cho 1/x+1/y+1/z=0.tính yz/x2+xz/y2+xy/x2.mẫu thức là mu 2 17/08/2021 Bởi Savannah cho 1/x+1/y+1/z=0.tính yz/x2+xz/y2+xy/x2.mẫu thức là mu 2
Đáp án: $\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3$ Giải thích các bước giải: Ta có: $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\rightarrow \dfrac{1}{x}+\dfrac{1}{y}=-\dfrac{1}{z}$ Đặt $A=\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}$ $\rightarrow A=xyz(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3})$ $\rightarrow A=xyz((\dfrac{1}{x}+\dfrac{1}{y})^3-3\dfrac{1}{xy}(\dfrac{1}{x}+\dfrac{1}{y})+\dfrac{1}{z^3})$ $\rightarrow A=xyz((\dfrac{-1}{z})^3-3\dfrac{1}{xy}(\dfrac{-1}{z})+\dfrac{1}{z^3})$ $\rightarrow A=xyz(3\dfrac{1}{xyz})$ $\rightarrow A=3$ $\rightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3$ Bình luận
Đáp án:
$\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3$
Giải thích các bước giải:
Ta có:
$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\rightarrow \dfrac{1}{x}+\dfrac{1}{y}=-\dfrac{1}{z}$
Đặt $A=\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}$
$\rightarrow A=xyz(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3})$
$\rightarrow A=xyz((\dfrac{1}{x}+\dfrac{1}{y})^3-3\dfrac{1}{xy}(\dfrac{1}{x}+\dfrac{1}{y})+\dfrac{1}{z^3})$
$\rightarrow A=xyz((\dfrac{-1}{z})^3-3\dfrac{1}{xy}(\dfrac{-1}{z})+\dfrac{1}{z^3})$
$\rightarrow A=xyz(3\dfrac{1}{xyz})$
$\rightarrow A=3$
$\rightarrow\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=3$