Câu hỏi của Tiểu Thư Kiêu Kì

Question

Cho $\dfrac{x}{a}$ + $\dfrac{y}{b}$ + $\dfrac{z}{c}$ = 1 và $\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0$

Tính A = $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}$

lê thị hương giang · Accepted Answer

$\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0$

$\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0$

$\Rightarrow ayz+bxz+cxy=0$

$\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$

$\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1$

$\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1$

$\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy+bxz+ayz}{abc}\right)=1$

$\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{0}{abc}\right)=1$

$\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+0=1$

$\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1$Câu trả lời: $\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0$

$\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0$

$\Rightarrow ayz+bxz+cxy=0$

$\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$

$\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1$

$\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{xz}{ac}+\dfrac{yz}{bc}\right)=1$

$\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{cxy+bxz+ayz}{abc}\right)=1$

$\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{0}{abc}\right)=1$

$\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+0=1$

$\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1$

nam do · Answer

Có:

$\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0$

$\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0$

$\Rightarrow ayz+bxz+cxy=0$

Có:

$\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$

$\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1^2$

$\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac}\right)=1$

$\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{ayz+bxz+cxy}{abc}\right)=1$

$\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(ayz+bxz+cxy=0\right)$Câu trả lời: Có:

$\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0$

$\Rightarrow\dfrac{ayz+bxz+cxy}{xyz}=0$

$\Rightarrow ayz+bxz+cxy=0$

Có:

$\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$

$\Rightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1^2$

$\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{xz}{ac}\right)=1$

$\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{ayz+bxz+cxy}{abc}\right)=1$

$\Rightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\left(ayz+bxz+cxy=0\right)$

Bài 7: Phép nhân các phân thức đại số