HELP ME! Надо вычислить.. Created by lerusikrom. algebra-ru

HELP ME! Надо вычислить...

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пусть (29+4*7^0,5)^(1/6)=А
пусть (1+2*7^0,5)^(1/3)=((1+2*7^0,5)^2)^(1/6)=((1+2*7^0,5)^2)^(1/6)=В

(1+2*7^0,5)^2=1+2*1*2*7^0,5+4*7=29+4*7^0,5

В=((1+2*7^0,5)^2)^(1/6)=(29+4*7^0,5)^(1/6)=А

((29+4*7^0,5)^(1/6)-(1+2*7^0,5)^(1/3))*(231+12^(1/7))^(1/10)=
=(A-B)*(231+12^(1/7))^(1/10)= 0*(231+12^(1/7))^(1/10) = 0 - это ответ

1 votes Thanks 1

van9l Пусть (29+4*7^0,5)^(1/6)=А
пусть (1+2*7^0,5)^(1/3)=((1+2*7^0,5)^2)^(1/6)=((1+2*7^0,5)^2)^(1/6)=В

(1+2*7^0,5)^2=1+2*1*2*7^0,5+4*7=29+4*7^0,5

В=((1+2*7^0,5)^2)^(1/6)=(29+4*7^0,5)^(1/6)=А

((29+4*7^0,5)^(1/6)-(1+2*7^0,5)^(1/3))*(231+12^(1/7))^(1/10)=
=(A-B)*(231+12^(1/7))^(1/10)= 0 *(231+12^(1/7))^(1/10) = 0 - это ответ

0 votes Thanks 1

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