Bài 1:

1/1.2+1/3.4+1/5.6+.....1/49.50=1/26+1/27+....1/50

Tính vế trái ta được:

1/1.2+1/3.4+1/5.6+.....1/49.50=1-1/2+1/3-1/4+1/5-1/6+....+1/49-1/50

=(1+1/3+1/5+...+1/49)-(1/2-1/4-1/6-..-1/50)

=(1+1/2+1/3+1/4+..+1/49+1/50)-2.(1/2+1/4+1/6+..+1/50)

=(1+1/2+1/3+1/4+..+1/49+1/50)-(1+1/2+1/3+..+1/25)

=1+1/2+1/3+1/4+..+1/49+1/50-1-1/2-1/3-..-1/25

=1/26+1/27+....1/50 (bằng vế phải)

=>1/1.2+1/3.4+1/5.6+.....1/49.50=1/26+1/27+....1/50(đfcm)

Vậy 1/1.2+1/3.4+1/5.6+.....1/49.50=1/26+1/27+....1/50

Bài 2

Ta có:

x+y=-6/5  (1)

x:y=3=>x=3y (2)

thay (2) vào (1) ta được:

3y+y=-6,5<=>4y=-6,5<=>y=-6,5/4=-1,625 (3)

Thay (3) vào (1) ta được:

x+(-1,625)=-6,5

<=>x=(-6,5)-(-1,625)=(-6,5)+1,625=-4,875

Vậy hai sso hữu tỉ x=-4,875 và y=-1,625

Ta có: 1 /1.2 + 1 /3.4 + 1 /5.6 + . . . + 1 /49.50

= 1 − 1/ 2 + 1/ 3 − 1 /4 + . . . + 1 /49 − 1/ 50

 = ( 1 + 1 /3 + 1 /5 + . . . + 1 /49 ) − ( 1 /2 + 1/ 4 + 1/ 6 + . . . + 1 /50 )

= ( 1 + 1/ 2 + 1 /3 + 1 /4 + . . . + 1 /49 + 1 /50 ) − 2 ( 1 /2 + 1 /4 + 1 /6 + . . . + 1/ 50 )

= ( 1 + 1/ 2 + 1 /3 + 1 /4 + . . . + 1 /50 ) − ( 1 + 1/ 2 + 1 3 + . . . + 1 25 )

= 1 /26 + 1 /27 + 1 /28 + . . . + 1/ 50

⇒ 1 /1.2 + 1 /3.4 + 1 /5.6 + . . . + 1 /49.50 = 1 26 + 1 27 + 1 28 + . . . + 1 50 (đpcm )

x:y=3

=>x=3y

vậy x+y=-6/5 ⇔3y+y=6/5

=>4y=6/5

=>y=3/10