a.

\(\Leftrightarrow x\left(y+1\right)^2=32y\Leftrightarrow x=\dfrac{32y}{\left(y+1\right)^2}\)

Do y và y+1 nguyên tố cùng nhau  \(\Rightarrow32⋮\left(y+1\right)^2\)

\(\Rightarrow\left(y+1\right)^2=\left\{4;16\right\}\)

\(\Rightarrow...\)

b.

\(2a^2+a=3b^2+b\Leftrightarrow2\left(a-b\right)\left(a+b\right)+a-b=b^2\)

\(\Leftrightarrow\left(2a+2b+1\right)\left(a-b\right)=b^2\)

Gọi \(d=ƯC\left(2a+2b+1;a-b\right)\)

\(\Rightarrow b^2\) chia hết \(d^2\Rightarrow b⋮d\) (1)

Lại có:

\(\left(2a+2b+1\right)-2\left(a-b\right)⋮d\)

\(\Rightarrow4b+1⋮d\) (2)

 (1);(2) \(\Rightarrow1⋮d\Rightarrow d=1\)

\(\Rightarrow2a+2b+1\) và \(a-b\) nguyên tố cùng nhau

Mà tích của chúng là 1 SCP nên cả 2 số đều phải là SCP (đpcm)

Cách lầy nèk

\(Q=\frac{a}{1+2a}+\frac{b}{1+2b}\le\frac{a}{2\sqrt{2a}}+\frac{b}{2\sqrt{2b}}=\frac{\sqrt{a}}{2\sqrt{2}}+\frac{\sqrt{b}}{2\sqrt{2}}\)

\(\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{2}}=\frac{\sqrt{\frac{a}{2}}+\sqrt{\frac{b}{2}}}{2\sqrt{2}.\frac{1}{\sqrt{2}}}\le\frac{\frac{a+\frac{1}{2}}{2}+\frac{b+\frac{1}{2}}{2}}{2}=\frac{a+b+1}{4}=\frac{2}{4}=\frac{1}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=\frac{1}{2}\)

Có a+b =1 

Áp dụng bất đẳng thức cô-si 

=> ab= \(\frac{\left(a+b\right)}{2}\) 

<=> ab= \(\frac{1}{2}\)

P=\(\frac{\left(ab+a+ab+b\right)}{\left(ab+a+b+1\right)}\)

= \(\frac{\left(2ab+1\right)}{\left(ab+2\right)}\)

=\(\frac{\left[\left(2ab+4\right)-3\right]}{\left(ab+2\right)}\)

=\(2+\left[\frac{-3}{\left(ab+2\right)}\right]\)
Có ab = \(\frac{1}{2}\)

\(ab+2\Leftarrow\frac{5}{2}\)

\(\frac{1}{\left(ab+2\right)}\ge\frac{2}{5}\)

\(\frac{-1}{\left(ab+2\right)}\Leftarrow\frac{-2}{5}\)

\(\frac{-3}{ \left(ab+2\right)}\Leftarrow\frac{-6}{5}\)

=> GTLN = \(\frac{-6}{5}+2=\frac{4}{5}\) tại \(a=b=\frac{1}{2}\)