Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a^3+b^3+c^3−3abc
=a^3+3ab(a+b)+b^3+c^3−3abc−3ab(a+b)
=(a+b)^3+c^3−3ab(a+b+c)
=(a+b+c)(a^2+2ab+b^2−ab−ac+c^2)−3ab(a+b+c)
=(a+b+c)(a^2+b^2+c^2−ab−bc−ca)
a^3 + b^3 + c^3 - 3abc
= ( a+ b)^3 - 3ab ( a+ b) - 3abc
= ( a+ b +c )^3 - 3 ( a + b ).c(a + b +c ) -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 - 3(a+b).c - 3ab ]
= ( a+ b + c ) [ a^2 + 2ab + b^2 + 2bc+ c^2 +2 ac - 3ac - 3bc - 3ab )
= ( a + b + c)(a^2 + b^2 + c^2 -ab - bc- ca)
Tick đúng nha
tìm số tự nhiên nhỏ nhất biết rằng khi chia cho 23 dư 21 khi chia cho 17 dư 16
a. \(x^4+3x^3-9x-9=x^3\left(x+1\right)-9\left(x+1\right)\)\(=\left(x+1\right)\left(x^3-9\right)\)
a3+b3+c3-3abc
=(a+b)3-3ab(a+b)-3abc+c3
=(a+b+c)3-3(a+b)c(a+b+c)-3ab(a+b+c)
=(a+b+c)[(a+b+c)2-3ab-3bc-3ca]
=(a+b+c)(a2+b2+c2+2ab+2bc+2ca-3ab-3bc-3ca)
=(a+b+c)(a2+b2+c2-ab-bc-ca)
=1/2(a+b+c)(2a2+2b2+2c2-2ab-2bc-2ca)
=1/2(a+b+c)(a2-2ab+b2+b2-2bc+c2+c2-2ca+a2)
=1/2(a+b+c)[(a-b)2+(b-c)2+(c-a)2]
Ta có:
\(x^3+x=0\)
\(\Rightarrow x\left(x^2+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=-1\left(vn\right)\end{cases}}\Rightarrow x=0\)
bn vũ quang vinh lam đúng rùi sao k tisk cho bn ấy, hay là không hiu j , hay là ích kỷ ,tui tisk cho bn z
a) \(x^2+2x+1=x^2+x+x+1=x\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x+1\right)=\left(x+1\right)^2\) *Câu này có thể áp dụng hằng đẳng thức \(a^2+2ab+b^2=\left(a+b\right)^2\) cho nhanh*
b) \(a^3-b^3+c^3+3abc=\left(a^3-3a^2b+3ab^2-b^2\right)+3a^2b-3ab^2+c^3+3abc\)
\(=\left(a-b\right)^3+c^3+\left(3a^2b-3ab^2+3abc\right)\)
\(=\left(a-b+c\right)\left[\left(a-b\right)^2-\left(a-b\right)c+c^2\right]+3ab\left(a-b+c\right)\)
\(=\left(a-b+c\right)\left(a^2-2ab+b^2-ac+bc+c^2+3ab\right)\)
\(=\left(a-b+c\right)\left(a^2+b^2+c^2-ac+bc+ab\right)\)
c) \(a^3-b^3-c^3-3abc=\left[a^3-3a^2b+3ab^2-b^3\right]+3a^2b-3ab^2-c^3-3abc\)
\(=\left[\left(a-b\right)^3-c^3\right]+3ab\left(a-b-c\right)=\left(a-b-c\right)\left[\left(a-b\right)^2+\left(a-b\right)c+c^2\right]+3ab\left(a-b-c\right)\)
\(=\left(a-b-c\right)\left[a^2-2ab+b^2+ac-bc+c^2+3ab\right]=\left(a-b-c\right)\left(a^2+b^2+c^2+ab+ac-bc\right)\)
a,(x+1)2
b,(a+c-b).{(a+c)^2+(a+c)b+b^2-3ac}
c,(a-c-b).{(a-c)^2+(a-c)b+b^2+3ac}
Bài 1:
a) \(x.\left(x^2-2xy+1\right)=x^3-2x^2y+x\)
b) \(\left(2x-3\right).\left(x+2\right)=2x^2+4x-3x-6=2x^2-x-6\)
Bài 2:
a) \(x^3-2x^2+x=x.\left(x^2-2x+1\right)=x.\left(x-1\right)^2\)
b) \(x^2-xy+2x-2y=\left(x^2-xy\right)+\left(2x-2y\right)=x.\left(x-y\right)+2.\left(x-y\right)=\left(x-y\right).\left(x+2\right)\)
c) Đề sai.
\(b.\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(a+b+c-a\right)\left(3a^2+b^2+c^2+3ab+2bc+3ac\right)-\left(b^3+c^3\right)\)
\(=\left(b+c\right)\left(3a^2+b^2+c^2+3ab+2bc+3ac\right)-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left(3a^2+b^2+c^2+3ab+2bc+3ac-b^2+bc-c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3ac+3bc\right)\)
\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
a ) \(a^3+b^3+c^3-3abc\)
\(=\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3a^2b-3ab^2-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)