tìm x a, (x+1)^x+10=(x+1)^x+4 2020^(x-2).(2x+3) (3/4)^x=2^8/3^4 [(-0,5)^3]^x=1/64 giúp minh với mai mình nộp rùi gạch chéo là phân số

+)  `(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 $x=2$