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\(a^3+b^3=c\left(3ab-c^2\right)\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left[2a^2+2b^2+2c^2-2ab-2bc-2ca\right]=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Rightarrow\orbr{\begin{cases}a+b+c=0\left(loai\right)\\a=b=c\end{cases}}\)
Mà a + b + c = 3 nên a = b = c = 1
Khi đó \(A=672.\left(1+1+1\right)+2=672.3+2=2018\)
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a = 2
b = 3
rồi tính ra nhé
ai k mình mình k lại cho
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a,= a\(^2\)+2a+b\(^2\)-2b-2ab+37
=a\(^2\)-2ab+b\(^2\)+2a-2b+37
=(a-b)\(^2\)+2(a-b)+37
⇒5\(^2\)+2.5+37= 25+10+37= 72
b,= a\(^3\)+a\(^2\)-b\(^3\)+b\(^2\)+ab-3a\(^2\)b+3ab\(^2\)-3ab-95
=a\(^3\)-3a\(^2\)b+3ab\(^2\)-b\(^3\)+a\(^2\)-2ab+b\(^2\)-95
=(a-b)\(^3\)+(a-b)\(^2\)-95
⇒5\(^3\)+5\(^2\)-95= 125+25-95= 60
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\(\left(a-b\right)\left(a^2+3ab+b^2\right)+\left(a+b\right)^3+ab\left(b-a\right)\)
\(=\left(a-b\right)\left(a^2+3ab+b^2-ab\right)+\left(a+b\right)^3\)
\(=\left(a-b\right)\left(a+b\right)^2+\left(a+b\right)^3=\left(a+b\right)^2\left(a-b+a+b\right)=2a\left(a+b\right)^2\)
\(\left(a-b\right)\left(a^2+3ab+b^2\right)+\left(a+b\right)^3+ab\left(b-a\right)\)
\(=\left(a-b\right)\left(a^2+3ab-ab+b^2\right)+\left(a+b\right)^3\)
\(=\left(a-b\right)\left(a^2+2ab+b^2\right)+\left(a+b\right)^3\)
\(=\left(a-b\right)\left(a+b\right)+\left(a+b\right)^3\)
\(=\left(a+b\right)^2\left(a+b+a-b\right)\)
\(=\left(a+b\right)^2.2a\)
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a, chắc bạn chép nhầm đề rồi đó nếu mà là 3ab thì k làm đc đâu
M=a3 + a2 - b3 + b2 + 3ab2 -2ab +3ab2
= (a-b)3 +(a-b)2
= 343+49=392
b, P= -(3x+4x2+1/4x-2014)
= - [ (2x)2 -4x+1 +x +1/4x - 2015]
= -[ (2x-1)2- (2x-1)2/4x +1 -2015]
Max P = 2014 X=1/2
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Ta có: a2 + b2 = (a + b)2 - 2ab = 62 - 2.4 = 28
a4 + b4 = (a2 + b2)2 - 2a2b2 = 282 - 2.42 = 752
ko bt tự làm đê!!!!