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\(N=a^3+b^3+3ab\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\)
=1
\(M=\left(a^2+b^2+2-a^2-b^2+2\right)\left[\left(a^2+b^2+2\right)^2+\left(a^2+b^2+2\right)\left(a^2+b^2-2\right)+\left(a^2+b^2-2\right)^2\right]-12\left(a^2+b^2\right)^2\\ M=4\left(a^4+b^4+4+4a^2+4b^2+2a^2b^2+\left(a^2+b^2\right)^2-4+a^4+b^4+4-4a^2-4b^2+2a^2b^2\right)-12\left(a^4+2a^2b^2+b^4\right)\\ M=4\left(3a^4+3b^4+4+6a^2b^2\right)-12\left(a^4+2a^2b^2+b^4\right)\\ M=4\left(3a^4+3b^4+4+6a^2b^2-3a^4-6a^2b^2-3b^4\right)\\ M=4\cdot4=164\)
b: Ta có: \(N=a^3+b^3+3ab\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\)
\(=1-3ab+3ab\)
=1
Ta có: a + b = 1
M = a3 + b3 + 3ab(a2 + b2) + 6a2b2(a + b)
= (a + b)3 - 3ab(a + b) + 3ab[(a + b)2 - 2ab] + 6a2 b2 (a + b)
= 1 - 3ab + 3ab(1 - 2ab) + 6a2 b2
= 1 - 3ab + 3ab - 6a2 b2 + 6a2 b2
= 1
nhwos tick nha :D
M=(a+b)(a2-ab+b2)+3ab(1-2ab)+6a2b2
M=a2-ab+b2+3ab
M=(a+b)2=1
Ta có: a + b = 1
M = a3 + b3 + 3ab(a2 + b2) + 6a2b2(a + b)
= (a + b)3 - 3ab(a + b) + 3ab[(a + b)2 - 2ab] + 6a2 b2 (a + b)
= 1 - 3ab + 3ab(1 - 2ab) + 6a2 b2
= 1 - 3ab + 3ab - 6a2 b2 + 6a2 b2
= 1
Nhóm vào , ta có :
\(\left(a+1\right)^3+\left(b+1\right)^3+a+b+1+1=0\)
Đến đây áp dụng HĐT là ra
\(a^3+3a^2+3a+1+b^3+3b^2+3b+1+a+b+2=0\)
\(\Leftrightarrow\left(a+1\right)^3+\left(b+1\right)^3+a+b+2=0\)
\(\Leftrightarrow\left(a+b+2\right)\left(\left(a+1\right)^2-\left(a+1\right)\left(b+1\right)+\left(b+1\right)^2\right)+a+b+2=0\)
\(\Leftrightarrow\left(a+b+2\right)\left(\left(a+1\right)^2-\left(a+1\right)\left(b+1\right)+\dfrac{\left(b+1\right)^2}{4}+\dfrac{3\left(b+1\right)^2}{4}+1\right)=0\)
\(\Leftrightarrow\left(a+b+2\right)\left(\left(a+1-\dfrac{b+1}{2}\right)^2+\dfrac{3\left(b+1\right)^2}{4}+1\right)=0\)
\(\Leftrightarrow a+b+2=0\) (ngoặc to phía sau luôn dương)
\(\Leftrightarrow a+b=-2\)
\(\Rightarrow M=2018\left(a+b\right)^2=2018.\left(-2\right)^2=8072\)