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\(b^2=ac\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}\)
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)
=> \(\frac{a}{c}=\left(\frac{a+2012b}{b+2012c}\right)^2\left(\text{đ}pcm\right)\)
\(\hept{\begin{cases}b^2=ac\\c^2=bd\end{cases}\Rightarrow\hept{\begin{cases}\frac{a}{b}=\frac{b}{c}\\\frac{b}{c}=\frac{c}{d}\end{cases}\Rightarrow}\frac{a}{b}=\frac{b}{c}=\frac{c}{d}}\)
=>\(\frac{a^3}{b^3}=\frac{2018b^3}{2018c^3}=\frac{2019c^3}{2019d^3}=\frac{a^3-2018b^3-2019c^3}{b^3-2018c^3-2019d^3}\left(1\right)\)
Mà \(\frac{a^3}{b^3}=\frac{a}{b}\cdot\frac{a}{b}\cdot\frac{a}{b}=\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{d}=\frac{a}{d}\left(2\right)\)
Từ (1) và (2) => đpcm
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
Sửa lại đề \(CM\)\(\frac{a}{c}=\frac{\left(a+20112b\right)^2}{\left(b+2012c\right)^2}\)
Có \(a,b,c\in R;a,b,c\ne0\)và \(b^2=ac\)
Ta có \(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\)
Lại có \(\frac{a}{b}=\frac{b}{c}=\frac{2012b}{2012c}\Rightarrow\frac{a}{b}=\frac{a+2012b}{b+2012c}\)
\(\Rightarrow\frac{a^2}{b^2}=\frac{\left(a+2012b\right)^2}{\left(b+2012c\right)^2}\Rightarrow\frac{a^2}{ac}=\frac{\left(a+2012b\right)^2}{\left(b+2012c\right)^2}\)
Hay \(\frac{a}{c}=\frac{\left(a+2012b\right)^2}{\left(b+2012c\right)^2}\)
\(\frac{\left(a+2012.b\right)^2}{\left(b+2012.c\right)^2}=\frac{a^2+2.2012.a.b+2012^2.b^2}{b^2+2.2012.b.c+2012^2.c^2}=\frac{a^2+2.2012.a.b+2012^2.a.c}{a.c+2.2012.b.c+2012^2.c^2}=\)
\(=\frac{a\left(a+2.2012.b+2012^2.c\right)}{c\left(a+2.2012.b+2012^2.c\right)}=\frac{a}{c}\)
Xem lại đề bài
Ta co:\(b^2=ac\Leftrightarrow\frac{a}{b}=\frac{b}{c}\)
\(=\frac{2007b}{2007c}=\frac{a+2007b}{b+2007c}\)
\(\Rightarrow\left(\frac{a+2007b}{b+2007c}\right)^2=\left(\frac{a}{b}\right)^2=\frac{a}{b}\times\frac{b}{c}=\frac{a}{c}\)
Vậy \(\frac{a}{c}=\left(\frac{a+2007b}{b+2007c}\right)^2\left(đpcm\right)\)
https://olm.vn/hoi-dap/question/61610.html
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có các câu hỏi tương tự, khá giống đó bạn ak