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

=a³+b³+3ab(a+b)+c³+3(a+b)c(a+b+c)

=a³+b³+c³+3(a+b)(ab+c(a+b+c))

=a³+b³+c³+3(a+b)(ab+ac+bc+c²)

=a³+b³+c³+3(a+b)(a+c)(b+c)

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

=a³+b³+3ab(a+b)+c³+3(a+b)c(a+b+c)

=a³+b³+c³+3(a+b)(ab+c(a+b+c))

=a³+b³+c³+3(a+b)(ab+ac+bc+c²)

=a³+b³+c³+3(a+b)(a+c)(b+c)

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

=a³+b³+3ab(a+b)+c³+3(a+b)c(a+b+c)

=a³+b³+c³+3(a+b)(ab+c(a+b+c))

=a³+b³+c³+3(a+b)(ab+ac+bc+c²)

=a³+b³+c³+3(a+b)(a+c)(b+c)

bạn phá dấu ngoặc rồi rút gọn nha

minh moi hok lop 6,00 nen 0 biet

mình chưa học đến

ta có: -(-a+b+c)+(b+c-1)= a-b-c+b+c-1=a-1   (1)

         (b-c+6)-(7-a+b)+c= b-c+6-7+a-b+c=a-1  (2)

Từ (1),(2) => -(-a+b+c)+(b+c-1)=(b-c+6)-(7-a+b)+c

Vế trái = -(-a+b+c)+(b+c-1)

= a-b-c+b+c-1

= a+(-b+b)+(-c+c)-1

= a+0+0-1

= a-1

Vế phải = (b-c+6)-(7-a+b)+c

= b-c+6-7+a-b+c

= (b-b)+(-c+c)+(6-7)+a

= 0+0-1+a

= a-1

- Vậy -(-a+b+c)+(b+c-1)=(b-c+6)-(7-a+b)+c

a)

Có: -(-a + b + c) + (b + c - 1) = a - b - c + b + c - 1

                                                = a - 1

Lại có: (b - c + 6) - (7 - a + b) + c = b - c + 6 - 7 + a - b + c 

                                                      = a - 1

Vì a - 1 = a - 1

nên -(-a + b + c) + (b + c - 1) = (b - c + 6) - (7 - a + b) + c (đpcm)

VT=\(-\left(-a+b+c\right)+\left(b+c-1\right)\)

\(=a-b-c+b+c-1\)

=a-1

\(VP=\left(b-c+6\right)-\left(7-a+b\right)+c\)

\(=b-c+6-7+a-b+c\)

=a-1

=>VT=VP

=>\(-\left(-a+b+c\right)+\left(b+c-1\right)=\left(b-c+6\right)-\left(7-a+b\right)+c\)

b) Ta có : a\(^2\)+ b\(^2\)+ c\(^2\) =ab+bc+ca

=> 2(a\(^2\)+b\(^2\)+c\(^2\))= 2(ab+bc+ca)

<=>2a\(^2\)+2b\(^2\)+2c\(^2\)=2ab+2bc+2ca

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

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

<=> (a\(^2\)-2ab+b\(^2\))+(b\(^2\)-2bc+b\(^2\))+(a\(^2\)-2ca+c\(^2\))

<=> (a-b)\(^2\)+(b-c)\(^2\)+(a-c)\(^2\) =a

<=> hoặc a-b=0 hoặc b-c=o hoặc a-c=o <=>a=b hoặc b=c hoặc a=c

=>a=b=c (đpcm)

a) Theo đề bài: \(a^2+b^2=ab\)

=>\(a^2+b^2-ab=0\)

=>\(a^2-2ab+b^2+ab=0\)

=>\(\left(a-b\right)^2+ab=0\)

Vì \(\left(a-b\right)^2\ge0\)  để \(\left(a-b\right)^2+ab=0\) <=> \(\left(a-b\right)^2=ab=0\)

(a-b)²=0 <=> a-b=0 <=> a=b (đpcm)

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

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

=>\(2a^2+2b^2+2c^2=2ab+2bc+2ac\)

=>\(2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

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

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

Vì \(\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(a-c\right)^2\ge0\end{cases}\) để \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

<=>\(\left(a-b\right)^2=\left(b-c\right)^2=\left(a-c\right)^2=0\)

<=>a-b=b-c=a-c=0

<=>a=b=c (đpcm)

http://olm.vn/hoi-dap/question/26002.html

ấn vô link này ****

-(-a+b+c)+(b+c-1)

=a-b-c+b+c-1

=a-1   (1)

(b-c+6)-(7-a+b)+c

=b-c+6-7+a-b+c

=-1+a

=a-1   (2)

Từ (1)(2) => -(-a+b+c)+(b+c-1)=(b-c+6)-(7-a+b)+c

Ta có:

-(-a+b+c)+(b+c-1)

=a-b-c+b+c-1

=(b-b)+(c-c)+a-1

=0+0+a-1

=a-1

(b-c+6)-(7-a+b)+c

=b-c+6-7+a-b+c

=(b-b)+(c-c)+a+[(-7)+6]

=0+0+a-1

=a-1

Vì a-1=a-1

=>-(-a+b+c)+(b+c-1)=(b-c+6)-(7-a+b)-c

Ta có:

-(-a+b+c)+(b+c-1)

=a-b-c+b+c-1

=(b-b)+(c-c)+a-1

=0+0+a-1

=a-1

(b-c+6)-(7-a+b)+c

=b-c+6-7+a-b+c

=(b-b)+(c-c)+a+[(-7)+6]

=0+0+a-1

=a-1

Vì a-1=a-1

=>-(-a+b+c)+(b+c-1)=(b-c+6)-(7-a+b)-c

thank nha bạm