cho a,b thỏa mãn: a^3+b^3+3.(a^2+b^2)+4.(a+b)+4
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Ta có : \(a^3+b^3+3\left(a^2+b^2\right)+4\left(a+b\right)+4=0\)
\(=>\left(a+1\right)^3+\left(b+1\right)^3+a+b+2=0\)
\(=>\left(a+b+2\right)\left[\left(a+1\right)^2-\left(a+1\right)\left(b+1\right)+\left(b+1\right)^2\right]+\left(a+b+2\right)=0\)
\(=>\left(a+b+2\right)\left(a^2+b^2+a+b-ab+2\right)=0\)
\(=>\left(a+b+2\right)2\left(a^2+b^2+a+b-ab+2\right)=0\)
\(=>\left(a+b+2\right)\left(2a^2+2b^2+2a+2b-2ab+4\right)=0\)
\(=>\left(a+b+2\right)\left[\left(a-b\right)^2+\left(a+1\right)^2+\left(b+1\right)^2+2\right]=0\)
Lại có : \(\left(a-b\right)^2\ge0;\left(a+1\right)^2\ge0;\left(b+1\right)^2\ge0\)
\(=>\left(a-b\right)^2+\left(a+1\right)^2+\left(b+1\right)^2+2\ge0\)
\(=>a+b+2=0=>a+b=-2=>M=2018.\left(-2\right)^2=8072\)
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Xét a4 - 2a3 \(\ge8a-16\)
<=> a4 -2a3 -8a +16\(\ge0\)
<=> (a4 - 2a3) - 8 (a-2) \(\ge0\)
<=> \(a^3\left(a-2\right)-8\left(a-2\right)\ge0\)
<=> \(\left(a-2\right)\left(a^3-8\right)\ge0\)
<=> \(\left(a-2\right)^2\left(a^2+2a+4\right)\ge0\) (luôn đúng)
Tương tự => \(\left\{{}\begin{matrix}b^4-2b^3\ge8b-16\\c^4-2c^3\ge8c-16\end{matrix}\right.\)
<=> \(a^4+b^4+c^4-2\left(a^3+b^3+c^3\right)\ge8\left(a+b+c\right)-48=0\)
<=> \(a^4+b^4+c^4\ge2\left(a^3+b^3+c^3\right)\)
Dấu "=" <=> a=b=c=2
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\(a^4+b^4\ge a^3+b^3\)
\(\Leftrightarrow2\left(a^4+b^4\right)\ge2\left(a^3+b^3\right)\)
\(\Leftrightarrow2a^4+2b^4\ge\left(a+b\right)\left(a^3+b^3\right)\)
\(\Leftrightarrow2a^4+2b^4\ge a^4+ab^3+a^3b+b^4\)
\(\Leftrightarrow a^4-a^3b+b^4-ab^3\ge0\)
\(\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\) đúng vì thằng thừa số 2 luôn \(\ge\)0
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1) ta có: A= x^3 -8y^3=> A=(x-2y)(x^2 +2xy+4y^2)=>A=5.(29+2xy) (vì x-2y=5 và x^2+4y^2=29) (1)
Mặt khác : x-2y=5(gt)=> (x-2y)^2=25=> x^2-4xy+4y^2=25=>29-4xy=25(vì x^2+4y^2=29)
=> xy=1 (2)
Thay (2) vào (1) ta đc: A= 5.(29+2.1)=155
Vậy gt của bt A là 155
2) theo bài ra ta có: a+b+c=0 => a+b=-c=>(a+b)^2=c^2=> a^2 +b^2+2ab=c^2=>c^2-a^2-b^2=2ab
=> \(\left(c^2-a^2-b^2\right)^2=4a^2b^2\)
=>\(c^4+a^4+b^4-2c^2a^2+2a^2b^2-2b^2c^2=4a^2b^2\)
=>\(a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2\)
=>\(2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2\)
=> \(a^4+b^4+c^4=\frac{1}{2}\left(a^2+b^2+c^2\right)^2\) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(a^3+b^3+3\left(a^2+b^2\right)+4\left(a+b\right)+4=0\)
<=> \(\left(a+b\right)^3-3ab\left(a+b\right)+3\left(a+b\right)^2-6ab+4\left(a+b\right)+4=0\)
<=> \(\left[\left(a+b\right)^3+2\left(a+b\right)^2\right]-3ab\left(a+b+2\right)+\left(a+b\right)^2+4\left(a+b\right)+4=0\)
<=> \(\left(a+b\right)^2\left(a+b+2\right)-3ab\left(a+b+2\right)+\left(a+b+2\right)^2=0\)
<=> \(\left(a+b+2\right)\left(\left(a+b\right)^2-3ab+a+b+2\right)=0\)
<=> \(\left(a+b+2\right)\left(a^2+b^2-ab+a+b+2\right)=0\)(1)
Có: \(a^2+b^2-ab+a+b+2=\frac{1}{2}\left[\left(a-b\right)^2+\left(a+1\right)^2+\left(b+1\right)^2\right]+1>0\)
=> (1) <=> a + b + 2 = 0 <=> a + b = -2
Thế vào tìm M .
Cố gắng học tốt giúp đỡ mọi người nhiều hơn nhé! :))))
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