Đáp án:
ta có
x+1/3=3.(x+1)/3.3= 3x+3/9
y+2/4=2.(y+2)/4.2=2y+4/8
z-3/5=4.(z-3)/5.4=4z-12/20
=> 3x+3/9=2y+4/8=4z-12/20
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
3x+3/9=2y+4/8=4z-12/20=3x+3+2y+4+4z-12/9+8+20=(3x+2y+4z)+(3+4-12)/37
= 47-5/37=42/37
*x+1/3=42/37=> 37.(x+1)=3.42
37.(x+1)=126
x+1=126:37
x+1=126/37
x=126/37-1=89/37
*y+2/4=42/37 ; * z-3/5=42/37
=>37.(y+2)=42.4 ; 37.(z-3)=5.47
37.(y+2)=168 ; 37.(z-3)=235
y+2=168:37 ; z-3=235:37
y+2=168/37 ; z-3=235/37
y=168/37-2=84/37 ; z=235/37 +3=346/37
Vậy x=89/37; y=84/37; z=346/37
