b)
`(3/5 -2/3x)^3=-64/125`
`⇒(3/5 -2/3x)^3=(-4/5)^3`
`⇒3/5-2/3x=-4/5`
`⇒-2/3x=-4/5-3/5`
`⇒-2/3x=-7/5`
`⇒x=-7/5:(-2/3)`
`⇒x=21/10 `
c)
`(x-3)^10=(x-3)^30`
`⇒(x-3)^30-(x-3)^10=0`
`⇒(x-3)^10 [(x-3)^20-1]=0`
`⇒`\(\left[ \begin{array}{l}(x-3)^{10}=0\\(x-3)^{20}-1=0\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x-3=0\\(x-3)^{20}=1\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x=3\\x-3=1\\x-3=-1\end{array} \right.\)
`⇒`\(\left[ \begin{array}{l}x=3\\x=4\\x=2\end{array} \right.\)
d)
`(x+1,5)^8+(2,7-y)^12=0`
Ta có:
`(x+1,5)^8≥0∀x`
`(2,7-y)^12≥0∀y`
`⇒(x+1,5)^8+(2,7-y)^12≥0`
Dấu "`=`" xảy ra khi `(x+1,5)^8=0` và `(2,7-y)^12=0`
`⇒x+1,5=0` và `2,7-y=0`
`⇒x=-1,5` và `y=2,7`
