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28 tháng 8 2021

e: Ta có: \(x^5-5x^3+4x\)

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

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

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

26 tháng 11 2017

bn gõ bài trong công thức trực quan ik, khó nhìn lắm, ko làm đc

29 tháng 11 2017

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

30 tháng 8

Ta có: \(a^2b + a^2c + ab^2 + ac^2 + b^2c + bc^2 + 2abc\)

\(= a^2(b + c) + a(b^2 + 2bc + c^2) + bc(b + c)\)

\(= a^2(b + c) + a(b + c)^2 + bc(b + c)\)

\(= (b + c)\left[ a^2 + a(b + c) + bc \right]\)

\(= (b + c)\left[ a^2 + ab + ac + bc \right]\)

=(b+c)(a+b)(a+c)

Ta có: \(a^2-4b^2-a^3+4a^2b-4ab^2\)

\(=\left(a^2-4b^2\right)-a\left(a^2-4ab+4b^2\right)\)

\(=\left(a-2b\right)\left(a+2b\right)-a\left(a-2b\right)^2\)

\(=\left(a-2b\right)\left(a+2b-a\left(a-2b\right)\right)\)

\(=\left(a-2b\right)\left(a+2b-a^2+2ab\right)\)

29 tháng 8 2025

12 tháng 10 2021

\(a,a^2\left(a-b\right)+ab\left(a-c\right)=a\left(a+b\right)\left(a-c\right)\\ c,=\left(x^2-2x+1\right)\left(x^2+2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\\ b,=\left(x-5\right)^2-9y^2=\left(x-5-3y\right)\left(x-5+3y\right)\\ d,=4\left(x^2-9x+14\right)=4\left(x-7\right)\left(x-2\right)\)