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Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\\\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\end{matrix}\right.\\ \RightarrowĐpcm\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\left(1\right)\)
Thay (1) vào từng vế của đề bài:
\(VT=\dfrac{a^2+ac}{c^2-ac}=\dfrac{bk\left(bk+dk\right)}{dk\left(dk-bk\right)}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\)
Vế phải đặt thừa số chung sẽ ra VT => đpcm.
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: \(\dfrac{a^2-c^2}{b^2-d^2}=\dfrac{ac}{bd}\)
Ta có: \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
mà \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{aa}{bb}=\dfrac{a^2+a^2}{b^2+b^2}\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{a^2.2}{b^2.2}\)
\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{a^2}{b^2}\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\left(đpcm\right)\)
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a; Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)
b: \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=k^2\)
Do đó: \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
c: \(\dfrac{7a^2-3ab}{11a^2-8b^2}=\dfrac{7b^2k^2-3\cdot bk\cdot b}{11b^2k^2-8b^2}=\dfrac{b^2\left(7k^2-3k\right)}{b^2\left(11k^2-8\right)}=\dfrac{7k^2-3k}{11k^2-8}\)
\(\dfrac{7c^2-3cd}{11c^2-8d^2}=\dfrac{7d^2k^2-3kd^2}{11d^2k^2-8d^2}=\dfrac{7k^2-3k}{11k^2-8}\)
Do đó: \(\dfrac{7a^2-3ab}{11a^2-8b^2}=\dfrac{7c^2-3cd}{11c^2-8d^2}\)
theo bài ra ta có:
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\\ \Rightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}=\dfrac{2ab}{2cd}\)
áp dụng tính chất dãy tỉ số bàng nhau ta có:
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}=\dfrac{2ab}{2cd}=\dfrac{a^2+b^2+2ab}{c^2+d^2+2cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(a+b\right)\left(a+b\right)}{\left(c+d\right)\left(c+d\right)}\\ \Rightarrow\dfrac{ab}{cd}=\dfrac{\left(a+b\right)\left(a+b\right)}{\left(c+d\right)\left(c+d\right)}\\ \Rightarrow\dfrac{c\left(a+b\right)}{a\left(c+d\right)}=\dfrac{b\left(c+d\right)}{d\left(a+b\right)}\\ \Rightarrow\dfrac{ca+cb}{ca+ad}=\dfrac{bc+bd}{ad+bd}\)áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{ca+cb}{ca+ad}=\dfrac{bc+bd}{ad+bd}=\dfrac{\left(ca+cb\right)-\left(bc+bd\right)}{\left(ca+ad\right)-\left(ad+bd\right)}=\dfrac{ca-bd}{ca-bd}=1\\ \Rightarrow ca+cb=ca+ad\\ \Rightarrow cb=ad\\ \Rightarrow ad=bc\left(đpcm\right)\)
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)=>\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
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