\(a,=\dfrac{2y^4}{3x\left(2x-3y\right)}\\ b,=-\dfrac{2y\left(3x-1\right)^2}{3x^2}\\ c,=\dfrac{5\left(4x^2-9\right)}{\left(2x+3\right)^2}=\dfrac{5\left(2x-3\right)\left(2x+3\right)}{\left(2x+3\right)^2}=\dfrac{5\left(2x-3\right)}{2x+3}\\ d,=\dfrac{5x\left(x-2y\right)}{-2\left(x-2y\right)^3}=-\dfrac{5x}{2\left(x-2y\right)^2}\)

Điều kiện \(\left\{{}\begin{matrix}\dfrac{4x-3x^2y-9xy^2}{x+3y}\ge0\\x+3y\ne0\end{matrix}\right.\)

Với \(3y\ge x\), hệ tương đương:

\(\left\{{}\begin{matrix}\left(x^4-2x^2+4\right)\left(x^2+2\right)=6x^5y\\\left(3y-x\right)^2=\dfrac{4x}{x+3y}-3xy\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^6+8=6x^5y\left(1\right)\\x^3+27y^3=4x\end{matrix}\right.\left(I\right)\)

Vì \(x=0\) thì hệ vô nghiệm nên \(x\ne0\), khi đó:

\(\left(I\right)\Leftrightarrow\left\{{}\begin{matrix}1+\dfrac{8}{x^6}=\dfrac{6y}{x}\\1+\dfrac{27y^3}{x^3}=\dfrac{4}{x^2}\end{matrix}\right.\)

Đặt \(\dfrac{3y}{x}=a,\dfrac{2}{x^2}=b\) ta được hệ:

\(\Leftrightarrow\left\{{}\begin{matrix}1+a^3=2b\\1+b^3=2a\end{matrix}\right.\)

Giải hệ này ta được \(a=b\Leftrightarrow\dfrac{3y}{x}=\dfrac{2}{x^2}\Leftrightarrow y=\dfrac{2}{3x}\)

\(\left(1\right)\Leftrightarrow x^6-4x^4+8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\\x=\sqrt{1+\sqrt{5}}\\x=-\sqrt{1+\sqrt{5}}\end{matrix}\right.\)

TH1: \(x=\sqrt{2}\Rightarrow y=\dfrac{\sqrt{2}}{3}\)

TH2: \(x=-\sqrt{2}\Rightarrow y=-\dfrac{\sqrt{2}}{3}\)

TH3: \(x=\sqrt{1+\sqrt{5}}\Rightarrow y=\dfrac{2}{3\sqrt{1+\sqrt{5}}}\)

TH4: \(x=-\sqrt{1+\sqrt{5}}\Rightarrow y=-\dfrac{2}{3\sqrt{1+\sqrt{5}}}\)

Đối chiếu với các điều kiện ta được \(\left(x;y\right)=\left(-\sqrt{1+\sqrt{5}};-\dfrac{2}{3\sqrt{1+\sqrt{5}}}\right)\)

1A,B,D

2 M=2

3 \(=\dfrac{3}{4x}\)

4 \(=\dfrac{4\left(x+y\right)}{x-y}=\dfrac{4x+4y}{x-y}\)

5 K rút gọn đc

6 \(=\dfrac{4\left(x-1\right)+2\left(x-1\right)}{6\left(x-1\right)}=\dfrac{6\left(x-1\right)}{6\left(x-1\right)}=1\)

\(\dfrac{20x^2+120x+180}{\left(3x+5\right)^2-4x^2}+\dfrac{5x^2-25}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{3\left(x^2+8x+15\right)}\)

\(=\dfrac{20\left(x^2+6x+9\right)}{\left(3x+5+2x\right)\left(3x+5-2x\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{3\left(x+3\right)\left(x+5\right)}\)

\(=\dfrac{20\left(x+3\right)^2}{5\left(x+1\right)\cdot\left(x+5\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{5\left(x+1\right)\left(x-5\right)}-\dfrac{\left(x+3\right)\cdot3\left(x+1\right)}{3\left(x+3\right)\left(x+5\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{\left(x+1\right)\left(x+5\right)}+\dfrac{x+5}{x+1}-\dfrac{x+1}{x+5}\)

\(=\dfrac{4\left(x+3\right)^2+\left(x+5\right)^2-\left(x+1\right)^2}{\left(x+1\right)\left(x+5\right)}\)

\(=\dfrac{4x^2+24x+36+x^2+10x+25-x^2-2x-1}{\left(x+1\right)\cdot\left(x+5\right)}\)

\(=\dfrac{4x^2+32x+60}{\left(x+1\right)\left(x+5\right)}=\dfrac{4\left(x^2+8x+15\right)}{\left(x+1\right)\left(x+5\right)}\)

\(=\dfrac{4\left(x+3\right)\cdot\left(x+5\right)}{\left(x+1\right)\left(x+5\right)}=\dfrac{4x+12}{x+1}\)

\(A=x^3.\left(-\dfrac{5}{4}x^2y\right).\left(\dfrac{2}{5}x^3y^4\right).\\ A=-\dfrac{1}{2}x^8y^5.\)

- Bậc: 8.

- Hệ số: \(-\dfrac{1}{2}.\)

- Biến: \(x;y.\)

\(B=\left(-\dfrac{3}{4}x^5y^4\right).\left(xy^2\right).\left(-\dfrac{8}{9}x^2y^3\right).\\ B=\dfrac{2}{3}x^8y^9.\)

- Bậc: 9.

- Hệ số: \(\dfrac{2}{3}.\)

- Biến: \(x;y.\)

\(a,\dfrac{21x^2y^3}{24x^3y^2}=\dfrac{7y}{8x}\)

\(b,\dfrac{15xy^3\left(x^2-y^2\right)}{20x^2y\left(x+y\right)^2}=\dfrac{15xy^3\left(x-y\right)\left(x+y\right)}{20x^2y\left(x+y\right)^2}=\dfrac{3y^2\left(x-y\right)}{4x\left(x+y\right)}=\dfrac{3xy^2-3y^3}{4x^2+4xy}\)

a) Ta có: \(\dfrac{21x^2y^3}{24x^3y^2}\)

\(=\dfrac{21x^2y^3:3x^2y^2}{24x^3y^2:3x^2y^2}\)

\(=\dfrac{7y}{8x}\)

a) ta có : \(\left(3x^5y^2+4x^3y^3-5x^2y^4\right):2x^2y^2=\dfrac{3x^5y^2}{2x^2y^2}+\dfrac{4x^3y^3}{2x^2y^2}-\dfrac{5x^2y^4}{2x^2y^2}\)

\(=\dfrac{3}{2}x^3+2xy-\dfrac{5}{2}y^2\)

b) ta có : \(\left(\dfrac{3}{5}a^6x^3+\dfrac{3}{7}a^3x^4-\dfrac{9}{10}ax^5\right):\dfrac{3}{5}ax^3\)

\(=\left(\dfrac{3}{5}a^6x^3+\dfrac{3}{7}a^3x^4-\dfrac{9}{10}ax^5\right)\dfrac{5}{3ax^3}\)

\(=\dfrac{3}{5}.\dfrac{5}{3}\dfrac{a^6x^3}{ax^3}+\dfrac{3}{7}.\dfrac{5}{3}\dfrac{a^3x^4}{ax^3}-\dfrac{9}{10}.\dfrac{5}{3}\dfrac{ax^5}{ax^3}\)

\(=a^5+\dfrac{5}{7}a^2x-\dfrac{3}{2}a^2\)