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17 tháng 11 2017

Ta có a3+b3+c3-3abc

=(a+b)3+c3-3ab(a+b+c)

=(a+b+c)[(a+b)2-c(a+b)+c2]-3ab(a+b+c)

=(a+b+c)(a2+b2+c2+2ab-3ab -ac-bc)

=(a+b+c)(a2+b2+c2-ab-bc-ac)

 =>a3+b3+c3-3abc / a2+b2+c2-ab-bc-ac

=a+b+c

16 tháng 11 2017

https://h.vn/hoi-dap/question/53588.html . Vào link này nha

9 tháng 7 2018

Bài 2:

a)  \(VP=\left(a+b\right)^3-3ab\left(a+b\right)\)

\(=a^3+b^3+3ab\left(a+b\right)-3ab\left(a+b\right)\)

\(=a^3+b^3=VT\)  (đpcm)

b)  \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(=a^3+ab^2+ac^2-a^2b-abc-a^2c+a^2b+b^3+bc^2-ab^2-b^2c-abc\)\(+a^2c+b^2c+c^3-abc-bc^2-ac^2\)

\(=a^3+b^3+c^3-3abc\)

9 tháng 7 2018

Bài 1:

\(N=\frac{x\left|x-2\right|}{x^2+8x-20}+12x-3\)

\(=\frac{x\left|x-2\right|}{\left(x-2\right)\left(x+10\right)}+12x-3\)

Nếu  \(x\ge2\)thì:     \(N=\frac{x\left(x-2\right)}{\left(x-2\right)\left(x+10\right)}+12x-3\)

                                      \(=\frac{x}{x+10}+12x+3\)  (lm tiếp nhé)

Nếu  \(x< 2\) thì:     \(N=\frac{x\left(2-x\right)}{\left(x-2\right)\left(x+10\right)}+12x-3\)

                                         \(=\frac{-x}{x+10}+12x-3\)  (lm tiếp nhé)

29 tháng 9 2018

\(1)\)

\(a)\)\(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)

\(A=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)

\(A=100+99+98+97+...+2+1\)

\(A=\frac{100\left(100+1\right)}{2}\)

\(A=5050\)

\(b)\)\(B=3\left(2^2+1\right)\left(2^4+1\right).....\left(2^{64}+1\right)+1\)

\(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right).....\left(2^{64}+1\right)+1\)

\(B=\left(2^4-1\right)\left(2^4+1\right).....\left(2^{64}+1\right)+1\)

\(B=\left(2^8+1\right).....\left(2^{64}+1\right)+1\)

\(............\)

\(B=\left(2^{64}-1\right)\left(2^{64}+1\right)+1\)

\(B=2^{128}-1+1\)

\(B=2^{128}\)

Chúc bạn học tốt ~ 

29 tháng 9 2018

\(1)\)

\(c)\)\(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)

\(C=\left(a+b\right)^2+2\left(a+b\right)c+c^2+\left(a+b\right)^2-2\left(a+b\right)c+c^2-2\left(a+b\right)^2\)

\(C=2\left(a+b\right)^2+2c^2-2\left(a+b\right)^2\)

\(C=2c^2\)

\(2)\)

\(a)\)\(VP=\left(a+b\right)^3-3ab\left(a+b\right)\)

\(VP=a^3+3a^2b+3ab^2+b^3-3ab\left(a+b\right)\)

\(VP=a^3+3ab\left(a+b\right)+b^3-3ab\left(a+b\right)\)

\(VP=a^3+b^3=VT\) ( đpcm ) 

\(b)\)\(VT=a^3+b^3+c^3-3abc\)

\(VT=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(VT=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(VT=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)

\(VT=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=VP\) ( đpcm ) 

Từ đó suy ra : 

\(i)\)\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow\)\(a^3+b^3+c^3-3abc=0\)\(\Rightarrow\)\(a+b+c=0\)

Hoặc \(a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow\)\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow\)\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Leftrightarrow\)\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow}a=b=c}\)

Chúc bạn học tốt ~ 

5 tháng 7 2018

Bài 1:

a)  \(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)

\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)

\(=100+99+98+97+...+2+1=5050\)

b)  \(B=3\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)

\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)

\(=2^{128}-1+1=2^{128}\)

c)  \(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)

\(=2c^2\)

16 tháng 7 2018

ket ban bang bang 2 ko ban

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)

27 tháng 7 2017

b) Xét VP ta có :

\(\left(a+b+c\right)\cdot\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(=a^3+ab^2+ac^2-ab^2-abc-ca^2+ba^2+b^3+bc^2-ab^2-bc^2-abc+ca^2+cb^2+c^3-abc-bc^2-c^2a\)

\(=a^3+b^3+c^3-abc-abc-abc\)

\(=a^3+b^3+c^3-3abc\)

\(=VT\)

Vậy đẳng thức đã được Cm

(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

23 tháng 9 2020

a) \(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)

\(=\left[x^2+\left(a+b\right)x+ab\right]\left(x+c\right)\)

\(=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\)

b) \(a^3+b^3+c^3-3abc\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)

\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ca-bc+c^2-3ab\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

c) \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=a^2\left(b-c\right)+b^2c-ab^2+c^2a-bc^2\)

\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b-c\right)\left(b+c\right)\)

\(=\left(b-c\right)\left(a^2+bc-ab-ca\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

23 tháng 9 2020

Nhầm đoạn cuối là \(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)