Đây là lần thứ 2 bn ghi cái đề này? Nhg chưa có lần nào đúng cả!

Sửa đề: \(\left\{{}\begin{matrix}a^3-3a^2+5a-17=0\\b^3-3b^2+5b+11=0\end{matrix}\right.\) (*)

Từ HPT (*) <=> \(\left\{{}\begin{matrix}a^3-3a^2+3a-1+2a-16=0\\b^3-3b^2+3b-1+12=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}\left(a-1\right)^3+2\left(a-8\right)=0\left(1\right)\\\left(b-1\right)^3+2\left(b+6\right)=0\left(2\right)\end{matrix}\right.\)

Cộng (1) với (2) vế theo vế ta có:

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

<=> \(\left(a+b-2\right)\left[\left(a-1\right)^2+\left(b-1\right)^2-\left(a-1\right)\left(b-1\right)+2\right]=0\)

Mà \(\left(a-1\right)^2+\left(b-1\right)^2-\left(a-1\right)\left(b-1\right)+2>0\)

=> \(a+b-2=0\)

=> \(a+b=2\)

Ta có: \(a^2+b^2+\left(a+b\right)^2=c^2+d^2+\left(c+d\right)^2\)

=> \(a^2+b^2+a^2+b^2+2ab=c^2+d^2+c^2+d^2+2cd\)

=> \(a^2+b^2+ab=c^2+d^2+cd\)

=> \(\left(a^2+b^2+ab\right)^2=\left(c^2+d^2+cd\right)^2\)

=> \(a^4+b^4+a^2b^2+2a^2b^2+2a^3b+2b^3a=c^4+d^4+c^2d^2\)

\(+2c^2d^2+2c^3d+2cd^3\)

=> \(2a^4+2b^4+6a^2b^2+4a^3b+4ab^3=2c^4+2d^4+6c^2d^2\)

\(+4c^3d+4cd^3\)

=> \(a^4+b^4+\left(a+b\right)^4=c^4+d^4+\left(c+d\right)^4\)

=> đpcm

Xem lại đề đi!

Ta có: \(\left\{{}\begin{matrix}a^3-3ab^2=19\\b^3-3a^2b=98\end{matrix}\right.\) => \(\left\{{}\begin{matrix}\left(a^3-3ab^2\right)^2=19^2=361\\\left(b^3-3a^2b\right)^2=98^2=9604\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}a^6-6a^4b^2+9a^2b^4=361\\b^6-6a^2b^4+9a^4b^2=9604\end{matrix}\right.\)

=> \(a^6+b^6+\left(9a^2b^4-6a^2b^4\right)+\left(9b^2a^4-6a^4b^2\right)=9965\)

=> \(a^6+3a^2b^4+3a^4b^2+b^6=9965\)

=> \(\left(a^2+b^2\right)^3=9965\)

=> \(a^2+b^2=\sqrt[3]{9965}\)

a) Ta có: \(P=\left(x+2\right)^3+\left(x-2\right)^3-2x\left(x^2+12\right)\)

\(P=x^3+6x^2+12x+8+x^3-6x^2+12x-8-2x^3-24x\)

\(P=0\)

=> P không phụ thuộc vào biến

b) Ta có: \(Q=\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)

=> \(Q=x^3-3x^2+3x-1-x^3-3x^2-3x-1+6x^2-6\)

=> \(Q=-8\)

=> Q không phụ thuộc vào biến