Answer:

\(\hept{\begin{cases}\frac{9}{x+y}+y=6\left(1\right)\\\frac{x}{x+1}+\frac{1}{y}=\frac{1}{xy+y}=\frac{1}{y.\left(x+1\right)}\left(2\right)\end{cases}}\)

\(\left(2\right)\Rightarrow xy+x+1=1\)

\(\Rightarrow xy+x=0\)

\(\Rightarrow x.\left(y+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\y+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\y=-1\end{cases}}\)

Trường hợp 1: Với x = 0 ta thay vào (1)

\(\Rightarrow\frac{9}{y}+y=6\)

\(\Rightarrow y^2-6y+9=0\)

\(\Rightarrow\left(y-3\right)^2=0\)

\(\Rightarrow y-3=0\)

\(\Rightarrow y=3\)

Trường hợp 2: Với y = - 1 ta thay vào (1)

\(\Rightarrow\frac{9}{x-1}=7\)

\(\Rightarrow7x=16\)

\(\Rightarrow x=\frac{16}{7}\)

b\(\left(a^2b-3ab^2\right):\left(\frac{1}{2}ab\right)+\left(6b^3-5ab^2\right):b^2\)

\(=ab\left(a-3b\right).\frac{2}{ab}+b^2\left(6b-5a\right).\frac{1}{b^2}\)

\(=\left(a-3b\right).2+\left(6b-5a\right)\)

\(=2a-6b+6b-5a\)

\(=-3a\)

a, Với \(a>0;a\ne1\)

\(M=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)

\(=\left(\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right).\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\frac{\sqrt{a}-1}{\sqrt{a}}\)

b, Ta có : \(M=\frac{\sqrt{a}-1}{\sqrt{a}}=1-\frac{1}{\sqrt{a}}< 1\)

Vậy M < 1