\(0,169\approx 0,2 \\34,3512\approx 34,4 \\3,44444\approx 3,4\)

a) P có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}a+1\ne0\\a-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\ne-1\\a\ne1\end{matrix}\right.\)

a) \(P\) có nghĩa \(\Leftrightarrow \left\{\begin{matrix} a+1\ne 0 \\ a-1\ne 0 \end{matrix} \right.\Leftrightarrow \left\{\begin{matric} a\ne -1 \\ a\ne 1 \end{matrix} \right.\)

b) \(P=\dfrac{2a^2}{(a-1)(a+1)}+\dfrac a{a+1}-\dfrac a{a-1}\)

\(=\dfrac{2a^2+a(a-1)-a(a+1)}{(a-1)(a+1)}\)

\(=\dfrac{2a^2+a^2-a-a^2-a}{(a-1)(a+1)}\)

\(=\dfrac{2a^2-2a}{(a-1)(a+1)}\)

\(=\dfrac{2a(a-1)}{(a-1)(a+1)}\)

\(=\dfrac{2a}{a+1}\)

c) Ta có: \(P=\dfrac{2a}{a+1}=\dfrac{2(a+1)-2}{a+1}=2-\dfrac 2{a+1}\)

\(P\in \mathbb{Z}\Leftrightarrow \dfrac 2{a+1}\in \mathbb{Z}\Leftrightarrow a+1\in Ư(2)=\left\{ \pm 1;\pm 2 \right\}\Leftrightarrow x\in \left\{ -3;-2;0;1 \right\}\)

Kết hợp với ĐKXĐ \(\Rightarrow a\in \left\{-3;-2;0 \right\}\)

\(3\left(x^2+y^2\right)-\left(x^3+y^3\right)+1\\ =3\left(x^2+2xy+y^2-2xy\right)-\left(x+y\right)\left(x^2+2xy+y^2-3xy\right)+1\\ =3\left[\left(x+y\right)^2-2xy\right]-2\left[\left(x+y\right)^2-3xy\right]+1\\ =3\left(x+y\right)^2-6xy-2\left(x+y\right)^2+6xy+1\\ =\left(x+y\right)^2+1\\ =2^2+1=5\)

\(\left(x^2-2x+1\right)-4=0\\ \Leftrightarrow\left(x-1\right)^2-2^2=0\\ \Leftrightarrow\left(x-1-2\right)\left(x-1+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Vậy \(x=3;x=-1\) là nghiệm của pt.

Biểu thức có nghĩa \(\Leftrightarrow x^2-1>0\Leftrightarrow x^2>1\Leftrightarrow\left|x\right|>1\Leftrightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)

ĐKXĐ: \(x\ge0\)

Xét hiệu: \(\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}-\dfrac{2}{3}=\dfrac{3\left(2-5\sqrt{x}\right)-2\left(\sqrt{x}+3\right)}{3\left(\sqrt{x}+3\right)}=\dfrac{6-15\sqrt{x}-2\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}=\dfrac{-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}\)

Vì \(-17\sqrt x\le 0;3(\sqrt x+3)>0\Rightarrow \dfrac{-17\sqrt x}{3(\sqrt x+3)}\le 0\)

\(x+4 \ \vdots \ x\)
\(\Rightarrow (x+4-x) \ \vdots \ x\)

\(\Rightarrow 4 \ \vdots \ x\)

\(\Rightarrow x\in Ư(4)=\left\{ \pm 1;\pm 2;\pm 4 \right\}\)

Mà \(n\in \mathbb{N}\Rightarrow x\in \left\{ 1;2;4 \right\}\)

Vậy \(x\in \left\{ 1;2;4 \right\}\) thì \((x+4) \ \vdots \ x\)

\(7(x-1)+3y=2xy \\\Leftrightarrow 14(x-1)+6y=4xy \\\Leftrightarrow 14x-4xy-21+6y+7=0 \\\Leftrightarrow 2x(7-2y)+3(7-2y)=-7 \\\Leftrightarrow (7-2y)(2x+3)=-7\)

Đến đấy thì ez rồi :v

hình đâu bạn :v