Question

(y+1)+(y+3)+(y+5)+...+(y+49)=700

Answer 1

#)Giải :

\(\left(y+1\right)+\left(y+3\right)+\left(y+5\right)+...+\left(y+49\right)=700\)

\(\left(y+y+y+...+y\right)+\left(1+3+5+...+49\right)=700\)(có 25 số hạng trong các ngoặc)

\(25y+\frac{\left(49+1\right)25}{2}=700\)

\(25y+625=700\)

\(\Rightarrow25y=75\)

\(\Rightarrow y=3\)

Answer 2

(y+1)+(y+3)+(y+5)+...+(y+49)=700

=>(y+y+y+...+y)+(1+3+5+...+49)=700

-> y=(49-1):2+1=25 số hạng

=>25y+[(49+1)×25÷2]=700

=>25y+625=700

=>25y=75

=>y=3