+) `(x+1)^{x+10}=(x+1)^{x+4}`
`<=>(x+1)^{x+10}-(x+1)^{x+4}=0`
`<=>(x+1)^{x+4}.[(x+1)^6-1]=0`
$⇔\left[\begin{array}{l}(x+1)^{x+4}=0\\(x+1)^6-1=0\end{array}\right.$
$⇔\left[\begin{array}{l}x+1=0\\(x+1)^6=1\end{array}\right.$
$⇔\left[\begin{array}{l}x=-1\\x+1=1\\x+1=-1\end{array}\right.$
$⇔\left[\begin{array}{l}x=-1\\x=0\\x=-2\end{array}\right.$
Vậy `x=-2;x=-1;x=0`
+) Chưa cho 2020^(x-2).(2x+3)=??
+) `(3/ 4)^x={2^8}/{3^4}`
`<=>{3^x}/{4^x}={2^8}/{3^4}`
`<=>3^x . 3^4 =4^x . 2^8 `
`<=>3^{x+4}=4^x . (2^2)^4`
`<=>3^{x+4}=4^{x+4}`
`<=>(3/ 4)^{x+4}=1`
`<=>x+4=0`
`<=>x=-4`
Vậy $x=-4$
+) `[(-0,5)^3]^x=1/{64}`
`<=>(-1/ 2)^{3x}=({-1}/2)^6`
`<=>3x=6`
`<=>x=6:3`
`<=>x=2`
Vậy $x=2$
