Câu 2:
a. Cho $\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}$
Chứng minh rằng:
$\frac{a}{x+2y+z}=\frac{b}{2x+y-z}=\frac{c}{4x-4y+z}$
b. Cho $a^{2012}+b^{2012}=a^{2013}+b^{2013}=a^{2014}+b^{2014}$.
Tính A = $a^{2016}+b^{2016}$
a.Đặt $\frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}=k$
$\Rightarrow \frac{x}{a+2b+c}=\frac{y}{2a+b-c}=\frac{z}{4a-4b+c}=k=\frac{2y}{4a+2b-2c}=\frac{x+2y+z}{9a}$
$\Rightarrow \frac{a}{x+2y+z}=\frac{1}{9k}$(1)
Chứng minh tương tự ta cũng có
$\Rightarrow \frac{b}{2x+y-z}=\frac{1}{9k}$ (2)
$\Rightarrow \frac{c}{4x-4y+z}=\frac{1}{9k}$ (3)
Từ (1),(2) & (3) ta có đpcm
b.
Nếu a=b=0 thì $a^{2016}+b^{2016}=0$
Nếu $a,b\neq 0$,ta có
$a^{2014}+b^{2014}=(a+b)(a^{2013}+b^{2013})-ab(a^{2012}+b^{2012})$ (*)
Do $a^{2012}+b^{2012}=a^{2013}+b^{2013}=a^{2014}+b^{2014}$ nên ta có
$(*)\Leftrightarrow 1=a+b-ab$
$\Leftrightarrow (a-1)(1-b)=0$
$\Leftrightarrow a=b=1$
$\Rightarrow a^{2016}+b^{2016}=2$
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