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b) \(\left(a+b+c\right)^2=3\left(ab+bc+ca\right)\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\) (chuyển vế qua)
\(\Leftrightarrow\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
Do VP >=0 với mọi a, b, c. Nên để đăng thức xảy ra thì a = b = c
tính giá trị của biểu thức (x2 - 5)(x+3)+(x+4)(x - x2) với x=0 ; 15 ;-15 ; 0,15
a3+b3+c3−3abc=(a+b)3+c3−3a2b−3ab2−3abc
=(a+b+c)3[(a+b)2−(a+b)c+c2]−3ab(a+b)−3abc
=(a+b+c)(a2+b2+2ab−ac−bc+c2)−3ab(a+b+c)
=(a+b+c)(a2+b2+2ab−ac−bc+c2−3ab)
=(a+b+c)(a2+b2+c2−ab−ac−bc)
=> a3 + b3 + c3 = ( a+b+ c)(a2+b2+c2 -ab - ac - bc ) + 3abc
Bài 1:
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b+c\right)^3-a^3\right]-\left(b^3+c^3\right)\)
\(=\left(a+b+c-a\right)\left[\left(a+b+c\right)^2+\left(a+b+c\right)a+a^2\right]-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left(a^2+b^2+c^2+2ab+2bc+2ac+a^2+ab+ac+a^2\right)-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3ac+2bc+b^2+c^2\right)-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3ac+2bc+b^2+c^2-b^2+bc-c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3ac+3bc\right)\)
\(=3\left(b+c\right)\left(a^2+ab+ac+bc\right)\)
\(=3\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]\)
\(=3\left(b+c\right)\left(a+b\right)\left(a+c\right)\)
b) \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
Bài 2:
Từ câu 1b ta đã chứng minh được:
\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Thay a + b + c = 0 vào ta được
\(a^3+b^3+c^3-3abc=0\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
a+b+c=0
=>(a+b+c)3=0
=>a3+b3+c3+3a2b+3ab2+3b2c+3bc2+3a2c+3ac2+6abc=0
=>a3+b3+c3+(3a2b+3ab2+3abc)+(3b2c+3bc2+3abc)+(3a2c+3ac2+3abc)-3abc=0
=>a3+b3+c3+3ab(a+b+c)+3bc(a+b+c)+3ac(a+b+c)=3abc
Do a+b+c=0
=>a3+b3+c3=3abc(ĐPCM)
(a+b+c)(a2+b2+c2-ab-bc-ca)
=(a+b+c)a2+(a+b+c)b2+(a+b+c)c2-(a+b+c)ab-(a+b+c)bc-(a+b+c)ca
=a3+a2b+a2c+ab2+b3+cb2+ac2+bc2+c3-a2b-ab2-abc-abc-b2c-bc2-a2c-abc-ac2
=(a3+b3+c3)+(a2b-a2b)+(a2c-a2c)+(ab2-ab2)+(cb2-cb2)+...-(abc+abc+abc)
=a3+b3+c3-3abc
=>đpcm
Bài 2:
a+b+c+d=0
nên b+c=-(a+d)
\(a^3+b^3+c^3+d^3\)
\(=\left(a+d\right)^3-3ad\left(a+d\right)+\left(b+c\right)^3-3bc\left(b+c\right)\)
\(=-\left(b+c\right)^3+3ad\left(b+c\right)+\left(b+c\right)^3-3bc\left(b+c\right)\)
\(=3ad\left(b+c\right)-3bc\left(b+c\right)\)
\(=\left(b+c\right)\left(3ad-3bc\right)\)
\(=3\left(b+c\right)\left(ad-bc\right)\)
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc=\left(a+b+c\right)^3\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2\right)-3ab\left(a+b+c\right)=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
Nhưng theo mình thấy a^3+b^3+c^3 không thể đổi thành (a+b+c)^3
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc\)
\(=\left(a+b+c\right)^3\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
A = a3 + b3 +c3 -3abc thành nhân tử.
Lời giải:
Từ (a+b)3= a3 + 3a2b +3ab2 + b3
= a3 + b3 + 3ab (a+b)
Ta suy ra: a3 + b3 = (a+b)3 - 3ab (a+b) (1)
áp dụng hằng đẳng thức (1) vào giải bài toán ta có:
A = (a3 + b3) + c3 - 3abc
= (a+b)3 - 3ab (a+b) + c3 - 3abc
= (a+b)3 + c3 - 3ab (a+b) - 3abc
= (a+b+c) (a2 +2ab + b2 -ac - bc + c2 - 3ab)
= (a+b+c) (a2+ b2 +c2 -ab - bc - ac) (*)
Do \(a+b+c=1\) nên BĐT cần chứng minh tương đương:
\(2\left(a^3+b^3+c^3\right)+3abc\ge\left(ab+bc+ca\right)\left(a+b+c\right)\)
\(\Leftrightarrow2\left(a^3+b^3+c^3\right)\ge ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\)
Thật vậy, ta có:
\(2\left(a^3+b^3+c^3\right)=\left(a^3+b^3\right)+\left(b^3+c^3\right)+\left(c^3+a^3\right)\)
\(=\left(a+b\right)\left(a^2+b^2-ab\right)+\left(b+c\right)\left(b^2+c^2-bc\right)+\left(c+a\right)\left(c^2+a^2-ca\right)\)
\(\ge\left(a+b\right)\left(2ab-ab\right)+\left(b+c\right)\left(2bc-bc\right)+\left(c+a\right)\left(2ca-ca\right)\)
\(=ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)