a: \(\frac{1}{x+1}=\frac{x-x^2}{\left(x-x^2\right)\left(x+1\right)}\)

\(\frac{6}{x-x^2}=\frac{6\left(x+1\right)}{\left(x-x^2\right)\left(x+1\right)}\)

b: \(\frac{y+5}{y^2+8y+16}=\frac{y+5}{\left(y+4\right)^2}=\frac{3\left(y+5\right)}{3\left(y+4\right)^2}\)

\(\frac{y}{3y+12}=\frac{y}{3\left(y+4\right)}=\frac{y\left(y+4\right)}{3\left(y+4\right)^2}\)

Ta có \(\frac{2}{x^3-y^3}=\frac{2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

\(\frac{2x-1}{x^2-y^2}=\frac{2x+1}{\left(x+y\right)\left(x-y\right)}\)

\(\frac{1}{x+y}\)  giữ nguyên

MTC: \(\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)

Các nhân tử phụ tương ứng là : \(\left(x+y\right);\left(x-y\right)\left(x^2+xy+y^2\right);\left(x^2+xy+y^2\right)\)

Ta có:

\(\frac{2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\frac{2.\left(x+y\right)}{\left(x-y\right)\left(x+y\right)\left(x^2+xy+y^2\right)}\)

\(\frac{1}{x+y}=\frac{1.\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)}\)

\(\frac{2x+1}{\left(x+y\right)\left(x-y\right)}=\frac{\left(2x+1\right)\left(x^2+xy+y^2\right)}{\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)}\)

a)  \(\left(x+y\right)^5-x-y=\left(x+y\right)^5-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^4-1\right]\)

= \(\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)     #áp dụng hàng đẳng thức#

c) \(x^9-x^7-x^6-x^5+x^4+x^3+x^2+1\)nhóm vào là đc

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

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

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

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

câu a ko phải -x-y mà là -x^5-y^5 bạn à