$\\$
`a,`
`(5x+1)^2=36/49`
TH1 :
`(5x+1)^2=(6/7)^2`
`->5x+1=6/7`
`->5x=(-1)/7`
`->x=(-1)/35`
TH2 :
`(5x+1)^2=( (-6)/7)^2`
`->5x+1=(-6)/7`
`->5x=(-13)/7`
`->x=(-13)/35`
Vậy `x=(-1)/35` hoặc `x=(-13)/35`
$\\$
`b,`
`(x-2/9)^3 = (2/3)^6`
`-> (x-2/9)^3=[(2/3)^2]^3`
`->(x-2/9)^3 = (4/9)^3`
`-> x-2/9=4/9`
`->x=2/3`
Vậy `x=2/3`
$\\$
`c,`
`(8x-1)^{2x+1}=5^{2x+1`
`-> 8x-1=5`
`->8x=6`
`->x=3/4`
Vậy `x=3/4`
$\\$
`d,`
`(x-3,5)^2 + (y-1/10)^4 ≤0`
Vì `(x-3,5)^2 ≥0∀x, (y-1/10)^4 ≥0∀y`
`-> (x-3,5)^2 + (y-1/10)^4 ≥0∀x,y`
Dấu "`=`" xảy ra khi :
`(x-3,5)^2=0, (y-1/10)^4=0`
`↔ x-3,5=0, y-1/10=0`
`↔x=7/2, y=1/10`
Vậy `x=7/2, y=1/10`
