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`a,`
`(3x-1/5)^{200}+(2/5y + 4/7)^{1968}=0`
Vì `(3x-1/5)^{200} ≥0∀x, (2/5 y+4/7)^{1968} ≥0∀y`
`-> (3x-1/5)^{200}+(2/5y + 4/7)^{1968} ≥0∀x,y`
Dấu "`=`" xảy ra khi :
`(3x-1/5)^{200}=0, (2/5y + 4/7)^{1968}=0`
`↔ 3x-1/5=0, 2/5 y+4/7=0`
`↔ x=1/15,y=(-10)/7`
Vậy `(x;y)=(1/15; (-10)/7)`
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`b,`
`(4/7 x-1)^2 +( (-2)/3 y+4)^{68} ≤0`
Vì `(4/7 x-1)^2 ≥0∀x,((-2)/3 y+4)^{68} ≥0∀y`
`->(4/7 x-1)^2+((-2)/3 y+4)^{68} ≥0∀x,y`
Dấu "`=`" xảy ra khi :
`(4/7 x-1)^2=0,((-2)/3 y+4)^{68}=0`
`↔ 4/7 x-1=0, (-2)/3 y+4=0`
`↔x=7/4, y=6`
Vậy `(x;y)=(7/4;6)`
