\(\frac{y^3}{y+1}+\frac{y^2}{y-1}+\frac{1}{y+1}-\frac{1}{y-1}\)

\(=\frac{y^3+1}{y+1}+\frac{y^2-1}{y-1}\)

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

\(=y^2-y+1+y+1=y^2+2\)

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

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

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

Nếu n\(\ge\)6 thì n! =1.2.3.4.5.6.....n chia hết cho9, n! chia hết cho 3

mà 105 chia hết cho 3

=> (n!+105) chia hết cho 3

n! + 105 là số chính phương => (n!+105) chia hết cho 9

mà n! chia hết cho 9

=> 105 chia hết cho 9( vô lý)(loạiI

+) Nếu n=5 thì n!+105 =5!+105=225=15² 

+) Nếu n=4 thì n!+105=4!+105=129 (không là SCP )(loại)

+) Nếu n=3 thì n!+105=3!+105=111(ko là SCP)(LOẠI)

+) Nếu n=2 thì n!+105=2!+105=107 (loại)

+) Nếu n=1 thì n!+105=106(loại)

+) Nếu n=0 thì n!+105=106 (loại) 

Vậy n=5

Ta có : 

\(P=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ba}{c^2}\)

\(\Leftrightarrow P=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}\)

\(\Leftrightarrow P=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)( 1 )

Biến đổi \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)ta được :

\(\frac{1}{a}+\frac{1}{b}=\frac{-1}{c}\)

\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)

\(\Leftrightarrow\frac{1}{a^3}+\frac{3}{a^2b}+\frac{3}{ab^2}+\frac{1}{b^3}=\frac{-1}{c^3}\)

\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)+\frac{1}{c^3}=0\)

\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{ab}\left(\frac{-1}{c}\right)=0\)

\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)( 2 )

Thay ( 2 ) vào ( 1 ) ta được

\(P=abc.\frac{3}{abc}=3\)

\(\frac{5}{2y^2+6y}-\frac{4y-3y^3}{y^3-9y}-3\)

ĐKXĐ : \(\hept{\begin{cases}y\ne0\\y\ne\pm3\end{cases}}\)

\(=\frac{5}{2y\left(y+3\right)}-\frac{y\left(4-3y^2\right)}{y\left(y^2-9\right)}-3\)

\(=\frac{5}{2y\left(y+3\right)}-\frac{4-3y^2}{y^2-9}-3\)

\(=\frac{5}{2y\left(y+3\right)}-\frac{4-3y^2}{\left(y-3\right)\left(y+3\right)}-3\)

\(=\frac{5\left(y-3\right)}{2y\left(y-3\right)\left(y+3\right)}-\frac{2y\left(4-3y^2\right)}{2y\left(y-3\right)\left(y+3\right)}-\frac{3\cdot2y\left(y-3\right)\left(y+3\right)}{2y\left(y-3\right)\left(y+3\right)}\)

\(=\frac{5y-15}{2y\left(y-3\right)\left(y+3\right)}-\frac{8y-6y^3}{2y\left(y-3\right)\left(y+3\right)}-\frac{6y\left(y^2-9\right)}{2y\left(y-3\right)\left(y+3\right)}\)

\(=\frac{5y-15-8y+6y^3-6y^3+54y}{2y\left(y-3\right)\left(y+3\right)}\)

\(=\frac{51y-15}{2y\left(y-3\right)\left(y+3\right)}=\frac{3\left(17y-5\right)}{2y\left(y-3\right)\left(y+3\right)}\)

\(A=\frac{x}{x^2-2x}-\frac{x^2+4x}{x^3-4x}-\frac{2}{x^2+2x}\)(ĐK: \(x\ne0,x\ne\pm2\))

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

\(A=\frac{x^2+2x-x^2-4x-2x+4}{x\left(x-2\right)\left(x+2\right)}\)

\(A=\frac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)