\right)+a\left(x-y\right)\)

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

\(B=7x^2-7xy-5x+5y\)

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

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

\(E=x^2+7x+12\)

\(=x^2+3x+4x+12\)

\(=x\left(x+3\right)+4\left(x+3\right)\)

\(=\left(x+3\right)\left(x+4\right)\)

\(F=x^2-9x+18\)

\(=x^2-3x-6x+18\)

\(=x\left(x-3\right)-6\left(x-3\right)\)

\(=\left(x-3\right)\left(x-6\right)\)

\(H=8x^2-2x-1\)

\(=8x^2-4x+2x-1\)

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

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

a ) x³ - 9x=0

<=> x (x² - 3 )= 0

<=> x(x+3)(x-3)

<=> x=0

     hoặc x=0-3=-3

      hoặc x=0+3=3

a: \(\Leftrightarrow x^2\left(9x^2-4\right)=0\)

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

hay \(x\in\left\{0;\dfrac{2}{3};-\dfrac{2}{3}\right\}\)

b: \(\Leftrightarrow2x^4-4x^2+3x^2-6=0\)

\(\Leftrightarrow x^2-2=0\)

hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)

d: \(\Leftrightarrow x^4-9x^2+6x^2-54=0\)

\(\Leftrightarrow x^2-9=0\)

=>x=3 hoặc x=-3

x^5-3*x^2-(7*x^4-9*x^3+x^2-1/4*x+5*x^4-x^5+x^2-2*x^3+3*x^2-1/4)=0
 

\(\Leftrightarrow x+\dfrac{1}{3}+x+\dfrac{1}{2}+x+\dfrac{3}{5}=x+\dfrac{1}{7}+x+\dfrac{1}{4}+x+\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{11}{10}=\dfrac{11}{28}\) (vô lý)

Vậy pt vô nghiệm