CRM
\(a^3+b^3+c^3=3abc\Leftrightarrow a+b+c=0\)
CMR :
\(a^3+b^3+c^3=3abc\Leftrightarrow a+b+c=0\)
Ta sẽ chứng minh \(a^3+b^3+c^3-3abc=0\Leftrightarrow a+b+c=0\)
Phân tích thành nhân tử : \(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-bc-ac\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)
Vì a + b + c = 0 nên \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\) hay \(a^3+b^3+c^3-3abc=0\Leftrightarrow a^3+b^3+c^3=3abc\)
Ta có : \(a+b+c=0\Leftrightarrow a+b=-c\Leftrightarrow\left(a+b\right)^3=-c^3\Leftrightarrow a^3+b^3+3ab\left(a+b\right)+c^3=0\)
\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)=-3ab.-c=3abc\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\)
1.CMR: nếu a + b + c = 0 thì ( a2 + b2 + c2 ) = 2( a4 + b4 + c4 )
2. Chứng minh đẳng thức : ( a + b + c )3 - ( b + c - a )3 - ( c + a - b )3 + ( a + b - c )3 = 24abc
3. CMR : a3 + b3 + c3 = 3abc \(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\)
\(a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)
Xét \(a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
\(\Leftrightarrow a=b=c\)
\(\RightarrowĐPCM\)
Đặt \(\left(b+c-a;c+a-b;a+b-c\right)\rightarrow\left(x,y,z\right)\)
\(\Rightarrow x+y+z=a+b+c\)
Ta có:\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(\left(x+y\right)^3-3\left(x+y\right)z\left(x+y+z\right)+z^3-x^3-y^3-z^3\)
\(=x^3+3xy\left(x+y\right)+y^3-3\left(x+y\right)z\left(x+y+z\right)+z^3-x^3-y^3-z^3\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
\(=3\cdot2a\cdot2b\cdot2c=24abc\)
Câu 1:
https://h.vn/hoi-dap/question/279693.html
Bạn chịu khó tự đánh nha.Mik cho bn link dc chớ hình như cái đường dẫn bị hỏng mà mik cũng nhác,bài này lm nhiều lần rồi nên chả muốn nhai lại:(((
Câu 3
mik nhầm dòng thứ 2 nha mn
\(\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\) nha!!
a+b+c=0.cmr a^3+b^3+c^3=3abc
em chứng minh thế này được không các thầy (cô) giáo
a+b+c=0
=>a+b=-c
=>a+b=3abc/-3ab
=>(a+b).(-3ab)=3abc
=>(a+b).(a^2-ab+b^2-a^2-2ab-b^2)=3abc
=>(a+b)(a^2-ab+b^2)-(a+b).(a^2+2ab+b^2)=3abc
=>a^3+b^3-(a+b)^3=3abc
mà a+b=-c=> a^3+b^3-(-c)^3=3abc
=>a^3+b^3+c^3=3abc
Được bạn nhé :"))))
Ủng hộ mình = cách theo dõi mình nha
a+b+c=0
\(\left(a+b+c\right)^3=0\)
\(\Leftrightarrow a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+6abc=0\)
\(\Leftrightarrow a^3+b^3+c^3+\left(3a^2b+3ab^2+3abc\right)+\left(3a^2c+3ac^2+3abc\right)+\left(3bc^2+3b^2c+3abc\right)-3abc=0\)\(\Leftrightarrow a^3+b^3+c^3+3ab\left(a+b+c\right)+3ac\left(a+b+c\right)+3bc\left(a+b+c\right)-3abc=0\)\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\)
mk ko chắc cách bn đúng nhưng cách của mk là phù hợp nhất đó
Không nên chứng minh như thế này nhé. Ở ngay phần \(a+b=\frac{3abc}{-3ab}\) đã sai sót vì bạn không tính đến trường hợp \(a=0\) hoặc $b=0$ đã thực hiện phép chia như vậy.
Sử dụng hằng đẳng thức: \((a+b)^3=a^3+b^3+3ab(a+b)\) ta có:
\(a^3+b^3+c^3=(a+b)^3-3ab(a+b)+c^3\)
Vì \(a+b+c=0\Rightarrow a+b=-c\). Thay vào biểu thức trên:
\((a+b)^3-3ab(a+b)+c^3=(-c)^3-3ab(-c)+c^3=-c^3+3abc+c^3=3abc\)
Do đó:
\(a^3+b^3+c^3=3abc\)
Cho (a+b+c)^2 = 3(ab+bc+ca). CMR: a=b=c
Cho a^3+b^3+c^3 = 3abc. CMR: a=b=c và a+b+c=0
Cho a+b+c=0. CMR: a^3+b^3+c^3 = 3abc
`(a+b+c)^2=3(ab+bc+ca)`
`<=>a^2+b^2+c^2+2ab+2bc+2ca=3(ab+bc+ca)`
`<=>a^2+b^2+c^2=ab+bc+ca`
`<=>2a^2+2b^2+2c^2=2ab+2bc+2ca`
`<=>(a-b)^2+(b-c)^2+(c-a)^2=0`
`VT>=0`
Dấu "=" xảy ra khi `a=b=c`
`a^3+b^3+c^3=3abc`
`<=>a^3+b^3+c^3-3abc=0`
`<=>(a+b)^3+c^3-3abc-3ab(a+b)=0`
`<=>(a+b)^3+c^3-3ab(a+b+c)=0`
`<=>(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0`
`**a+b+c=0`
`**a^2+b^2+c^2=ab+bc+ca`
`<=>a=b=c`
1, cho a+b+c=0
CMR: a3+b3+c3=3abc
2, cho a+b-c=0
CMR: a3+b3-c3=-3abc
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: \(a+b-c=0\)
\(\Leftrightarrow a+b=c\)
\(\Leftrightarrow\left(a+b\right)^3=c^3\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=c^3\)
\(\Leftrightarrow a^3+b^3+3abc=c^3\)
\(\Leftrightarrow a^3+b^3-c^3=-3abc\)
C/m rằng
a) Nếu a+b+c = 0 thì a3+b3+c3 =3abc
b) Nếu a3+b3+c3 = 3abc thì a+b+c = 0 hoặc a=b=c
a/ \(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^3=0\)
\(\Leftrightarrow\left[\left(a+b\right)+c\right]^3=0\)
\(\Leftrightarrow\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3=0\)
\(\Leftrightarrow a^3+b^3+c^3+3a^2b+3ab^2+3bc^2+3b^2c+3a^2c+3ac^2+6abc=0\)
\(\Leftrightarrow a^3+b^3+c^3+\left(3a^2b+3ab^2+3abc\right)+\left(3bc^2+3b^2c+3abc\right)+\left(3ac^2+3a^2c+3abc\right)-3abc=0\)
\(\Leftrightarrow a^3+b^3+c^3+3abc\left(a+b+c\right)+3bc\left(a+b+c\right)+3ac\left(a+b+c\right)-3abc=0\)
Mà \(a+b+c=0\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\left(đpcm\right)\)
b/ \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{matrix}\right.\)
+) Nếu : \(a^2+b^2+c^2-ab-bc-ac=0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\) \(\Leftrightarrow a=b=c\)
Vậy \(a^3+b^3+c^3=3abc\) \(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
a, Cmr : ( a + b + c ). ( a^2 + b^2 + c^2 -ab - ac - bc ) = a^3 + b^3 + c^3 -3abc
b, Áp dụng :
a+b+c= 0 thì
a^3 + b^3 + c^3 = 3abc
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
b,
Ta có:
\(\left(a+b+c\right)^3=0\Rightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Rightarrow a^3+b^3+c^3-3.\left(-c\right)\left(-a\right)\left(-b\right)=0\)
1) c/m :nếu a+b+c = 0 thì a3 + b3+c3 =3abc
2) nếu a,b,c>0 thì a3 +b3+c3 > = 3abc. dấu '' ='' xảy ra khi a=b=c
1) Phân tích đa thức thành nhân tử: \(a^3+b^3+c^3-3abc\)
2) Cho a, b, c thỏa mãn a+b+c=0. Chứng minh \(a^3+b^3+c^3=3abc\).
3) Cho a, b, c ≠ 0 thỏa mãn \(a^3+b^3+c^3=3abc\). Chứng minh a=b=c.
1. \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left[\left(abc\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2+c^2-ac-bc\right)-3ab\left(a+b+c\right)\)
\(\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc+2ab-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
2. \(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
3.Còn có a + b + c = 0 nữa mà bn.
\(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-ac-bc=0\end{matrix}\right.\)
+ \(a^2+b^2+c^2-ab-bc-ac=0\)
\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\ \left(c-a\right)^2=0\end{matrix}\right.\)
\(\Rightarrow a=b=c\)