Question

! СРОЧНО ! Даю 90 баллов ​

Answer 1

1.14

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$

$\frac{bc}{abc}+\frac{ac}{abc}+\frac{ab}{abc}=0$

$\frac{bc+ac+ab}{abc}=0$

$bc+ac+ab=0$

$a+b+c=1\ \ \ |()^2$

$(a+b+c)^2=1$

$[a+(b+c)]^2=1$

$a^2+2a(b+c)+(b+c)^2=1$

$a^2+2ab+2ac+b^2+2bc+c^2=1$

$a^2+b^2+c^2+2(ab+ac+bc)=1$

$a^2+b^2+c^2+2\cdot0=1$

$a^2+b^2+c^2=1$

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1.15

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

$\frac{ayz}{xyz}+\frac{bxz}{xyz}+\frac{cxy}{xyz}=0$

$\frac{ayz+bxz+cxy}{xyz}=0$

$ayz+bxz+cxy=0$

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

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

$\left[\frac{x}{a}+ \left(\frac{y}{b}+\frac{z}{c} \right) \right]^2 =1$

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

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

$\frac{x^2}{a^2}+\frac{y^2}{b^2} +\frac{z^2}{c^2} + \frac{2xy}{ab}+\frac{2xz}{ac}+\frac{2yz}{bc} =1$

$\frac{x^2}{a^2}+\frac{y^2}{b^2} +\frac{z^2}{c^2} + \frac{2cxy}{abc}+\frac{2bxz}{abc}+\frac{2ayz}{abc} =1$

$\frac{x^2}{a^2}+\frac{y^2}{b^2} +\frac{z^2}{c^2} + \frac{2cxy+2bxz+2ayz}{abc} =1$

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

$\frac{x^2}{a^2}+\frac{y^2}{b^2} +\frac{z^2}{c^2} + \frac{2\cdot 0}{abc} =1$

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