Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có :
\(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\)
\(\Leftrightarrow\dfrac{a\left(bz-cy\right)}{a^2}=\dfrac{b\left(cx-az\right)}{b^2}=\dfrac{c\left(ay-bx\right)}{c^2}\)
\(\Leftrightarrow\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-bcx}{c^2}\)
Theo t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-bcx}{c^2}=\dfrac{abc-acy-bcx-abz-acy-bcx}{a^2+b^2+c^2}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{bz-cy}{a}=0\\\dfrac{cx-az}{b}=0\\\dfrac{ay-bx}{c}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}bz=cy\\cx=az\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b}{y}=\dfrac{c}{z}\\\dfrac{c}{z}=\dfrac{a}{x}\end{matrix}\right.\) \(\Leftrightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\left(đpcm\right)\)
Phương Ann Nhã Doanh Đinh Đức Hùng Mashiro Shiina
Nguyễn Thanh Hằng Nguyễn Huy Tú Lightning Farron
Akai Haruma Võ Đông Anh Tuấn
mấy anh chị cm cho e thêm cái : \(\dfrac{ay+bx}{c}=\dfrac{bz+cy}{a}=\dfrac{cx+az}{b}\)
\(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\)
\(=\dfrac{bxz-cxy}{ax}=\dfrac{cyx-ayz}{by}=\dfrac{azy-bxz}{cz}\)
\(=\dfrac{bxz-cxy+cyx-ayz+azy-bxz}{ax+by+cz}=0\)
\(\Rightarrow bz-cy=0\Rightarrow bz=cy\Rightarrow\dfrac{y}{b}=\dfrac{z}{c}\)
Tương tự...
\(\Rightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\left(dpcm\right)\)
Bài 1:
1) \(a\left(b-c\right)+b\left(c-a\right)+c\left(a-b\right)\)
\(=ab-ac+bc-ba+ca-cb\)
\(=0\)
2) \(a\left(bz-cy\right)+b\left(cx-az\right)+c\left(ay-bx\right)\)
\(=abz-acy+bcx-baz+cay-cbx\)
\(=0\)
Bài 2:
Ta có:
\(\dfrac{x^2+ax+ab+bx}{3bx-a^2-ax+3ab}\)
\(=\dfrac{\left(x^2+bx\right)+\left(ax+ab\right)}{\left(3bx-ax\right)+\left(3ab-a^2\right)}\)
\(=\dfrac{x\left(x+b\right)+a\left(x+b\right)}{x\left(3b-a\right)+a\left(3b-a\right)}\)
\(=\dfrac{\left(x+a\right)\left(x+b\right)}{\left(x+a\right)\left(3b-a\right)}\)
\(=\dfrac{x+b}{3b-a}\)
Ta có :
+) \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\)
\(\Leftrightarrow\dfrac{ayz+bxz+cxy}{xyz}=0\)
\(\Leftrightarrow ayz+bxz+cxy=0\)
+) \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)
\(\Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\)
\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{zc}\right)=1\)
\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{ayz+bxz+cxy}{abc}\right)=1\)
\(\Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(đpcm\right)\)