\(< =>42-x-1\cdot3=9< =>42-x-3=9< =>42-x=12< =>x=10\)

`+`

`1^3+2^3`
`=1+8`
`=9`
`1^5+2^3+3^3`
`=1+8+27`
`=9+27`

`=36`

`1^3+2^3+3^3`
`=1+8+27`

`=9+27`

`=36`
 

kết quả là bao nhiêu bạn

[ 7600 - ( 40 : 2 + 35³ ) ] : 5

= [ 7600 - ( 20 + 1225 ) ] : 5

= [ 7600 - 1245 ] : 5

= 6355 : 5

= 1271

`[7600-(40:2+35^3)]:5`

`=[7600-(40:2+42875)]:5`

`=[7600-(20+42875)]:5`

`=[7600-42895]:5`
`=-35295:5`

`=-7059`
 

Giải dùm tui iii

a) Ta có : 2021⁰ = 1 ; 1²⁰²¹ = 1

Vậy 1 = 1 nên 2021⁰ = 1²⁰²¹ 

b) Ta có : 21¹⁵ = ( 3 . 15 )¹⁵ = 3¹⁵ . 15¹⁵ 

27⁵ . 49⁸ = ( 3³ )⁵ . ( 7² )⁸ = 3¹⁵ . 14⁸

Vì 15¹⁵ > 14⁸ nên 3¹⁵ . 15¹⁵ > 3¹⁵ . 14⁸ hay 21¹⁵ > 27⁵ . 49⁸ 

c) Ta có : 3²ⁿ = ( 3² )ⁿ = 9ⁿ

2³ⁿ = ( 2³ )ⁿ = 8ⁿ

Vì 8 < 9 nên 9ⁿ > 8ⁿ hay 3²ⁿ > 2³ⁿ 

1) \(x^2+33=69\)

\(x^2=36\)

\(x=\pm6\)

2) \(\left(x-2\right)^2+14=63\)

\(\left(x-2\right)^2=49\)

\(x-2=\pm7\)

\(\left[{}\begin{matrix}x=9\\x=-5\end{matrix}\right.\)

3) \(15+\left(x+1\right)^3=42\)

\(\left(x+1\right)^3=27\)

\(x+1=3\)

\(x=2\)

câu cuối chưa hết đề nha e

.

\(A=27+x;x\inℕ\)

a.

\(A⋮3\Leftrightarrow27+x⋮3\Rightarrow x⋮3;x\inℕ\)

b

\(A⋮̸3\) \(\Leftrightarrow27+x⋮̸3\) \(\Rightarrow x⋮̸3\) 

Ta có:

\(x^2=x\\ \Leftrightarrow x^2-x=0\\ \Leftrightarrow x\left(x-1\right)=0\\ \left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Đs....