Đáp án:
$\\$
`a,`
Xét vế trái có :
`32^8 × 625^{10}`
`= (2^5)^8 × (5^4)^{10}`
`= 2^{40} × 5^{40}`
`= (2×5)^{40}`
`= 10^{40}` (Bằng với vế phải)
`-> 32^8 × 625^{10} = 10^{40}` (đpcm)
$\\$
`b,`
`A = (4^2 × 25^2 + 32 × 125)/(2^3 × 5^2)`
`-> A = ( (2^2)^2 × (5^2)^2 + 2^5 × 5^3)/(2^3 × 5^2)`
`->A = (2^4 × 5^4 + 2^5 × 5^3)/(2^3 × 5^2)`
`-> A = (2^3 × 5^2 × (2 × 5^2 + 2^2 × 5) )/(2^3 × 5^2)`
`-> A = 2 × 5^2 + 2^2 × 5`
`-> A= 2 × 25 + 4×5`
`->A = 50 + 20`
`->A=70`
Vậy `A=70`
