Question

Cho \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\);  \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\).  Tính  \(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\)

Answer 1

ta có: 
a² + b² + c² = (a + b + c)² - 2(ab + bc + ca). (*) 
rất dể cm: ta khai triển hằng đẳng thức (a + b + c)² rồi rút gọn là ra (*) 
AD (*): 
x²/a² + y²/b² + z²/c² = (x/a + y/b + z/c)² - 2(xy/ab + yz/bc + zx/ca) = 
= 1² - 2(xyz/abc)(c/z + a/x + b/y) = 1 + 0 = 1

Answer 2

làm sao mà ra dc -2(xyz/abc)(c/z+a/x+b/y)

Answer 3

\(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=\frac{x\cdot x}{a\cdot a}+\frac{y\cdot y}{b\cdot b}+\frac{z\cdot z}{c\cdot c}=\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\)1

Answer 4

bỏ 1 số 1

Answer 5

\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\Rightarrow ayz+bxz+cxy=0\)

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

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

\(=1-2.\frac{cxy+bxz+ayz}{abc}=1-2.0=1\)