已知xy的值滿足x+1/2=y+3/4=x+y/5,求代數(shù)式3x+2y+1/x+2y+3的值是多少

數(shù)學(xué)人氣：864 ℃時(shí)間：2019-08-20 10:51:03

優(yōu)質(zhì)解答

已知(x+1)/2=(y+3)/4=(x+y)/5,
得出,
(x+1)/2=(y+3)/4
(2x+2)/4=(y+3)/4
則,2x+2=y+3
2x-y=1 .(1)
再將已知條件進(jìn)行通分,
(x+1)/2=(y+3)/4=(x+y)/5
(10x+10)/20=(5y+15)/20=(4x+4y)/20
得出,(10x+10)+(5y+15)=2*(4x+4y)
10x+5y+25=8x+8y
2x-3y=-25 .(2)
(1)-(2)得出,
2y=26,y=13
代入(1)得出,
x=(y+1)/2=(13+1)/2=7
將x=7,y=13代入(3x+2y+1)/(x+2y+3)
(3x+2y+1)/(x+2y+3)
=(3*7+2*13+1)/(7+2*13+3)
=48/36
=4/3