Gọi 3 số tự nhiên liên tiếp đó là: \(k-1;k;k+1\)

Theo đề bài, ta có: \(k.\left(k+1\right)-k.\left(k-1\right)=50\)

                               \(k^2+k-k^2+k=50\)

                                    \(2k=50\Rightarrow k=\frac{50}{2}=25\)

\(\Rightarrow k-1=25-1=24\)

     \(k+1=25+1=26\)

Vậy 3 số tự nhiên liên tiếp đó là 24;25;26

a, \(n\left(n+5\right)-\left(n-3\right)\left(n+2\right)=n^2+5n-n^2-2n+3n+6=6n+6=6\left(n+1\right)⋮6\) (đpcm)

b, \(\left(n-1\right)\left(n+1\right)-\left(n-7\right)\left(n-5\right)=n^2-1-n^2+5n+7n-35=12n-36=12\left(n-3\right)⋮12\) (đpcm)

\(N=3\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.615}\)

Đặt  \(\frac{1}{105}=a;\)\(\frac{1}{651}=b\)

Khi đó:

  \(N=3.\left(3a\right)b-a.3\left(1-b\right)-4\left(3a\right).b\)

\(=9ab-3a+3ab-12ab=-3a\)

Vậy  \(N=-3.\frac{1}{105}=-\frac{1}{35}\)

ai choi bang bang 2 ket ban voi minh

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

\(=bc\left(b+c\right)+ca\left(c-a\right)-ab\left[\left(b+c\right)-\left(c-a\right)\right]\)

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

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

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

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

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

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

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

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

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

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

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

a, sửa (x³+y³+z³) thành (x+y+z)³

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

\(=x^3+3x^2\left(y+z\right)+3x\left(y+z\right)^2+\left(y+z\right)^3-a^3-b^3-c^3\)

\(=3\left(y+z\right)\left[x^2+x\left(y+z\right)\right]+y^3+3y^2z+3yz^2+z^3-y^3-z^3\)

\(=3\left(y+z\right)\left(x^2+xy+xz\right)+3yz\left(y+z\right)\)

\(=3\left(y+z\right)\left(x^2+xy+xz+yz\right)\)

\(=3\left(y+z\right)\left[x\left(x+y\right)+z\left(x+y\right)\right]\)

\(=3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

b, \(x^3+x^2z+y^2z-xyz+y^3\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)+z\left(x^2-xy+y^2\right)\)

\(=\left(x+y+z\right)\left(x^2-xy+y^2\right)\)

c, không phân tích được

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

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

\(=x^3+y^3+z^3+3xy\left(x+y\right)+3\left(x+y\right)z\left(x+y+z\right)\)

\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)

\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)