\(a,\text{Vì }x,y\in N\Leftrightarrow x+2\ge2;y+3\ge3\\ \Leftrightarrow\left(x+2\right)\left(y+3\right)=6=2\cdot3=3\cdot2\\ \Leftrightarrow\left\{{}\begin{matrix}x+2=2\\y+3=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(0;0\right)\)

\(b,\Leftrightarrow\left(x-3\right)\left(y+1\right)=7\cdot1=1\cdot7\\ \left\{{}\begin{matrix}x-3=7\\y+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=0\end{matrix}\right.\\ \left\{{}\begin{matrix}x-3=1\\y+1=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=6\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\left(10;0\right);\left(4;6\right)\right\}\)

Bài 2: 

a: =>x=0 hoặc x+3=0

=>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

Bài 1

a) \(x=x^5\)

\(x^5-x=0\)

\(x\left(x^4-1\right)=0\)

\(x=0\) hoặc \(x^4-1=0\)

* \(x^4-1=0\)

\(x^4=1\)

\(x=1\)

Vậy x = 0; x = 1

b) \(x^4=x^2\)

\(x^4-x^2=0\)

\(x^2\left(x^2-1\right)=0\)

\(x^2=0\) hoặc \(x^2-1=0\)

*) \(x^2=0\)

\(x=0\)

*) \(x^2-1=0\)

\(x^2=1\)

\(x=1\)

Vậy \(x=0\); \(x=1\)

c) \(\left(x-1\right)^3=x-1\)

\(\left(x-1\right)^3-\left(x-1\right)=0\)

\(\left(x-1\right)\left[\left(x-1\right)^2-1\right]=0\)

\(x-1=0\) hoặc \(\left(x-1\right)^2-1=0\)

*) \(x-1=0\)

\(x=1\)

*) \(\left(x-1\right)^2-1=0\)

\(\left(x-1\right)^2=1\)

\(x-1=1\) hoặc \(x-1=-1\)

**) \(x-1=1\)

\(x=2\)

**) \(x-1=-1\)

\(x=0\)

Vậy \(x=0\); \(x=1\); \(x=2\)

Bài 2: 

a: =>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

a) x=3 y=13

x=16 y=0

x=4 y=5

x=9 y=1

a) \(\left(x-2\right)\left(y+1\right)=14\)

Do \(x,y\in N\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2=1\\y+1=14\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=14\\y+1=1\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=2\\y+1=7\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=7\\y+1=2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\left(tm\right)\\y=13\left(tm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=16\left(tm\right)\\y=0\left(tm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\left(tm\right)\\y=6\left(tm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=9\left(tm\right)\\y=1\left(tm\right)\end{matrix}\right.\end{matrix}\right.\)

a \(\dfrac{2}{3}x+\dfrac{1}{3}=\dfrac{1}{5}\\ \dfrac{2}{3}x=\dfrac{1}{5}-\dfrac{1}{3}\\ \dfrac{2}{3}x=\dfrac{-2}{15}\\ x=-\dfrac{2}{15}:\dfrac{2}{3}\\ x=-\dfrac{1}{5}\)   b) \(\dfrac{4}{5}-\dfrac{5}{3}x=-2\\ \dfrac{5}{3}x=\dfrac{4}{5}+2\\ \dfrac{5}{3}x=\dfrac{14}{5}\\ x=\dfrac{14}{5}:\dfrac{5}{3}\\ x=\dfrac{42}{25}\)c) \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\\ \dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}\\ \dfrac{5}{3}:x=\dfrac{3}{10}\\ x=\dfrac{5}{3}:\dfrac{3}{10}\\ x=\dfrac{50}{9}\)d) \(\dfrac{5}{7}:x-3=-\dfrac{2}{7}\\ \dfrac{5}{7}:x=3-\dfrac{2}{7}\\ \dfrac{5}{7}:x=\dfrac{19}{7}\\ x=\dfrac{5}{7}:\dfrac{19}{7}\\ x=\dfrac{5}{19}\)

a, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x+y}{3+5}=\dfrac{16}{8}=2\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=10\end{matrix}\right.\\ b,x:2=y:\left(-5\right)\Rightarrow\dfrac{x}{2}=\dfrac{y}{-5}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{2}=\dfrac{y}{-5}=\dfrac{x-y}{2-\left(-5\right)}=\dfrac{-7}{7}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-2\\y=5\end{matrix}\right.\)

\(a,\) Vì \(x,y\in Z\) nên \(\left(3x+2\right):3R2;R1\)

Mà \(\left(3x+2\right)\left(y-8\right)=12\) nên \(3x+2\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)

Do đó \(3x+2\in\left\{-4;-1;2\right\}\)

\(\Rightarrow x\in\left\{-2;-1;0\right\}\)

Với \(x=-2\Rightarrow\left(-4\right)\left(y-8\right)=12\Rightarrow y-8=-3\Rightarrow y=5\)

Với \(x=-1\Rightarrow\left(-3\right)\left(y-8\right)=12\Rightarrow y-8=-4\Rightarrow y=4\)

Với \(x=0\Rightarrow2\left(y-8\right)=12\Rightarrow y-8=6\Rightarrow y=14\)

Vậy PT có nghiệm \(\left(x;y\right)\) là \(\left(-2;5\right);\left(-1;4\right);\left(0;14\right)\)

\(b,\) Vì \(x,y\in Z\) nên \(\left(5x-4\right):5R1;R4\)

Mà \(\left(5x-4\right)\left(y+3\right)=-18\)

\(\Rightarrow5x-4\inƯ\left(-18\right)=\left\{-18;-9;-6;-3;-2;-1;1;2;3;6;9;18\right\}\\ \Rightarrow5x-4\in\left\{-9;1;6\right\}\\ \Rightarrow x\in\left\{-1;1;2\right\}\)

Với \(x=-1\Rightarrow-9\left(y+3\right)=-18\Rightarrow y+3=2\Rightarrow y=-1\)

Với \(x=1\Rightarrow y+3=18\Rightarrow y=15\)

Với \(x=2\Rightarrow6\left(y+3\right)=18\Rightarrow y+3=3\Rightarrow y=0\)

Vậy PT có nghiệm \(\left(x;y\right)\) là \(\left(-1;-1\right);\left(1;15\right);\left(2;0\right)\)