\(f\left(2012\right)=2012^2a+2012b+c=2013\Rightarrow c\text{ lẻ.}\)

\(f\left(2014\right)=2014^2a+2014b+c=2014\Rightarrow c\text{ chẵn.}\)

2 điều trên mâu thuẫn nên ta có đpcm.

chán ghê hk ai giúp hết

Áp dụng tính chất của dãy tỉ số bằng nhau , ta có :

\frac{a}{2012}=\frac{b}{2013}=\frac{c}{2014}=\frac{a-b}{2012-2013}=\frac{b-c}{2013-2014}=\frac{c-a}{2014-2012}2012a​=2013b​=2014c​=2012−2013a−b​=2013−2014b−c​=2014−2012c−a​

\Rightarrow\frac{a-b}{-1}=\frac{b-c}{-1}=\frac{c-a}{2}⇒−1a−b​=−1b−c​=2c−a​

\Rightarrow\left(\frac{a-b}{-1}\right)\left(\frac{b-c}{-1}\right)=\left(\frac{c-a}{2}\right)^2⇒(−1a−b​)(−1b−c​)=(2c−a​)2

hay \left(a-b\right)\left(b-c\right)=\frac{\left(c-a\right)^2}{4}(a−b)(b−c)=4(c−a)2​

\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2⇒4(a−b)(b−c)=(c−a)2

Áp dụng tính chất của dãy tỉ số bằng nhau , ta có :

\frac{a}{2012}=\frac{b}{2013}=\frac{c}{2014}=\frac{a-b}{2012-2013}=\frac{b-c}{2013-2014}=\frac{c-a}{2014-2012}

\Rightarrow\frac{a-b}{-1}=\frac{b-c}{-1}=\frac{c-a}{2}

\Rightarrow\left(\frac{a-b}{-1}\right)\left(\frac{b-c}{-1}\right)=\left(\frac{c-a}{2}\right)^2

hay \left(a-b\right)\left(b-c\right)=\frac{\left(c-a\right)^2}{4}

\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2

Đặt \(\frac{a}{2012}=\frac{b}{2013}=\frac{c}{2014}=k\)

\(\Rightarrow a=2012k,b=2013k,c=2014k\)

\(\Rightarrow4\left(a-b\right)\left(b-c\right)=4\left(2012k-2013k\right)\left(2013k-2014k\right)\)

\(=4\left(-k\right)\left(-k\right)\)

\(=4k^2\left(1\right)\)

Mặt khác:\(\left(a-c\right)^2=\left(2012-2014\right)^2\)

\(=\left(2k\right)^2\)

\(=4k^2\left(2\right)\)

Từ (1),(2) suy ra.......

Bài 3. 

\(\left\{{}\begin{matrix}a\left(a+b+c\right)=-\dfrac{1}{24}\left(1\right)\\c\left(a+b+c\right)=-\dfrac{1}{72}\left(2\right)\\b\left(a+b+c\right)=\dfrac{1}{16}\left(3\right)\end{matrix}\right.\)

Dễ thấy \(a,b,c\ne0\Rightarrow a+b+c\ne0\)

Chia (1) cho (2), ta được \(\dfrac{a}{c}=3\Rightarrow a=3c\left(4\right)\)

Chia (2) cho (3) ta được: \(\dfrac{c}{b}=-\dfrac{2}{9}\Rightarrow b=-\dfrac{9}{2}c\left(5\right)\).

Thay (4), (5) vào (2), ta được: \(-\dfrac{1}{2}c^2=-\dfrac{1}{72}\)

\(\Rightarrow c=\pm\dfrac{1}{6}\).

Với \(c=\dfrac{1}{6}\Rightarrow\left\{{}\begin{matrix}a=3c=\dfrac{1}{2}\\b=-\dfrac{9}{2}c=-\dfrac{3}{4}\end{matrix}\right.\)

Với \(c=-\dfrac{1}{6}\Rightarrow\left\{{}\begin{matrix}a=3c=-\dfrac{1}{2}\\b=-\dfrac{9}{2}c=\dfrac{3}{4}\end{matrix}\right.\)

Vậy: \(\left(a;b;c\right)=\left\{\left(\dfrac{1}{2};-\dfrac{3}{4};\dfrac{1}{6}\right);\left(-\dfrac{1}{2};\dfrac{3}{4};-\dfrac{1}{6}\right)\right\}\)

Bài 2)

Ta có \(\frac{a}{b}< \frac{c}{d}\)

\(\Rightarrow ad< bc\)

Xét \(\frac{a}{b}< \frac{a+c}{b+d}\)

\(\Rightarrow a\left(b+d\right)< b\left(a+c\right)\)

\(\Rightarrow ab+ad< ab+bc\)

\(\Rightarrow ad< bc\) ( thỏa mãn đề bài )

Vậy \(\frac{a}{b}< \frac{a+c}{b+d}\) (1)

Xét \(\frac{a+c}{b+d}< \frac{c}{d}\)

\(\Rightarrow d\left(a+c\right)< c\left(b+d\right)\)

\(\Rightarrow ad+cd< bc+cd\)

\(\Rightarrow ad< bc\) ( thỏa mãn đề bài )

Vậy \(\frac{a+c}{b+d}< \frac{c}{d}\) (2)

Từ (1) và (2)

\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\) (đpcm)

Lời giải:

Đặt $\frac{a}{2012}=\frac{b}{2013}=\frac{c}{2014}=k$

$\Rightarrow a=2012k; b=2013k; c=2014k$. Khi đó:

$A=4(a-b)(b-c)(c-a)=4(2012k-2013k)(2013k-2014k)(2014k-2012k)$

$=4(-k)(-k)(2k)=8k^3$