\(B=x^8+2x^5-2x^4+x^2-2x-100+10x\left(x^4+x\right)+\left(5x-1\right)^2\)

\(=x^8+2x^5-2x^4+x^2-2x-100+10x^5+25x^2-10x+1\)

\(=x^8+12x^5-2x^4+36x^2-12x-99\)

\(=x^8+6x^5+9x^4+6x^5+36x^2+54x-11x^4-66x-99\)

\(=x^4\left(x^4+6x+9\right)+6x\left(x^4+6x+9\right)-11\left(x^4+6x+9\right)\)

\(=\left(x^4+6x+9\right)\left(x^4+6x-11\right)\)

\(P\left(x\right)=2x^4-7x^3-2x^2+13x+6\)

           \(=2x^4-4x^3-3x^3+6x^2-8x^2+16x-3x+6\)

           \(=2x^3\left(x-2\right)-3x^2\left(x-2\right)-8x\left(x-2\right)-3\left(x-2\right)\)

           \(=\left(2x^3-3x^2-8x-3\right)\left(x-2\right)\)

           \(=\left[2x^3-6x^2+3x^2-9x+x-3\right].\left(x-2\right)\)

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

           \(=\left[\left(2x^2+3x+1\right)\left(x-3\right)\right]\left(x-2\right)\)

           \(=\left(2x+1\right)\left(x+1\right)\left(x-3\right)\left(x-2\right)\)

a, \(\Delta QAE\infty\Delta QDC\left(g.g\right)\Rightarrow\frac{QA}{QD}=\frac{QE}{QC}\Rightarrow QA.QC=QD.QE\)

b, \(\Delta ABC\infty\Delta DQC\left(g.g\right)\Rightarrow\frac{AB}{AC}=\frac{DQ}{DC}\) (1)

\(\Delta QAE\infty\Delta QDC\left(cmt\right)\Rightarrow\frac{AQ}{AE}=\frac{DQ}{DC}\)(2)

Từ (1) và (2) \(\Rightarrow\frac{AB}{AC}=\frac{AQ}{AE}\Rightarrow AB.AE=AQ.AC\)

Mình hướng dẫn ý thôi. Bạn tự trình bày nhé. Chúc bạn học tốt.