Bài về tỉ lệ thức và dãy tỉ số

Q

quynhanhtron_1

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N

nhuquynhdat

Từ $\dfrac{a+b}{b+c}=\dfrac{c+d}{d+a} \Longrightarrow (a+b)(d+a)=(c+d)(b+c)$

$\Longrightarrow ad+a^2+bd+ab=c^2+bc+bd+cd$

$\Longrightarrow a^2+ad+ab-c^2-bc-cd=0$

$\Longrightarrow a^2+ad+ab+ac-c^2-bc-cd-ac=0$

$\Longrightarrow a(a+b+c+d)-c(a+b+c+d)=0$

$\Longrightarrow (a-c)(a+b+c+d)=0$

$\Longrightarrow \left[\begin{matrix} a-c=0\\ a+b+c+d=0\end{matrix}\right. \Longrightarrow \left[\begin{matrix}a=c\\ a+b+c+d=0\end{matrix}\right.$
 
D

demon311

2)

Đặt: $x=\dfrac{ a}{2013}=\dfrac{ b}{2014}=\dfrac{ c}{2015} \\
\rightarrow \begin{cases}
a=2013 x \\
b=2014 x \\
c=2015 x
\end{cases} \\
4(a-b)(b-c)=4(2013x-2014x)(2014x-2015x)=4x^2 \\
(c-a)^2=(2015x-2013x)^2=4x^2 \\
\rightarrow 4(a-b)(b-c)=(c-a)^2$
 
P

pinkylun

Câu 2:
$\dfrac{a}{2013}=\dfrac{b}{2014}=\dfrac{c}{2015}$
$=\dfrac{a-b}{-1}=\dfrac{b-c}{-1}=\dfrac{c-a}{2}$
$=-a+b=-b+c=\dfrac{c-a}{2}$

=>$[-(a-b)]^2=(\dfrac{c-a}{2})^2$
=>$4(a-b)^2=(c-a)^2$
=>$4(a-b)(b-c)=(c-a)^2$

Xong!
 
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