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\(a^2+b^2+c^2+d^2+1=a\left(b+c+d+1\right)\)
\(\Leftrightarrow4a^2+4b^2+4c^2+4d^2+4=4ab+4ac+4ad+4a\)
\(\Leftrightarrow a^2-4ab+4b^2+a^2-4ac+4c^2+a^2-4ad+4d^2+a^2-4a+4=0\)
\(\Leftrightarrow\left(a-2b\right)^2+\left(a-2c\right)^2+\left(a-2d\right)^2+\left(a-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=2b\\a=2c\\a=2d\\a=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=c=d=1\end{matrix}\right.\).
Vậy \(\left(a,b,c,d\right)=\left(2,1,1,1\right)\)
Trước hết, với \(a+b+c=1\) ta có:
\(a^2+b^2+c^2=\left(a^2+b^2+c^2\right)\left(a+b+c\right)\)
\(=\left(a^3+ab^2\right)+\left(b^3+bc^2\right)+\left(c^3+ca^2\right)+a^2b+b^2c+c^2a\)
\(\ge2a^2b+2b^2c+2c^2a+a^2b+b^2c+c^2a\)
Hay \(a^2+b^2+c^2\ge3\left(a^2b+b^2c+c^2a\right)\)
Từ đó:
\(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}=\dfrac{a^4}{a^2b}+\dfrac{b^4}{b^2c}+\dfrac{c^4}{c^2a}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^2b+b^2c+c^2a}\)
\(\ge\dfrac{3\left(a^2b+b^2c+c^2a\right)\left(a^2+b^2+c^2\right)}{a^2b+b^2c+c^2a}=3\left(a^2+b^2+c^2\right)\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
Có : a + b + c = 0
=> (a + b)5 = (-c)5
a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5 = -c5
a5 + b5 + c5 = -5a4b - 10a3b2 - 10a2b3 - 5ab4
a5 + b5 + c5 = -5ab(a3 + 2a2b + 2ab2 + b3)
a5 + b5 + c5 = -5ab[(a3 + b3) + (2a2b + 2ab2)]
a5 + b5 + c5 = -5ab[(a + b)(a2 - ab + b2) + 2ab(a + b)]
a5 + b5 + c5 = -5ab(a + b)(a2 + b2 + ab)
a5 + b5 + c5 = 5abc(a2 + b2 + ab) (do a+b+c=0=> a+b=-c)
2(a5 + b5 + c5) = 5abc(2a2 + 2b2 + 2ab)
2(a5 + b5 + c5) = 5abc[a2 + b2 +(a2 + 2ab + b2)]
2(a5 + b5 + c5) = 5abc[a2 + b2 + (a + b)2]
2(a5 + b5 + c5) = 5abc(a2 + b2 + c2) (do a+b=-c=> (a +b )2 = c2
\(\Leftrightarrow\) \(a^5+b^5+c^5=\dfrac{5}{2}abc\left(a^2+b^2+c^2\right)\)
Vậy...
b: (3x-2)^5+(5-x)^5+(-2x-3)^5=0
Đặt a=3x-2; b=-2x-3
Pt sẽ trở thành:
a^5+b^5-(a+b)^5=0
=>a^5+b^5-(a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5)=0
=>-5a^4b-10a^3b^2-10a^2b^3-5ab^4=0
=>-5a^4b-5ab^4-10a^3b^2-10a^2b^3=0
=>-5ab(a^3+b^3)-10a^2b^2(a+b)=0
=>-5ab(a+b)(a^2-ab+b^2)-10a^2b^2(a+b)=0
=>-5ab(a+b)(a^2-ab+b^2+2ab)=0
=>-5ab(a+b)(a^2+b^2+ab)=0
=>ab(a+b)=0
=>(3x-2)(-2x-3)(5-x)=0
=>\(x\in\left\{\dfrac{2}{3};-\dfrac{3}{2};5\right\}\)
\(\left(ad+bc\right)\left(a^2d^2+b^2c^2\right)=0\)
\(\Rightarrow a^3d^3+adb^2c^2+bca^2d^2+b^3c^3=0\)
\(\Rightarrow a^3d^3+abcd\left(bc+ad\right)+b^3c^3=0\)
\(\Rightarrow a^3d^3+abcd.0+b^3c^3=0\)
\(\Rightarrow a^3d^3+b^3c^3=0\)
\(a+b=c+d\Leftrightarrow a=c+d-b\Leftrightarrow a^2=b^2+c^2+d^2-2bc+2cd-2bd\Leftrightarrow a^2+b^2+c^2+d^2=\left(b^2-2bc+c^2\right)+\left(c^2+2cd+d^2\right)+\left(d^2-2bd+b^2\right)\Leftrightarrow a^2+b^2+c^2+d^2=\left(b-c\right)^2+\left(c+d\right)^2+\left(b-d\right)^2\)Vì a,b,c thuộc tập số nghuyên nên ta có điều phải chứng minh.