\(B=\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3\)

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

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

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

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

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

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

\(=\left(c+a-2b\right)\left[\left(2a-b-c\right)\left(5c-a-4b\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left[\left(b+c-2a\right)\left(a+4b-5c\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left(b+c-2a\right)\left(a+4b-5c-b-c+2a\right)\)

\(=\left(c+a-2b\right)\left(b+c-2a\right)\left(3a+3b-6c\right)\)

\(=3\left(c+a-2b\right)\left(b+c-2a\right)\left(a+b-2c\right)\)

\(B=\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3\)

Đặt: \(a+b-2c=x;b+c-2a=y;c+a-2b=z\)

\(\Rightarrow B=x^3+y^3+z^3=\left(x+y+z\right)^3-3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

Ta thấy: \(x+y+z=a+b-2c+b+c-2a+c+a-2b=0\)

\(x+y=a+b-2c+b+c-2a=2b-a-c\)

\(y+z=b+c-2a+c+a-2b=2c-a-b\)

\(z+x=c+a-2b+a+b-2c=2a-b-c\)

Thay vào B \(\Rightarrow B=0-3\left(2b-a-c\right)\left(2c-a-b\right)\left(2a-b-c\right)\)

Vậy \(B=-3\left(2b-a-a\right)\left(2c-a-b\right)\left(2a-b-c\right).\)

m nghĩ máy tính cầm tay sinh ra để làm j ?

(123456789)³=1881676372_x10²⁴

a) \(\left(3x-4\right)^2+2\left(3x-4\right)\left(2x+4\right)+\left(2x+4\right)^2\)\(=\left(3x-4+2x+4\right)^2=\left(5x\right)^2=25x^2\)

b)\(\left(3x+4\right)^2+\left(7+3x\right)^2-\left(6x+8\right)\left(3x+7\right)\)

\(=\left(3x+4\right)^2-2\left(3x+4\right)\left(7+3x\right)+\left(7+3x\right)^2\)

\(=\left[3x+4-\left(7+3x\right)\right]^2=\left(3x+4-7-3x\right)^2=\left(-3\right)^2=9\)

c)\(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)

\(=\left(2x+1\right)^2+2\left(\left(2x\right)^2-1^2\right)+\left(2x-1\right)^2\)

\(=\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x-1\right)^2\)

\(=\left(2x+1+2x-1\right)^2=\left(4x\right)^2=14x^2\)

xong rồi đấy,bạn k cho mình nhé

đó là HĐT số 6

(A+B)^3=A^3+3A^2B+3AB^2+B^3 

99^3+1+3(99^2+99)=99^3+3*99^2*1+3*99*1^2+1^3=(99+1)^3=1000000

=(99+1)^3=100^3=1000000

Mình xin lỗi đề bài là Tìm min

tìm min là gì

Ta có:\(4x^2-12x+1018=\left(\left(2x\right)^2-2.2x.3+\left(3\right)^2\right)+1011\)

                                                 \(=\left(2x-3\right)^2+1011\ge1011\)

\(\Rightarrow\frac{1}{\left(2x-3\right)^2+1011}\le\frac{1}{1011}\)

lại nhân 2018 vào hai vế ta được:

\(\frac{2018}{\left(2x-3\right)^2+1011}\le\frac{2018}{1011}\)

Vậy \(Max_{\left(A\right)}=\frac{2018}{1011}\)Dấu \("="\Leftrightarrow\left(2x-3\right)^2=0\)

                                                               \(\Leftrightarrow2x-3=0\)

                                                              \(\Leftrightarrow x=\frac{2}{3}\)

Xong rồi đấy,chúc bạn học tốt

Gọi số nguyên đầu tiên là a

số nguyên tiếp theo là a+1;a+2;...a+k-1

thực hiện phép chia a cho k ta được 

a=kq+r với r=0;1;2;...k-1

từ đó ta có đpcm