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SuperStar1111111
@SuperStar1111111
September 2021
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Представте многочлен в виде квадрата двучлена 1) 4с^2+4сd+d^2 2) t^2z^2+2tz+1
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Гоша68
Verified answer
1)=(2c)^2+2*(2c)*d+(d)^2=(2c+d)^2
2)=(tz)^2+2*(t)*(z)+1=(t*z+1)^2
3 votes
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SuperStar1111111
спасибо!
SVETA1951TIT
Verified answer
= (2С + Д)^2
= (tz + 1)^2
0 votes
Thanks 1
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Answers & Comments
Verified answer
1)=(2c)^2+2*(2c)*d+(d)^2=(2c+d)^22)=(tz)^2+2*(t)*(z)+1=(t*z+1)^2
Verified answer
= (2С + Д)^2= (tz + 1)^2