x + y * 1 * 0 = 9

<=> x + 0 = 9

=> x = 9

Tôi đồng ý với Help you solve math

\(\)20+30+40+50+60=

(20+60)+(30+50)+4=

80+80+40=

160+40=

200

10+20+3+40+50+60=

(10+60)+(20+50)+(30+40)=

70+70+70=

=210

20+30+40+50+60+70+80=

(20+80)+(30+70)+(40+60)+50=

100+100+100+50=

300+50=

350

20+30+40+50+60+70=

=(20+70)+(30+60)+(40+50)

=90+90+90

=170

20+70)+(30+60)+(40+50)

1+2+3+4+5+6+7+8=

=(1+8)+(2+7)+(3+6)+(4+5)

=9+9+9+9

=36

10+20+30+40+60+50=

=(10+60)+(20+50)+(30+40)=

=70+70+70

=210

x+20+30+10+40+50=170

x=170-(20-30)-(10-40)-50

x=170-50-50-50

x=20

We have: $84=2\cdot 3\cdot 14$ is the least distinct set of possible integers.

So their sum is: $2+3+14=19$

$\mathrm{Answer:19}$

$2+3+$

x+10+20+30+40+50=a

x=a-50-40-30-20-10

x=a-150

10+20+30+40+50+60+70+80+90=

=(10+90)+(20+80)+(30+70)+(40+60)+50

=100+100+100+100+50

=450

I play soccer and play games

6 numbers

20+30+40+50+60+70

=(20+70)+(30+60)+(40+50)

=90 + 90 +90

=270

10+20+30+40+50+60

=(1+2+3+4+5+6) x10

=21x10

=210

=

we have:

  \(\dfrac{10+10+10}{3}\) = \(\dfrac{30}{3}\)= 10

answer:10

We have:

\(\dfrac{6+8+16}{3}\)=10

answer: 10

10+20+30+40+50+60+70+80+90

=(10+90)+(20+80)+(30+70)+(40+60)+50

=100+100+100+100+50

=400+50

=450

10+20+60+40+80+90

=(10+90)+(20+80)+(60+40)

=100+100+100

=300