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Ta co :a+b+c=0
a3+b3+c3+3a2b+3ab2+3b2c+3bc2+3a2c+3ac2+6abc=0
a3+b3+c3+3a2b+3ab2+3b2c+3bc2+3a2c+3ac2+3abc+3abc+3abc-3abc=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
Vi : a+b+c=0
\(\Rightarrow\)a3+b3+c3+0+0+0=3abc
\(\Rightarrow\)a3+b3+c3=3abc
\(\Rightarrow\)dpcm
nho k nha
a+b+c = 0
=> a+b = -c
=> (a+b)^3 = (-c) ^3
=> a^3 + b^3 + 3ab(a+b) = -c^3
=> a^3 +b^3 +c^3 = - 3ab(a+b)
=> a^3 + b^3 + c^3 = 3abc ( vì a+b = -c) ( đpcm)
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
Bài 1:
a)Từ \(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Rightarrow\left[\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{matrix}\right.\) (Điều phải chứng minh)
b)Ngược lại ta cũng có : nếu \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\)
Bài 2:
a)\(\frac{3m^2+7m+1}{m-3}=\frac{3m\left(m-3\right)+16m+1}{m-3}=\frac{3m\left(m-3\right)}{m-3}+\frac{16m+1}{m-3}=3m+\frac{16m+1}{m-3}\in Z\)
Suy ra \(16m+1⋮m-3\)
\(\frac{16m+1}{m-3}=\frac{16\left(m-3\right)+49}{m-3}=\frac{16\left(m-3\right)}{m-3}+\frac{49}{m-3}=16+\frac{49}{m-3}\in Z\)
Suy ra 49 chia hết m-3....
b)tương tự
\(a+b+c=0\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)=0
\(\Leftrightarrow\)\(a^3+ab^2+ac^2-a^2b-a^2c-abc+a^2b+b^3+bc^2-ab^2-\)
\(abc-b^2c+ca^2+bc^2+c^3-abc-ac^2-bc^2\)=0
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\Leftrightarrow a^3+b^3-3abc=-c^3\)
Lời giải:
a)
\((a+b+c)^3-a^3-b^3-c^3=[(a+b)+c]^3-a^3-b^3-c^3\)
\(=(a+b)^3+3(a+b)^2c+3(a+b)c^2+c^3-a^3-b^3-c^3\)
\(=a^3+3ab(a+b)+b^3+3(a+b)^2c+3(a+b)c^2+c^3-a^3-b^3-c^3\)
\(=3ab(a+b)+3(a+b)^2c+3(a+b)c^2\)
\(=3(a+b)[ab+c(a+b)+c^2]\)
\(=3(a+b)(ab+ca+bc+c^2)=3(a+b)[a(b+c)+c(b+c)]\)
\(=3(a+b)(a+c)(b+c)\)
b)
Áp dụng kết quả phần a: Nếu $a+b+c=0$ thì:
\(0^3-a^3-b^3-c^3=3(0-c)(0-a)(0-b)\)
\(\Leftrightarrow -(a^3+b^3+c^3)=-3abc\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\)
a +b +c = 0 => a + b = - c
Ta có
a^3 + b^3 + c^3 = ( a + b)^3 - 3ab( a+b) + c^3
Thay a+ b= -c ta có
a^3 + b^3 + c^3 = -c^3 - 3ab.-c + c^3 = 3abc
=> ĐPCM
cách khác:
\(a+b+c=0\)
\(\Rightarrow\)\(a+b=-c\)
\(\Rightarrow\)\(\left(a+b\right)^3=c^3\)
Ta có: \(a^3+b^3+c^3\)
\(=a^3+b^3-\left(a+b\right)^3\)
\(=-3ab\left(a+b\right)\)
\(=-3ab\left(-c\right)=3abc\)
Vậy \(a^3+b^3+c^3=3abc\)
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc=\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow a^3+b^3+c^3=3abc\left(đpcm\right)\)
1) Có: \(a+b+c=0\)
\(\Leftrightarrow a+b=-c\)
\(\Leftrightarrow\left(a+b\right)^3=-c^3\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Leftrightarrow a^3+b^3-3abc=-c^3\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\)
2)Có: \(a+b-c=0\)
\(\Leftrightarrow a+b=c\)
\(\Leftrightarrow\left(a+b\right)^3=c^3\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=c^3\)
\(\Leftrightarrow a^3+b^3+3abc=c^3\)
\(\Leftrightarrow a^3+b^3-c^3=-3abc\)