2

Ta có:

1, \(VT=a^2+b^2=a^2+b^2+2ab-2ab=\left(a+b\right)^2-2ab=VP\left(đpcm\right)\)

Ta có: a+b=1(1)

=> (a+b)³=1

<=> \(a^3+3a^2b+3ab^2+b^3=1\)

<=> \(a^3+b^3+3ab\left(a+b\right)=1\)(2)

Từ (1)(2)=> \(a^3+b^3+3ab=1\)

<=> \(a^3+b^3=1-3ab\)(đpcm)

Đề có thiếu không ạ?

     \(a^3+b^3=3ab-1\)

\(\Rightarrow a^3+b^3+1-3ab=0\)

\(\Rightarrow\left(a+b\right)^3+1-3ab\left(a+b\right)-3ab=0\)

\(\Rightarrow\left(a+b+1\right)\left(a^2+2ab+b^2-a-b+1\right)-3ab\left(a+b\right)=0\)

\(\Rightarrow\left(a+b+1\right)\left(a^2-ab+b^2-a-b+1\right)=0\)

Mà \(a,b>0\Rightarrow a+b+1>0\)

\(\Rightarrow a^2-ab+b^2-a-b+1=0\)

\(\Rightarrow2a^2-2ab+2b^2-2a-2b+2=0\)

\(\Rightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2=0\)

\(\Rightarrow a=b=1\Rightarrow a^{2018}+b^{2019}=1+1=2\)

1) \(\left(y+3\right)^3-\left(y-1\right)^3\)

=(y+3-y+1)\(\left[\left(y+3\right)^2+\left(y+3\right)\left(y-1\right)+\left(y-1\right)^2\right]\)

=4.(\(y^2+6y+9\)+\(y^2-y+3y-3\)+\(y^2-2y+1\))

=4(\(3y^2+6y+7\))

=\(12y^2+24y+28\)

3.

\(a^3+b^3=\left(a+b\right).\left(a^2-ab+b^2\right)\)

\(=1.\left(a^2+b^2-ab\right)\) (1)

Lại có : \(a^2+b^2=\left(a+b\right)^2-2ab=1-2ab\) thay vào (1) có :

\(a^3+b^3=1.\left(1-2ab-ab\right)\)

\(=1-3ab\left(đpcm\right)\)

2.

\(A=\left(m+n\right)^3+2m^2+4mn+2n^2\)

\(=\left(m+n\right)^3+\left(\sqrt{2}m\right)^2+2.\sqrt{2}m.\sqrt{2}n+\left(\sqrt{2}n\right)^2\)

\(=\left(m+n\right)^3+\left(\sqrt{2}m+\sqrt{2}n\right)^2\)

\(=7^3+\left(\sqrt{2}.7\right)^2=343+98=441\)

( Do \(m+n=7\) )