=(x-y)+(x²-y²)

=(x-y)+(x-y).(x-y)

=(x-y).(x-y+1)

x - y + x² - y²

= (x - y) + (x² + y²)

= (x - y) + (x + y) (x - y)

= (x - y) (x + y + 1)

#Hoc tot!!!

P/s : bài này khá khó nên mình thử thôi ! 

Không mất tính tổng quát , ta giả sử : \(a\ge b\ge c\)

Đặt \(M=ab+bc+ca-12\left(a^3+b^3+c^3\right)\left(a^2b^2+b^2c^2+c^2a^2\right)\)

      \(N=a\left(b+c\right)-12\left[a^3+\left(b+c\right)^3\right]\left[a^2\left(b+c\right)^2\right]\)

Ta có : \(ab+ac+bc\ge a\left(b+c\right)\)hay \(a^2b^2+b^2c^2+c^2a^2\le a^2\left(b+c\right)^2\)

\(\Rightarrow M\ge N\)

Tiếp , ta sẽ chứng minh \(N\ge0\)

\(\Leftrightarrow a\left(b+c\right)-12\left[a^3+\left(b+c\right)^3\right]\left[a^2\left(b+c\right)^2\right]\ge0\)

\(\Leftrightarrow a\left(b+c\right)\left\{1-12a\left(b+c\right)\left[a^3+\left(b+c\right)^3\right]\right\}\ge0\)

\(\Leftrightarrow1-12a\left(b+c\right)\left[a^3\left(b+c\right)^3\right]\ge0\)

\(\Leftrightarrow1-12a\left(b+c\right)\left[\left(a+b+c\right)^3-3a\left(b+c\right)\left(a+b+c\right)\right]\ge0\)

\(\Leftrightarrow1-12a\left(b+c\right)\left[1-3a\left(b+c\right)\right]\ge0\left(1\right)\)

Đặt x = a ; y = b + c ta có : \(x+y=1\Rightarrow xy\le\frac{1}{4}\)

Theo bất đẳng thức AM - GM , ta có :

\(12xy\left(1-3xy\right)\le\frac{1}{4}.12xy\left(4-12xy\right)\le\frac{1}{4}\left(\frac{12xy+4-12xy}{2}\right)^2=1\)

=> Bất đẳng thức ( 1 ) luôn đúng 

\(\Rightarrow N\ge0\)

Vậy \(M\ge0\)\(\Leftrightarrow ab+bc+ca\ge12\left(a^3+b^3+c^3\right)\left(a^2b^2+b^2c^2+c^2a^2\right)\)

Đẳng thức xảy ra với bộ \(\left(\frac{3+\sqrt{3}}{6};\frac{3-\sqrt{3}}{6};0\right)\)và các hoán vị của chúng .

WLOG: \(c=min\left\{a,b,c\right\}\)

Let \(p=a+b+c;ab+bc+ca=q;abc=r\) so p = 1; \(r\ge0\)and \(3\ge q\ge ab\left(\text{vì }c\ge0\right)\)

Need: \(q\ge12\left(p^3-3pq+3r\right)\left(q^2-2pr\right)\)

Have: \(VP=12\left(1-3q+3r\right)\left(q^2-2r\right)=\frac{2}{3}.\left(1-3q+3r\right).18\left(q^2-2r\right)\)

\(\le\frac{1}{6}\left[1-3q+3r+18\left(q^2-2r\right)\right]=\frac{1}{6}\left[18q^2-3q+1-33r\right]\)

\(\le\frac{1}{6}\left(18q^2-3q+1\right)=3q^2-\frac{1}{2}q+\frac{1}{6}\)

Hence, we need to prove: \(q\ge3q^2-\frac{1}{2}q+\frac{1}{6}\)

\(\Leftrightarrow3q^2-\frac{3}{2}q+\frac{1}{6}\le0\Leftrightarrow\frac{1}{6}\le q\le\frac{1}{3}\)

Which it is obvious because:

\(q=ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=\frac{1}{3}\)

\(q-\frac{1}{6}=ab+bc+ca-\frac{1}{6}=ab+c-\frac{1}{6}+c\left(a+b-1\right)\)\(=ab-\frac{1}{6}+1-\left(a+b\right)-c\left[1-\left(a+b\right)\right]\)

\(=ab-\frac{1}{6}+\left[1-\left(a+b\right)\right]\left(1-c\right)\ge0\)

Ta có: 

\(6x^2+10xz+9x+15z\)

\(=\left(6x^2+10xz\right)+\left(9x+15z\right)\)

\(=2x.\left(3x+5z\right)+3.\left(3z+5z\right)\)

\(=\left(2x+3\right).\left(3x+5z\right)\)

vậy thừa số cần tìm là: \(3x+5z\)

Ta có:

6x²+10xz+9x+15z

=(6x²+9x)+(10xz+15z)

=3x(2x+3)+5z(2x+3)

=(2x+3)(3x+5z)

Anh(chị) xem lại đề nhé!!!

Em mới lớp 7 thôi nên có sai sót thì đừng k sai cho em nhé!!!

\(a+b+c=0\)

\(\Rightarrow\hept{\begin{cases}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{cases}}\)

Thay vào A , ta có 

\(A=\frac{a^2}{\left(b+c\right)^2-b^2-c^2}\)\(+\frac{b^2}{\left(a+c\right)^2-a^2-c^2}\)\(+\frac{c^2}{\left(a+b\right)^2-a^2-b^2}\)

=> \(A=\frac{a^2}{b^2+2bc+c^2-b^2-c^2}+\frac{b^2}{a^2+2ac+c^2-a^2-c^2}\)\(+\frac{c^2}{a^2+2ab+b^2-a^2-b^2}\)

=> \(A=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{a^3+b^3+c^3}{2abc}\)

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

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

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

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

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

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

=> \(A=\frac{3abc}{2abc}=\frac{3}{2}\)

Vậy A ko phụ thuộc vào a,b,c

\(x^5-3x^4-x^3-x^2+3x+1\)

\(=\left(x^5-x^2\right)-\left(3x^4-3x\right)-\left(x^3-1\right)\)

\(=x^2\left(x^3-1\right)-3x\left(x^3-1\right)-\left(x^3-1\right)\)

\(=\left(x^3-1\right)\left(x^2-3x-1\right)\)

\(=\left(x-1\right)\left(x^2+x+1\right)\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}-1\right)\)

\(=\left(x-1\right)\left(x^2+x+1\right)\left[\left(x-\frac{3}{2}\right)^2-\frac{13}{4}\right]\)

\(=\left(x-1\right)\left(x^2+x+1\right)\left(x-\frac{3}{2}-\frac{\sqrt{13}}{2}\right)\left(x-\frac{3}{2}+\frac{\sqrt{13}}{2}\right)\)

\(x^5-3x^4-x^3-x^2+3x+1\)\(1\)\(=\left(x^5-x^4\right)-\left(2x^4-2x^3\right)-\left(3x^3-3x^2\right)-\left(4x^2-4x\right)-\left(x-1\right)\)

  \(=x^4\left(x-1\right)-2x^3\left(x-1\right)-3x^2\left(x-1\right)-4x\left(x-1\right)-\left(x-1\right)\)

   \(=\left(x-1\right)\left(x^4-2x^3-3x^2-4x-1\right)\)

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

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

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

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

\(=3x^2y+3xy^2+3x^2z+6xyz+3zy^2+3xz^2+3yz^2\)

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

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

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

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

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

\(x^4-5x^2+4\)

\(=x^4-4x^2-x^2+4\)

\(=\left(x^4-4x^2\right)-\left(x^2-4\right)\)

\(=x^2\left(x^2-4\right)-\left(x^2-4\right)\)

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

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

\(x^4-5x^2+4\)

\(=x^4-4x^2-x+4\)

\(=x\left(x^3-1\right)-\left(4x^2-4\right)\)

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

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

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

\(=\left(x-1\right)\left(x^3+x^2+x-4x+1\right)\)

\(=\left(x-1\right)\left(x^3+x^2-3x+1\right)\)

P/s : Có thể sai vì mk chưa soát lại bài , nên sai thông cảm !