\(9x^2-9xy-4y^2\)

\(=9x^2-12xy+3xy-4y^2\)

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

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

\(\dfrac{9x^2+3xy-2y^2}{9x^2+9xy+2y^2}=\dfrac{9x^2-3xy+6xy-2y^2}{9x^2+3xy+6xy+2y^2}\\ =\dfrac{3x\left(3x-y\right)+2y\left(3x-y\right)}{3x\left(3x+y\right)+2y\left(3x+y\right)}=\dfrac{\left(3x-y\right)\left(3x+2y\right)}{\left(3x+y\right)\left(3x+2y\right)}\\ =\dfrac{3x-y}{3x+y}\)

9x^2+3xy-2y^2 :9x^2+9xy+2y^2

=3/9

2

BẬC  LÀ 6

 BẠN THAM KHẢO NHA

      ĐKXĐ:   \(x\ne\pm\frac{2}{3}\)

             \(\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)

\(\Leftrightarrow\)\(\frac{\left(3x+2\right)^2}{\left(3x-2\right)\left(3x+2\right)}-\frac{6\left(3x-2\right)}{\left(2+3x\right)\left(3x-2\right)}=\frac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Rightarrow\)\(\left(3x+2\right)^2-6\left(3x-2\right)=9x^2\)

\(\Leftrightarrow\)\(9x^2+12x+4-18x+12-9x^2=0\)

\(\Leftrightarrow\)\(6x=16\)

\(\Leftrightarrow\)\(x=\frac{8}{3}\)  (t/m ĐKXĐ)

Vậy...

Đặt y = \(x+1=\sqrt[3]{8+2\sqrt{14}}+\sqrt[3]{8-2\sqrt{14}}\)

=> \(y^3=8+2\sqrt{14}+8-2\sqrt{14}+3\sqrt[3]{\left(8+2\sqrt{14}\right)\left(8-2\sqrt{14}\right)}.y\)

<=> \(y^3=16+6y\)

=> \(\left(x+1\right)^3=16+6\left(x+1\right)\)

=> \(x^3+3x^2+3x+1=6x+32\)

<=> \(x^3+3x^2-3x-5=26\)

Ta có: 

\(x^6+3x^5-3x^4-2x^3+9x^2-9x+2018\)

= \(x^6+3x^5-3x^4-5x^3+3x^3+9x^2-9x-15+2033\)

= \(\left(x^3+3x^2-3x-5\right)\left(x^3+3\right)+2033\)

= \(26x^3+2111\)

\(=26\left(\sqrt[8]{8+2\sqrt{14}}+\sqrt[8]{8-2\sqrt{14}}-1\right)^3+2033\)

\(9x^2-9xy-4y^2\)

\(=9x\left(x-y\right)-4y^2\)

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