Tính A=$16^{3}$.$3^{10}$+$120^{6}$:$4^{6}$.$3^{12}$+$6^{11}$ 09/07/2021 Bởi Isabelle Tính A=$16^{3}$.$3^{10}$+$120^{6}$:$4^{6}$.$3^{12}$+$6^{11}$
Đáp án: `A=2^6.(3^10).32965` Giải thích các bước giải: `A=16^3.(3^10)+120^6:(4^6).3^12+6^11` `A=(2^4)^3.(3^10)+(2^3.(3).5)^6:(2^2)^6.(3^12)+(2.3)^11` `A=2^12.(3^10)+2^6.(3^18).5+2^11.(3^11)` `A=2^6.(3^10)(2^6+3^8.5+2^5.(3))` `A=2^6.(3^10).32965` Bình luận
Đáp án:
`A=2^6.(3^10).32965`
Giải thích các bước giải:
`A=16^3.(3^10)+120^6:(4^6).3^12+6^11`
`A=(2^4)^3.(3^10)+(2^3.(3).5)^6:(2^2)^6.(3^12)+(2.3)^11`
`A=2^12.(3^10)+2^6.(3^18).5+2^11.(3^11)`
`A=2^6.(3^10)(2^6+3^8.5+2^5.(3))`
`A=2^6.(3^10).32965`
Đáp án:
A=(4^3)^2+2.4^3.3^12+(3^11)^2
A=(4^3+3^11)^2
A=31403738521