Bạn cần làm gì với những biểu thức này thì bạn cần nêu rõ trong đề bài.

Đề bài: Phân tích đa thức thành nhân tử lớp 8 bằng nhiều phương pháp

Ta có :

\(\dfrac{ab+bc+ca}{xy+yz+zx}=\dfrac{ab}{xy}=\dfrac{bc}{yz}=\dfrac{ca}{zx}\)

\(\Rightarrow\dfrac{a}{x}.\dfrac{b}{y}=\dfrac{b}{y}.\dfrac{c}{z}=\dfrac{c}{z}.\dfrac{a}{x}\)

Mà \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)

\(\Rightarrow\dfrac{a^2}{x^2}=\dfrac{b^2}{y^2}=\dfrac{c^2}{z^2}=\dfrac{a^2+b^2+c^2}{x^2+y^2+z^2}\left(\text{Đ}PCM\right)\)

\(ab\left(a^2-b^2\right)+bc\left(b^2-c^2\right)+ca\left(c^2-a^2\right)\)

\(=a^3b-ab^3+b^3c-bc^3+c^3a-ca^3\)

\(=a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)

\(=a^3\left(b-a+a-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)

\(=\left(a-b\right)\left(c^3-a^3\right)+\left(c-a\right)\left(b^3-a^3\right)\)

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

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

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

\(=\left(a-b\right)\left(c-a\right)\left[-a\left(b-c\right)-\left(b-c\right)\left(b+c\right)\right]\)

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

\(16x^2+8x+100=\left(4x\right)^2+2.4x.1+1^2+99\\ =\left(4x+1\right)^2+99>=99>0\left(DPCM\right)\)

\(16x^2+8x+100>0\)

\(\Leftrightarrow\left(4x\right)^2+2.4x.1+1+99>0\)

\(\Leftrightarrow\left(4x+1\right)^2+99>0\left(\forall x\in R\right)\)

\(x^2-9x+20=\left(x-\dfrac{9}{2}\right)^2-\left(\dfrac{9}{2}\right)^2+20=\left(x-\dfrac{9}{2}\right)^2-\dfrac{1}{4}>=-\dfrac{1}{4}\)

\(\widehat{A}+\widehat{B}+\widehat{C}=180^o=>60^o+2.\widehat{C}+\widehat{C}=180^o\\ < =>3.\widehat{C}=120^o\\ < =>\widehat{C}=40^o\\=>\widehat{B}=2.40^o=80^o\)

Giải giúp em với ạ huhu