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Ioanna1Kvit
@Ioanna1Kvit
November 2021
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TARTILLO
Verified answer
1) 2^(1/2)=√2
5^(1/3)=∛5
3^(3/4)=⁴√(3)³
2) а) [ m-n^(1/2)]²+ [ m+n^(1/2)]²=m²-2m·n^(1/2)+n+m²+2m·n^(1/2)+n=
=2m²+2n
б) [ m^(1/3)+2n^(1/2)]² - [ m^(1/3)-2n^(1/2)]²=
={[ m^(1/3)+2n^(1/2)] -[ m^(1/3)-2n^(1/2)]}·
·{[ m^(1/3)+2n^(1/2)]+[ m^(1/3)-2n^(1/2)]}=
=4n^(1/2)·2m^(1/3)=8n^(1/2)·m^(1/3)
в)[ m^(1/4)-n^(1/2)]· [ m^(1/4)+n^(1/2)]=m^(1/2)-n
г) [ m^(1/2)+n]· [ m-m^(1/2)n+n²)]=m^(3/2)+n³
3) [7^(1/2)-3^(1/2)]²+[7^(1/2)+3^(1/2)]²=
=7-2[7^(1/2)·3^(1/2)]+3 +7+2[7^(1/2)·3^(1/2)]+3=14+6=20
4)(x-y)/[x^(1/2)-y^(1/2)]=[x^(1/2)+y^(1/2)]
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Answers & Comments
Verified answer
1) 2^(1/2)=√25^(1/3)=∛5
3^(3/4)=⁴√(3)³
2) а) [ m-n^(1/2)]²+ [ m+n^(1/2)]²=m²-2m·n^(1/2)+n+m²+2m·n^(1/2)+n=
=2m²+2n
б) [ m^(1/3)+2n^(1/2)]² - [ m^(1/3)-2n^(1/2)]²=
={[ m^(1/3)+2n^(1/2)] -[ m^(1/3)-2n^(1/2)]}·
·{[ m^(1/3)+2n^(1/2)]+[ m^(1/3)-2n^(1/2)]}=
=4n^(1/2)·2m^(1/3)=8n^(1/2)·m^(1/3)
в)[ m^(1/4)-n^(1/2)]· [ m^(1/4)+n^(1/2)]=m^(1/2)-n
г) [ m^(1/2)+n]· [ m-m^(1/2)n+n²)]=m^(3/2)+n³
3) [7^(1/2)-3^(1/2)]²+[7^(1/2)+3^(1/2)]²=
=7-2[7^(1/2)·3^(1/2)]+3 +7+2[7^(1/2)·3^(1/2)]+3=14+6=20
4)(x-y)/[x^(1/2)-y^(1/2)]=[x^(1/2)+y^(1/2)]