Question

已知xy的值满足x+1/2=y+3/4=x+y/5，求代数式3x+2y+1/x+2y+3的值是多少

Answer 1

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
则,2x+2=y+3
2x-y=1 ......(1)
再将已知条件进行通分,
(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
将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

Answer 2

解：根据已知条件可知：
x+1/2=x+y/5
即得：y=5/2
代入x+1/2=y+3/4
得x=11/4
把x、y的数值代入所求代数式得
3x+2y+1/x+2y+3
=3*11/4+2*5/2+4/11+2*5/2+3
=21又27/44

Answer 3

x+1/2=y+3/4=x+y/5
x = 11/4
y = 5/2
3x+2y+1/x+2y+3 = 21 + 27/44

Answer 4

x+1/2=x+y/5，解得y=5/2，x=5/2+3/4-1/2=11/4
3x+2y+1/x+2y+3
=3x11/4+2x5/2+4/11+2x5/2+3
=21+27/44