Đáp án:

Bài 1:

`a) x=5`

`b) x=1`

`c) x=4` hoặc `x=5`

`d)` `x∈{2; 3; 4; 5}`

Bài 2:

`a)` `2/3`

`b)` `1/32`

Giải thích các bước giải:

Bài 1:

`a)` `3^(x-2)+5*3^(x-2)=162`

`⇒ 3^(x-2)*(1+5)       =162`

`⇒ 3^(x-2)                 =162:6`

`⇒ 3^(x-2)                 =27`

`⇒ 3^(x-2)                 =3^3`

`⇒     x-2                   =3`

`⇒    x                       =3+2`

`⇒    x                       =5`

Vậy `x=5`

`b)` `1^x=x^15`

`⇒        x=1`

Vậy `x=1`

`c)` `(x-4)^2021=(x-4)^2018`

`(x-4)^2021-(x-4)^2018=0`

`⇒(x-4)^2018*[(x-4)^3-1]=0`

`⇒`\(\left[ \begin{array}{l}(x-4)^{2018}=0\\(x-4)^3-1=0\end{array} \right.\) `⇒`\(\left[ \begin{array}{l}(x-4)^{2018}=0^{2018}\\(x-4)^3=1\end{array} \right.\) `⇒`\(\left[ \begin{array}{l}x-4=0\\(x-4)^3=1^3\end{array} \right.\) `⇒`\(\left[ \begin{array}{l}x=4\\x-4=1\end{array} \right.\) `⇒`\(\left[ \begin{array}{l}x=4\\x=5\end{array} \right.\) 

Vậy `x=4` hoặc `x=5`

`d) 8≤x^3<216`

`⇒2^3≤x^3<6^3`

`⇒2≤x<6`

`⇒x∈{2; 3; 4; 5}`

Vậy `x∈{2; 3; 4; 5}`

Bài 2:

`a)` `(3^5*7+3^7)/(9^4-27^2)`

`=(3^5*(7+3^2))/((3^2)^4-(3^3)^2)`

`=(3^5*(7+9))/(3^8-3^6)`

`=(3^5*16)/(3^5*(3^3-3))`

`=16/(27-3)`

`=16/24`

`=2/3`

`b)` `(4^7*2^8)/(3*2^15*16^2-5*2^22)`

`=((2^2)^7*2^3)/(3*2^15*(2^4)^2-5*2^22)`

`=(2^(2*7+3))/(3*2^(15+4*2)-5*2^22)`

`=(2^17)/(3*2^23-5*2^22)`

`=(2^17)/(2^22*(3*2-5))`

`=(1)/(2^(22-17)*1)`

`=1/(2^5)`

`=1/32`

_______________________

#phamuyen15032008