Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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)
\(b^2=ac\Rightarrow\frac{b}{c}=\frac{a}{b}=\frac{2010a}{2010b}=\frac{2011b}{2011c}=\frac{2010a+2011b}{2010b+2011c}\)
\(\Rightarrow\frac{b}{c}.\frac{a}{b}=\left(\frac{2010a+2011b}{2010b+2011c}\right).\left(\frac{2010a+2011b}{2010b+2011c}\right)\)
\(\Rightarrow\frac{a}{c}=\frac{\left(2010a+2011b\right)^2}{\left(2010b+2011c\right)^2}\)
\(b^2=ac\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}\)
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: \(\left\{{}\begin{matrix}\dfrac{a}{b}.\dfrac{b}{c}=\dfrac{a}{c}=t^2\\\left(\dfrac{a+2018b}{b+2018c}\right)^2=t^2\end{matrix}\right.\Leftrightarrowđpcm\)
\(b^2=ac\)
\(\Rightarrow\frac{a}{b}=\frac{b}{c}\)
Đặt \(\frac{a}{b}=\frac{b}{c}=k\), ta có: \(a=bk;b=ck\)
\(\frac{a}{c}=\frac{bk}{c}=\frac{ck\times k}{c}=k^2\) (1)
\(\left(\frac{a+2012b}{b+2012c}\right)^2=\left(\frac{bk+2012b}{ck+2012}\right)^2=\left(\frac{b\left(k+2012\right)}{c\left(k+2012\right)}\right)^2=\left(\frac{b}{c}\right)^2=k^2\) (2)
Từ (1) và (2)
=> \(\frac{a}{c}=\left(\frac{a+2012b}{b+2012c}\right)^2\left(\text{đ}pcm\right)\)
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