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`a,`
`(5x+1)^2 = 36/49`
`->` \(\left[ \begin{array}{l}(5x+1)^2=(\dfrac{6}{7})^2\\(5x+1)^2=(\dfrac{-6}{7})^2\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}5x+1=\dfrac{6}{7}\\5x+1=\dfrac{-6}{7}\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}5x=\dfrac{-1}{7}\\5x=\dfrac{-13}{7}\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}x=\dfrac{-1}{35}\\x=\dfrac{-13}{35}\end{array} \right.\)
Vậy `x=(-1)/35` hoặc `x=(-13)/35`
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`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`
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`c,`
`(x-1,5)^2=9`
`->` \(\left[ \begin{array}{l}(x-1,5)^2=3^2\\(x-1,5)^2=(-3)^2\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}x-1,5=3\\x-1,5=-3\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}x=4,5\\x=-1,5\end{array} \right.\)
Vậy `x=4,5` hoặc `x=-1,5`
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`d,`
`(x-2)^3=64`
`-> (x-2)^3=4^3`
`->x-2=4`
`->x=6`
Vậy `x=6`
