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![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
Ta có: \(b^2=ac\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\)
Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=k\)
+) \(k^2=\dfrac{a}{b}.\dfrac{b}{c}=\dfrac{a}{c}\) (1)
+) \(k=\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{2011b}{2011c}=\dfrac{a+2011b}{b+2011c}\) ( t/c dãy tỉ số bằng nhau )
\(\Rightarrow k^2=\left(\dfrac{a+2011b}{b+2011c}\right)^2=\dfrac{\left(a+2011b\right)^2}{\left(b+2011c\right)^2}\) (2)
Từ (1), (2) \(\Rightarrow\dfrac{a}{c}=\dfrac{\left(a+2011b\right)^2}{\left(b+2011c\right)^2}\left(đpcm\right)\)
Giải:
Từ hằng đẳng thức: \(\left(a+b\right)^2=a^2+2ab+b\) ta có:
\(VP=\dfrac{\left(a+2011b\right)^2}{\left(b+2011c\right)^2}=\dfrac{a^2+2.2011ab+\left(2011b\right)^2}{b^2+2.2011bc+\left(2011c\right)^2}\)
\(=\dfrac{a^2+2.2011ab+2011^2ac}{ac+2.2011bc+2011^2c^2}\)
\(=\dfrac{a\left(a+2.2011b+2011^2c\right)}{c\left(a+2.2011b+2011^2c\right)}=\dfrac{a}{c}=VT\)
Vậy \(\dfrac{a}{c}=\dfrac{\left(a+2011b\right)^2}{\left(b+2011c\right)^2}\) (Đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(b^2=ac\Rightarrow\frac{b}{c}=\frac{a}{b}=\frac{2010a}{2010b}=\frac{2011b}{2011c}=\frac{2010a+2011b}{2010b+2011c}\)
\(\Rightarrow\frac{b}{c}.\frac{a}{b}=\left(\frac{2010a+2011b}{2010b+2011c}\right).\left(\frac{2010a+2011b}{2010b+2011c}\right)\)
\(\Rightarrow\frac{a}{c}=\frac{\left(2010a+2011b\right)^2}{\left(2010b+2011c\right)^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$b.b=ac\Rightarrow \frac{b}{c}=\frac{a}{b}$.
Đặt $\frac{b}{c}=\frac{a}{b}=k\Rightarrow b=ck; a=bk$.
Khi đó:
$\frac{a}{c}=\frac{bk}{c}=\frac{ck.k}{c}=k^2(1)$
Và:
$\frac{(a+2011b)^2}{(b+2011c)^2}=\frac{(bk+2011b)^2}{(ck+2011c)^2}$
$=\frac{b^2(k+2011)^2}{c^2(k+2011)^2}=\frac{b^2}{c^2}=\frac{(ck)^2}{c^2}=k^2(2)$
Từ $(1);(2)$ ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
B1:
Từ \(b=\frac{a+c}{2}\Rightarrow2b=a+c\left(1\right)\)
Từ \(c=\frac{2bd}{b+a}\)thay vào (1) ta được:
\(2b=a+\frac{2bd}{b+a}\)
\(\Leftrightarrow2b\left(b+a\right)=a\left(b+a\right)+2bd\)
\(\Leftrightarrow2b^2+2ab=ab+a^2+2bd\)
\(\Leftrightarrow2b^2+ab-a^2-2bd=0\)
\(\Leftrightarrow2b\left(b-d\right)+a\left(b-a\right)=0\)
\(\Leftrightarrow2b\left(b-d\right)=a\left(a-b\right)\Leftrightarrow\frac{2b}{a}=\frac{a-b}{b-d}\)
B2: Từ \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}hay2ab=c\left(a+b\right)\)
\(\Rightarrow ab+ab=ac+bc\Rightarrow ab-bc=ac-ab\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)
Do đó: \(\frac{a-c}{c-b}=\frac{a}{b}\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(b^2=ac\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}\)
Đặt: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{2018b}{2018c}=t\)
tính chất dãy tỉ số bằng nhau: \(\dfrac{a}{b}=\dfrac{2018b}{2018c}=\dfrac{a+2018b}{b+2018c}\)
Ta có: \(\left\{{}\begin{matrix}\dfrac{a}{b}.\dfrac{b}{c}=\dfrac{a}{c}=t^2\\\left(\dfrac{a+2018b}{b+2018c}\right)^2=t^2\end{matrix}\right.\Leftrightarrowđpcm\)
b^2 = a.c
=> a/b = b/c
Đặt a/b = b/c = k
=> a=bk ; b=ck
=> a = c.k.k = c.k^2 => a/c = k^2
Lại có : (a+2011b)^2/(b+2011c)^2
= (bk+2011b)^2/(ck+2011c)^2
= [b.(k+2011)]^2/[c.(k+2011)]^2
= b^2.(k+2011)^2/c^2.(k+2011)^2
= b^2/c^2
= (b/c)^2
= k^2
=> a/c = (a+2011)^2/(b+2011c)^2
Tk mk nha