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Đề bài đúng phải là : Cho a,b,c thỏa mãn a+b+c=0 . CMR : \(2\left(a^5+b^5+c^5\right)=5abc\left(a^2+b^2+c^2\right)\)
a) Từ \(a+b+c=0\Rightarrow b+c=-a\Rightarrow\left(b+c\right)^5=-a^5\)
\(\Rightarrow b^5+5b^4c+10b^3c^2+10b^2c^3+5bc^4+c^5=-a^5\)
\(\Rightarrow\left(a^5+b^5+c^5\right)+5bc\left(b^3+2b^2c+2bc^2+c^3\right)=0\)
\(\Rightarrow\left(a^5+b^5+c^5\right)+5bc\left[\left(b+c\right)\left(b^2-bc+c^2\right)+2bc\left(b+c\right)\right]=0\)
\(\Rightarrow\left(a^5+b^5+c^5\right)+5bc\left(b+c\right)\left(b^2+bc+c^2\right)=0\)
\(\Rightarrow2\left(a^5+b^5+c^5\right)-5abc\left[\left(b^2+2bc+c^2\right)+b^2+c^2\right]=0\)
\(\Rightarrow2\left(a^5+b^5+c^5\right)=5abc\left[\left(b+c\right)^2+b^2+c^2\right]\)
Vậy : \(2\left(a^5+b^5+c^5\right)=5abc\left(a^2+b^2+c^2\right)\)
a, Ta có: \(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)
\(=n^3+3n^2-n+2n^2+6n-2-n^3+2\)
\(=5n^2+5n=5\left(n^2+n\right)⋮5\)
\(\Rightarrowđpcm\)
b, \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)
\(=6n^2+31n+5-6n^2-7n+5\)
\(=24n+10=2\left(12n+5\right)⋮2\)
\(\Rightarrowđpcm\)
\(a+b+c=0\Rightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^5=-c^5\)
\(\Rightarrow a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=-c^5\)
\(\Rightarrow a^5+b^5+c^5+5ab\left[a^3+2a^2b+2ab^2+b^3\right]=0\)
\(\Rightarrow a^5+b^5+c^5+5ab\left[\left(a+b\right)\left(a^2-ab+b^2\right)+2ab\left(a+b\right)\right]=0\)
\(\Rightarrow a^5+b^5+c^5+5ab\left(a+b\right)\left(a^2+ab+b^2\right)=0\)
\(\Rightarrow2\left(a^5+b^5+c^5\right)+5ab\left(-c\right)\left[2a^2+2ab+2b^2\right]=0\)
\(\Rightarrow2\left(a^5+b^5+c^5\right)-5abc\left[\left(a^2+2ab+b^2\right)+a^2+b^2\right]=0\)
\(\Rightarrow2\left(a^5+b^5+c^5\right)-5abc\left[a^2+b^2+c^2\right]=0\)
\(\Rightarrow2\left(a^5+b^5+c^5\right)=5abc\left(a^2+b^2+c^2\right)\)
Chúc bạn học tốt.
\(x^2-x+1=x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)
\(-x^2+4x-5=-\left(x^2-2.x.2+2^2\right)-1=-\left(x-2\right)^2-1< 0\forall x\)
\(a\left(2a-3\right)-2a\left(a+1\right)=a\left(2a-3-2a-2\right)=-5a⋮5\forall a\inℤ\)
Bài 2:Tìm x biết
\\(\\left(4x+3\\right)^3+\\left(5-7x\\right)^3+\\left(3x-8\\right)^3=0\\)
\\(\\Leftrightarrow\\left[\\left(4x\\right)^3+3.\\left(4x\\right)^2.3+3.4x.3^2+3^3\\right]+\\left[5^3-3.5^2.7x+3.5.\\left(7x\\right)^2-\\left(7x\\right)^3\\right]+\\left[\\left(3x\\right)^3-3.\\left(3x\\right)^2.8+3.3x.8^2-8^3\\right]=0\\)
\\(\\Leftrightarrow64x^3+144x^2+108x+27+125-525x+735x^2-343x^3+27x^3-216x^2+576x-512=0\\)
\\(\\Leftrightarrow-252x^3+663x^2+159x-360=0\\)
\\(\\Leftrightarrow3\\left(-84x^3+221x^2+53x-120\\right)=0\\)
Câu trả lời hay nhất: Do a+b+c=0 =>a+b= -c
Ta có (a+b)^5=c^5
<=>a^5+5a^4b+10a^3b^2+10a^2b^3 + 5ab^4 + b^5=-c^5
<=>a^5+b^5+c^5= -5ab(a^3+2a^2b+2ab^2+b^3)
<=>a^5+b^5+c^5= -5ab( a^2(a+b)+ab(a+b)+b^2(a+b))
<=>a^5+b^5+c^5= -5ab(-c)(a^2+ab+b^2) Vì a+b= -c
<=>2(a^5+b^5+c^5)=5abc2(a^2+ab+b^2)
<=>2(a^5+b^5+c^5)=5abc(a^2+b^2+(a+b)^2)
<=>2(a^5+b^5+c^5)=5abc(a^2+b^2+(-c)^2)
<=>2(a^5+b^5+c^5)=5abc(a^2+b^2+c^2) (đpcm)
a)
a) n2−3n+5 : n−2 = n - 1 (R=3) . Để phép chia hết nên suy ra: n-1 thuộc Ư(3) . Suy ra : n = { 4 ; -2 ; 0 ; 2 }