(a+1) + (a+2)+(a+3) +(a+4) = 50
\(\Rightarrow\)a + 1 + a + 2 + a + 3 + a + 4 = 50
\(\Leftrightarrow\) 4a + 10 = 50
\(\Leftrightarrow\)4a = 40
\(\Leftrightarrow\) a = 10
Ta có: \(\left(a+1\right)+\left(a+2\right)+\left(a+3\right)+\left(a+4\right)=50\)
⇒a+a+a+a+(1+2+3+4)=50
ax4+10=50
ax4=50-10
ax4=40
a=40:4
a=10
(a+1) + (a+2)+(a+3) +(a+4) = 50
⇒a + 1 + a + 2 + a + 3 + a + 4 = 50
⇒(a+a+a+a)+(1+2+3+4)=50
⇔ 4a + 10 = 50
⇔4a = 40
⇔ a = 10