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\(\dfrac{a^2+b^2}{b^2+c^2}\)
\(=\dfrac{a^2+ac}{ac+c^2}\)(vì b2=ac)
\(=\dfrac{a\left(a+c\right)}{c\left(a+c\right)}\)(đặt a,c ra ngoài)
\(=\dfrac{a}{c}\)(rút gọn a+c)
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\(\dfrac{a}{c}=\dfrac{a^2+b^2}{b^2+c^2}\)
\(VP=\dfrac{a^2+ac}{ac+c^2}=\dfrac{a\left(a+c\right)}{c\left(a+c\right)}=\dfrac{a}{c}=VT\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét \(\left(a^2+b^2\right).C-\left(b^2+c^2\right).a=a^2c+b^2a\)=\(b^2a-c^2a=a^2c+ac.c-ac.a=0\)
(thay \(b^2=ac\))
\(\Rightarrow\left(a^2+b^2\right).c=\left(b^2+c^2\right).a\Rightarrow\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a}{c}\)
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Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\\\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\end{matrix}\right.\\ \RightarrowĐpcm\)
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Có: \(a+b+c=1\Leftrightarrow\left(a+b+c\right)^2=1\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}\)
\(\Rightarrow\dfrac{x^2}{a^2}=\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}=\dfrac{\left(x+y+z\right)^2}{\left(a+b+c\right)^2}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\)
\(\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2\) (do \(\left(a+b+c\right)^2=a^2+b^2+c^2=1\))
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Đặt :
\(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow a=bk;c=dk\)
\(VT=\dfrac{ac}{bd}=\dfrac{bkdk}{bd}=\dfrac{bdk^2}{bd}=k^2\left(1\right)\)
\(VP=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrowđpcm\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) =>\(a=bk,c=dk\)
=> \(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k.k=k^2\left(1\right)\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}\)
=\(\dfrac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)
Từ (1)và(2)=>\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
Chúc Bạn Học Tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{a^2-c^2}{b^2-d^2}=k^2\)
\(\dfrac{ac}{bd}=k^2\)
Do đó: \(\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{ac}{bd}\)
\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a^2+ac}{ac+c^2}=\dfrac{a\left(a+c\right)}{c\left(a+c\right)}=\dfrac{a}{c}\)