51.
a) $10^{2}$. $3^{2}$ + $5^{3}$ = $100^{}$. $9^{}$ + $125^{}$ = $900^{}$ + $125^{}$ = $1025^{}$
b) $12^{4}$ : $12^{3}$ + $7^{3}$ : $7^{}$ = $12^{1}$ + $7^{2}$ = $12^{}$ + $49^{}$ = $61^{}$
c) $(21 ^{}$ + $19)^{4}$ : $40^{2}$ + $31^{}$ - $1600^{}$
= $40^{4}$ : $40^{2}$ + $31^{}$ - $1600^{}$
= $40^{2}$ + $31^{}$ - $1600^{}$
= $1600^{}$ + $31^{}$ - $1600^{}$
= $31^{}$
d) $(572^{}$ - $72)^{2}$ + $3^{3}$. $2^{}$ - $20000^{}$
= $500^{2}$ + $27^{}$. $2^{}$ - $20000^{}$
= $250000^{}$ + $54^{}$ - $20000^{}$
= $230054^{}$
52.
a) $46^{}$.$(2020^{}$ + $40^{}$. $102)^{}$ + $54^{}$. $(2020^{}$ + $40^{}$. $102)^{}$
= $(2020^{}$ + $40^{}$. $102^{})$. $(46^{}$ + $54)^{}$
= $(2020^{}$ + $4080^{})$. $100^{}$
= $6100^{}$. $100^{}$
= $610000^{}$
b) $2345^{2}$ + $[11^{2}$ - $(575^{}$ - $572)^{3}$ + $6]^{2}$
= $2345^{}$ + $[121^{}$ - $3^{3}$ + $6]^{2}$
= $2345^{}$ + $[121^{}$ - $27^{}$ + $6]^{2}$
= $2345^{}$ + $100^{2}$
= $2345^{}$ + $10000^{}$
= $12345^{}$
c) $102^{}$ + $101^{}$ - $100^{}$ - $99^{}$ + $98^{}$ + $97^{}$ - $96^{}$ - $95^{}$ + ... +$6^{}$ + $5^{}$ - $4^{}$ - $3^{}$ + $2^{}$ + $1^{}$
= $(102^{}$ + $101^{}$ - $100^{}$ - $99)^{}$ + $(98^{}$ + $97^{}$ - $96^{}$ - $95)^{}$ + ... +$(6^{}$ + $5^{}$ - $4^{}$ - $3)^{}$ + $2^{}$ + $1^{}$
= $4^{}$ + $4^{}$ + ... + $4^{}$ + $3^{}$ (25 số hạng 4)
= $4^{}$. $25^{}$ + $3^{}$
= $100^{}$ + $3^{}$
= $103^{}$
Vậy $18576^{}$ : `\{`$105^{0}$ + $[2^{}$. $(102^{}$ + $101^{}$ - $100^{}$ - $99^{}$ + $98^{}$ + $97^{}$ - $96^{}$ - $95^{}$ + ... +$6^{}$ + $5^{}$ - $4^{}$ - $3^{}$ + $2^{}$ + $1)^{}$ - $201]^{}$`\}`$^{3}$
= $18576^{}$ : `\{`$1^{}$ + $2^{}$. $103^{}$ - $201^{}$`\}`$^{3}$
= $18576^{}$ : `\{`$1^{}$ + $206^{}$ - $201^{}$`\}`$^{3}$
= $18576^{}$ : $6^{3}$
= $18576^{}$ : $216^{}$
= $86^{}$
Chúc học tốt! ^v^
- Mong bn cho câu trả lời hay nhất + 5* + tym.
