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Theo đề bài: ab+bc+ca=0
=> \(\frac{1}{c}+\frac{1}{b}+\frac{1}{a}=0\)(chia 2 vế cho abc)
<=> \(\frac{1}{c^3}+\frac{1}{b^3}+\frac{1}{a^3}=3\cdot\frac{1}{abc}\)(1)
( Áp dụng tính chất x+y+z=0 suy ra \(x^3+y^3+z^3=3zxy\)- Bạn tự Cm)
Ta có: P=\(\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}=\)\(\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)(2)
Từ (1)(2)=> P=abc\(\cdot3\cdot\frac{1}{abc}\)=3
\(S=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc}{2a^2+2b^2+2c^2-2ab-2bc-2ac}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{2a^2+2b^2+2c^2-2ab-2bc-2ac}\)
\(=\dfrac{3\cdot\left(2a^2+2b^2+2c^2-2ab-2bc-2ac\right)\cdot\dfrac{1}{2}}{2a^2+2b^2+2c^2-2ab-2bc-2ac}=\dfrac{3}{2}\)
\(S=-1^2+2^2-3^2+4^2-...+2016^2\)
\(=\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(2016-2015\right)\left(2016+2015\right)\)
\(=3+7+..+4031\)
\(=2033136\)
\(A=\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)-\frac{1}{15}\times4^{64}\)
\(15A=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)-4^{64}\)
\(15A=\left(4^4-1\right)\left(4^4+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)-4^{64}\)
\(15A=\left(4^{16}-1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)-4^{64}\)
\(15A=\left(4^{32}-1\right)\left(4^{32}+1\right)-4^{64}\left(4^{32}\right)\)
\(15A=4^{64}-1-4^{64}\)
\(A=-\frac{1}{15}\)
\(M=\dfrac{\left(a-b\right)^3-c^3+3ab\left(a-b\right)-3abc}{\left(a+b\right)^2+\left(b-c\right)^2+\left(c+a\right)^2}\)
\(=\dfrac{\left(a-b-c\right)\left(a^2-2ab+b^2+ac-bc+c^2+3ab\right)}{2a^2+2b^2+2c^2+2ab-2bc+2ac}\)
\(=\dfrac{\left(a-b-c\right)\cdot\left(a^2+b^2+c^2-ab-bc+ac\right)}{2\cdot\left(a^2+b^2+c^2+ab-bc+ac\right)}=\dfrac{2}{2}=1\)
Ta có: \(\left\{\begin{matrix}a^4+b^2\ge2\sqrt{a^4b^2}=2a^2b\\b^4+a^2\ge2\sqrt{b^4a^2}=2b^2a\end{matrix}\right.\)
Do đó \(S\le\frac{1}{2a^2b+2ab^2}+\frac{1}{2b^2a+2a^2b}\)\(=\frac{1}{a^2b+ab^2}\)
\(\le\frac{1}{4ab}\cdot\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{2ab}\le\frac{1}{8}\left(\frac{1}{a}+\frac{1}{b}\right)^2=\frac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=1\)
bài này mà tìm Min thì hay quá .