(x+1)(y+1) = 5

=> x+1 và y + 1 thuộc Ư(5) = {-1;1;-5;5}

ta có bảng :

a.

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

b.

\(\Leftrightarrow2x^2+2x-x-1-2x^2-3x+1=0\)

\(\Leftrightarrow-2x=0\)

\(\Leftrightarrow x=0\)

Ta có x - y = 1 => x = y + 1 

\(\dfrac{x+2}{9}=\dfrac{1}{y+2}\Rightarrow\left(x+2\right)\left(y+2\right)=9\)

\(\Leftrightarrow\left(3+y\right)\left(y+2\right)=9\Leftrightarrow y^2+5y-3=0\Leftrightarrow y=\dfrac{-5\pm\sqrt{37}}{2}\)

thay vào tìm x 

ps nhưng số xấu quá bạn ạ, kiểm tra lại đề nhé 

ĐKXĐ:\(y\ne-2\)

\(\left\{{}\begin{matrix}\dfrac{x-1}{9}+\dfrac{1}{3}=\dfrac{1}{y+2}\\x-y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{y+1-1}{9}+\dfrac{3}{9}=\dfrac{1}{y+2}\\x=y+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{y+3}{9}=\dfrac{1}{y+2}\\x=y+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(y+3\right)\left(y+2\right)=9\\x=y+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y^2+5y+6-9=0\\x=y+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y^2+5y-3=0\\x=y+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}y=\dfrac{-5+\sqrt{37}}{2}\\y=\dfrac{-5-\sqrt{37}}{2}\end{matrix}\right.\\x=y+1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{-3+\sqrt{37}}{2}\\y=\dfrac{-5+\sqrt{37}}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{-3-\sqrt{37}}{2}\\y=\dfrac{-5-\sqrt{37}}{2}\end{matrix}\right.\end{matrix}\right.\)

Ta có:

\(x-y=1\Rightarrow x=1+y\)

Thay vào 

\(\dfrac{x-1}{9}+\dfrac{1}{3}=\dfrac{1}{y}+2\) \(\left(đk:y\ne0\right)\)

\(\dfrac{x+2}{9}=\dfrac{2y+1}{y}\)

\(\Leftrightarrow\dfrac{y+3}{9}=\dfrac{2y+1}{y}\)

\(\Leftrightarrow y^2+3y=18y+9\)

\(\Leftrightarrow y^2-15y-9=0\)

\(\Leftrightarrow\)\(\left(y-\dfrac{15}{2}\right)^2=\dfrac{261}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}y-\dfrac{15}{2}=\dfrac{\sqrt{261}}{2}\\y-\dfrac{15}{2}=-\dfrac{\sqrt{261}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{\sqrt{261}+15}{2}\\y=\dfrac{15-\sqrt{261}}{2}\end{matrix}\right.\)

$\large A=\frac{2\sqrt{x}-1}{\sqrt{x}+1}=2-\frac{3}{\sqrt{x}+1}$

Ta có: $\large \sqrt{x}+1\ge1\Leftrightarrow -\frac{3}{\sqrt{x}+1}\ge-3$

Do đó: $\large A \ge 2-3=-1$

Vậy $A_{min}=-1$

Dấu $"="$ xảy ra khi $x=0$

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

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=1\\x+\dfrac{1}{2}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+\sqrt{x^2+1}=a>0\\y+\sqrt{y^2+1}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2+1}=a-x\\\sqrt{y^2+1}=b-y\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2ax=a^2-1\\2by=b^2-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{a^2-1}{2a}\\y=\dfrac{b^2-1}{2b}\end{matrix}\right.\)

 \(\Rightarrow\left(\dfrac{a^2-1}{2a}+\sqrt{\left(\dfrac{b^2-1}{2b}\right)+1}\right)\left(\dfrac{b^2-1}{2b}+\sqrt{\left(\dfrac{a^2-1}{2a}\right)+1}\right)=1\)

\(\Rightarrow\left(\dfrac{a^2-1}{2a}+\dfrac{b^2+1}{2b}\right)\left(\dfrac{b^2-1}{2b}+\dfrac{a^2+1}{2a}\right)=1\)

\(\Rightarrow\left(\dfrac{a+b}{2}+\dfrac{a-b}{2ab}\right)\left(\dfrac{a+b}{2}-\dfrac{a-b}{2ab}\right)=\dfrac{4ab}{4ab}=\dfrac{\left(a+b\right)^2}{4ab}-\dfrac{\left(a-b\right)^2}{4ab}\)

\(\Rightarrow\dfrac{\left(a+b\right)^2}{4}-\dfrac{\left(a+b\right)^2}{4ab}-\dfrac{\left(a-b\right)^2}{4\left(ab\right)^2}+\dfrac{\left(a-b\right)^2}{4ab}=0\)

\(\Rightarrow\dfrac{\left(a+b\right)^2}{4}\left(1-\dfrac{1}{ab}\right)+\dfrac{\left(a-b\right)^2}{4ab}\left(1-\dfrac{1}{ab}\right)=0\)

\(\Rightarrow\left(1-\dfrac{1}{ab}\right)\left(\dfrac{\left(a+b\right)^2}{4}+\dfrac{\left(a-b\right)^2}{4ab}\right)=0\)

\(\Rightarrow1-\dfrac{1}{ab}=0\Rightarrow ab=1\)

\(\Rightarrow\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=1\)

\(\Rightarrow x+y=0\Rightarrow y=-x\)

\(P=2\left(x^2+\left(-x\right)^2\right)+0=4x^2\ge0\)

Dấu "=" xảy ra khi \(x=y=0\)

\(xy\le\dfrac{\left(x+y\right)^2}{4}=\dfrac{1}{4}\)

\(\Rightarrow P=xy+\dfrac{1}{xy}=xy+\dfrac{1}{16xy}+\dfrac{15}{16xy}\ge2\sqrt{xy.\dfrac{1}{16xy}}+\dfrac{15}{16.\dfrac{1}{4}}=\dfrac{1}{2}+\dfrac{15}{4}=\dfrac{17}{4}\)

\(min_P=\dfrac{17}{4}\Leftrightarrow x=y=\dfrac{1}{2}\)

sorry pạn ( nãy mik làm sai )

Bài 2: 

a: Ta có: \(x^2+4x+7\)

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

\(=\left(x+2\right)^2+3\ge3\forall x\)

Dấu '=' xảy ra khi x=-2

Chắc đề đúng là số dương, vì ko tồn tại x;y nguyên dương thỏa mãn x+y=1

\(A=\dfrac{y^2}{xy+y}+\dfrac{x^2}{xy+x}\ge\dfrac{\left(x+y\right)^2}{x+y+2xy}\ge\dfrac{\left(x+y\right)^2}{x+y+\dfrac{1}{2}\left(x+y\right)^2}=\dfrac{2}{3}\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)