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2. Ta có:
+) Nếu p = 2 => 2 + 10 = 12 (không là số nguyên tố), 2 + 14 = 16 (không là số nguyên tố) => loại p = 2
+) Nếu p = 3 => 3 + 10 = 13 (là số nguyên tố), 3 + 14 = 17 (là số nguyên tố) => chọn p = 3
+) Nếu p > 3 => p = 3k + 1. p = 3k + 2 (k \(\in\) N*)
=> p = 3k + 1 => p + 10 = 3k + 12 chia hết cho 3 => loại p = 3k + 1
=> p = 3k + 2 => p + 14 = 3k + 15 chia hết cho 3 => loại p = 3k + 2.
Vậy p = 3.
a,\(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)
\(\Rightarrow\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}\)
\(\Rightarrow\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}=\dfrac{x+3+63}{63}+\dfrac{x+4+62}{62}\)
\(\Rightarrow\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)
\(\Rightarrow\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)
\(\Rightarrow x+66=0\) ( vì \(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}>0\) )
\(\Rightarrow x=-66\)
\(b,\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)
\(\Rightarrow\dfrac{x-12}{77}-1+\dfrac{x-11}{78}-1=\dfrac{x-74}{15}-1+\dfrac{x-73}{16}-1\)
\(\Rightarrow\dfrac{x-12-77}{77}+\dfrac{x-11-78}{78}=\dfrac{x-74-15}{15}+\dfrac{x-73-16}{16}\)
\(\Rightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}-\dfrac{x-89}{15}-\dfrac{x-89}{16}=0\)
\(\Rightarrow\left(x-89\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)
\(\Rightarrow x-89=0\)
\(\Rightarrow x=89\)
b.
\(\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\\ \Rightarrow\left(\dfrac{x-12}{77}-1\right)+\left(\dfrac{x-11}{78}-1\right)=\left(\dfrac{x-74}{15}-1\right)+\left(\dfrac{x-73}{16}-1\right)\\ \Leftrightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}=\dfrac{x-89}{15}+\dfrac{x-89}{16}\\ \Leftrightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}-\dfrac{x-89}{15}-\dfrac{x-89}{16}=0\\ \\ \Leftrightarrow\left(x-89\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\\ \Leftrightarrow x-89=0\\ \Leftrightarrow x=89\)