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 !

dễ thôi bạn ơi 

CHÚNG TA CỐ GẮNG GHÉP VỀ CÁC TỔNG BÌNH PHƯƠNG LÀ XONG

lili À,cái phân tích tổng bình phương thì em mình cũng biết bạn ạ.Quan trọng là phân tích như thế nào ấy.còn dễ thì không đăng lên đây làm gì rồi

\(a.=4x\left(x^2-2xy+y^2\right)\)

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

\(b.=4x\left(x-2y\right)-7\left(x-2y\right)\)

\(=\left(4x-7\right)\left(x-2y\right)\)