*_*!!!
a, \(9x^2+6xy+y^2=9x^2+3xy+3xy+y^2\)
\(=\left(9x^2+3xy\right)+\left(3xy+y^2\right)\)
\(=3x.\left(3x+y\right)+y.\left(3x+y\right)\)
\(=\left(3x+y\right)^2\)
b, \(x^4+2x^3+x^2=x^4+x^3+x^3+x^2\)
\(=\left(x^4+x^3\right)+\left(x^3+x^2\right)\)
\(=x^3.\left(x+1\right)+x^2.\left(x+1\right)\)
\(=\left(x+1\right).\left(x^3+x^2\right)=\left(x+1\right).x^2.\left(x+1\right)\)
\(=\left(x+1\right)^2.x^2\)
Chúc bạn học tốt!!!
a)\(9x^2+6xy+y^2=\left(3x+y\right)^2\)
b)\(x^4+2x^3+x^2\)
\(=x^2\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)^2\)
c)\(5x^2-10xy+5xy^2-5z^2\)
\(=5\left(x^2-2xy+xy^2-z^2\right)\)
a) \(9x^2+6xy+y^2\)
= \(\left(3x-y\right)^2\)
b) \(x^4+2x^3+x^2\)
= \(x^4+x^3+x^3+x^2\)
= \(x^3.\left(x+1\right)+x^2.\left(x+1\right)\)
= \(\left(x+1\right).\left(x^3+x^2\right)\)
Sửa đề
\(5x^2-10xy+5y^2-5z^2\)
\(=5x^2-5xy-5xy+5y^2-5z^2\)
\(=\left(5x^2-5xy\right)-\left(5xy-5y^2\right)-5z^2\)
\(=5x.\left(x-y\right)-5y.\left(x-y\right)-5z^2\)
\(=5\left(x-y\right)^2-5z^2=5.\left[\left(x-y\right)^2-z^2\right]\)
\(=5.\left\{\left[\left(x-y\right)-z\right].\left[\left(x-y\right)+z\right]\right\}\)
\(=5.\left[\left(x-y-z\right).\left(x-y+z\right)\right]\)
Chúc bạn học tốt!!!
c) \(5x^2-10xy+5xy^2-5z^2\)
= \(5\left(x^2-z^2\right)-5\left(2xy-xy^2\right)\)
= \(5\left(x^2-z^2-2xy+xy^2\right)\)
nhờ mn là giúp mình với ạ , minh đang cần gấp :( 
