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HollywooD2016
@HollywooD2016
July 2022
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3 и 6 плиз... срочно
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3) x²-4x-4>0
x²-4x-4=0
x1=2-√(4+4)=2-2√2 x1=2+√(4+4)=2+2√2
+ - +
-------------(2-2√2)---------------------(2+2√2)---------------
x∈(-∞;2-2√2 )∪(2+2√2;∞)
6) f(x)=√(2x² -x-1)
область определения функции (2x² -x-1)≥0
2x² -x-1=0
x1=(1-√(1+8))/4=(1-3)/4=-1/2
x2=(1+√(1+8))/4=(1+3)/4=1
+ - +
----------------------(-1/2]--------------------[1)------------
x∈(-∞;-1/2]∪[1;∞)
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Answers & Comments
Verified answer
3) x²-4x-4>0x²-4x-4=0
x1=2-√(4+4)=2-2√2 x1=2+√(4+4)=2+2√2
+ - +
-------------(2-2√2)---------------------(2+2√2)---------------
x∈(-∞;2-2√2 )∪(2+2√2;∞)
6) f(x)=√(2x² -x-1)
область определения функции (2x² -x-1)≥0
2x² -x-1=0
x1=(1-√(1+8))/4=(1-3)/4=-1/2
x2=(1+√(1+8))/4=(1+3)/4=1
+ - +
----------------------(-1/2]--------------------[1)------------
x∈(-∞;-1/2]∪[1;∞)