$#dhlv$

`9.`

`2^x + 3^x = 5^x`

`(2^x + 3^x)/5^x = 5^x/5^x = 1`

`=> (2/5)^x + (3/5)^x = 1`

`***` TH`1`: Nếu `x > 1`

`=> (2/5)^x > (2/5)^1`

`=> (3/5)^x > (3/5)^1`

`=> (2/5)^x + (3/5)^x > 2/5 + 3/5 > 1`

`=>` Loại 

`***` TH`2`: Nếu `x < 1`

`=> (2/5)^x  < (2/5)^1`

`=> (3/5)^x < (3/5)^1`

`=> (2/5)^x + (3/5)^x < 2/5 + 3/5 < 1` 

`=>` Loại

Vậy `x` thoả mãn có giá trị là `1`

`--------------`

`10.`

`2^y. 3^2x = 6^(x+1)`

`2^y. 3^2x = 3^(x+1). 2^(x+1)`

`=> x + 1 = y`

`x + 1 = 2x -> x = 1`

`=> y = 1 + 1 = 2`

Vậy `x = 1; y= 2`