a]={23;24;25;26....;35}

b]={4;8;14;...}

c]{4}

Tim x thuoc N=thi ko biet nha!

Ta có: \(\left(x-3\right)^2=25\)

\(\Leftrightarrow\left(x-3\right)^2=5^2\left(x\inℕ^∗\right)\)

\(\Leftrightarrow x-3=5\)

\(\Leftrightarrow x=5+3\)

\(\Leftrightarrow x=8\)

hok tốt!!

(x-3)² = 25

(x-3)² = 5²

x-3 = 5

x= 5+3

x=8

\(\left(x-3\right)^2=25\)

\(\Leftrightarrow\left(x-3\right)^2=5^2\)

\(\Leftrightarrow x-3=5\)

\(\Leftrightarrow x=5+3\)

\(\Leftrightarrow x=8\)

Vậy x = 8

\(\frac{5}{n+1}=\frac{n+1}{5}\)

\(\Leftrightarrow\left(n+1\right)^2=5^2\)

\(\Leftrightarrow\sqrt{\left(n+1\right)^2}=\sqrt{5^2}\)

\(\Leftrightarrow n+1=5\)

\(\Leftrightarrow n=5-1\)

\(\Leftrightarrow n=4\)

\(\frac{2x+1}{5}=\frac{x+2}{8}\)

\(\Leftrightarrow8\left(2x+1\right)=5\left(x+2\right)\)

\(\Leftrightarrow16x+8=5x+10\)

\(\Leftrightarrow11x=2\)

\(\Leftrightarrow x=\frac{2}{11}\)

\(x\cdot y=8\)

\(\Rightarrow x;y\inƯ\left(8\right)\)

\(\Rightarrow x;y\in\left\{-1;1;-2;2;-4;4;-8;8\right\}\)

\(\left(x-2\right)\left(y+3\right)=15\)

\(\Rightarrow\left(x-2\right);\left(y+3\right)\inƯ\left(15\right)\)

\(\Rightarrow\left(x-2\right);\left(y+3\right)\in\left\{-1;1;-3;3;-5;5-15;15\right\}\)

ta có bảng :