Question

BT1: Chứng minh rằng nếu:

\(a^3+b^3+c^3=3abc\)

Và \(a,b,c\) dương thì a=b=c

BT2: Nếu \(a^4+b^4+c^4+d^4=4abcd\)

Và \(a,b,c,d>0\). Chứng minh a=b=c=d

BT3: Cho \(a^2+b^2=1\), \(c^2+d^2=1\), \(ac+bd=0\)

Chứng minh: \(ab+cd=0\)

Answer 1

Lm dc hết chưa bn ơi.

Answer 2

câu 2

a^4 + b^4 + c^4 + d^4 = 4abcd

<=> \(a^4-2a^2b^2+b^4+c^4-2c^2d^2+d^4+2a^2b^2-4abcd+2b^2d^2=0\)

<=> \(\left(a^2-b^2\right)^2+\left(c^2-d^2\right)^2+2\left(ab-cd\right)^2=0\)

<=> \(\left\{{}\begin{matrix}a^2=b^2\\c^2=d^2\\ab=cd\end{matrix}\right.\Leftrightarrow a=b=c=d\)

Answer 3

Ta có \(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

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

\(=\left(a+b+c\right)[\left(a+b\right)^2-c\left(a+b\right)+c^2]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-bc-ca+c^2-3ba\right)\)

hay \(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

Ta lại có \(a^3+b^3+c^3=3abc\) khi \(\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

khi a,b,c dương thì a+b+c ko thể =0 nên a=b=c