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Xét TS
Có a^3 + b^3 + c^3 - 3abc = a^3 + 3a^2b + 3ab^2 + b^2 + c^3 - 3abc - 3a^2b - 3ab^2 = (a + b)^3 + c^3 - 3ab(a + b + c) = (a + b + c)( (a+b)^2 + (a + b)c + c^2 - 3abc) = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac)
Rút gọn TS/MS được kết quả = a + b + c = 2009 => điều phải chứng minh
Ta có a3 + b3 + c3 - 3abc
=[ (a+ b)3 + c3 ] - [3ab(a+b) + 3abc] = (a + b+ c)3 - 3(a + b).c(a + b + c) - 3ab.(a + b + c)
= (a + b+ c). [(a + b + c)2 - 3c(a + b) - 3ab]
= (a + b+ c).(a2 + b2 + c2 + 2ab + 2bc + 2ca - 3ac - 3bc - 3ab)
= (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
=> \(\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-ac-bc}=a+b+c=2009\)
Vậy.......
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
b,
Ta có:
\(\left(a+b+c\right)^3=0\Rightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Rightarrow a^3+b^3+c^3-3.\left(-c\right)\left(-a\right)\left(-b\right)=0\)
Xét : \(a^3+b^3+c^3=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-bc-ac\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)
Suy ra : \(\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ac}=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{a^2+b^2+c^2-ab-bc-ac}=a+b+c=2016\)
Vậy ta có điều phải chứng minh.
chào bạn còn nhớ mình ko bai nay o vong 15 luyen thi phai ko. Bạn phân tích từ số thành nhân tử
B=(a+b+c)(a^2 + b^2 + c^2 -ab-bc-ac)/a^2 +b^2 +c^2 -ab-bc-ac
suy ra B=a+b+c. suy ra B=2016
Ta có
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ac+ab}{abc}=0\Rightarrow ab+bc+ac=0.\)
\(A=\frac{\left(bc\right)^3+\left(ac\right)^3+\left(ab\right)^3}{\left(abc\right)^2}\)
Ta có
\(\left(ab\right)^3+\left(bc\right)^3+\left(ac\right)^3-3\left(abc\right)^2=\)
\(=\left(ab+bc+ac\right)\left[\left(ab\right)^2+\left(bc\right)^2+\left(ac\right)^2-abbc-bcac-abac\right]=0\)
\(\Rightarrow\left(ab\right)^3+\left(bc\right)^3+\left(ac\right)^3=3\left(abc\right)^2\)
\(\Rightarrow A=\frac{3\left(abc\right)^2}{\left(abc\right)^2}=3\)
a) Xét vế trái: \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3a^2bc+3abc^2+c^3-a^3-b^3-c^3\)
\(=a^3+b^3+3ab\left(a+b\right)+3\left(a+b\right)^2c+3\left(a+b\right)c^2-a^3-b^3\)
\(=3ab\left(a+b\right)+3\left(a+b\right)^2c+3\left(a+b\right)c^2\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
b) \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
Chúc bạn học tốt và nhớ k cho mình với nhá!
Ta có :
\(\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-ac-bc}\)
\(=\frac{\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3a^2b-3ab^2-3abc}{a^2+b^2+c^2-ab-ac-bc}\)
\(=\frac{\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)}{a^2+b^2+c^2-ab-ac-bc}\)
\(=\frac{\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)}{a^2+b^2+c^2-ab-ac-bc}\)
\(=\frac{\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)}{a^2+b^2+c^2-ab-ac-bc}\)
\(=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)}{a^2+b^2+c^2-ab-ac-bc}\)
\(=a+b+c=2009\)(đpcm)