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a) Ta có\(\frac{3a-b}{3a+b}=\frac{3c-d}{3c+d}\)
=> (3a - b)(3c + d) = (3a + b)(3c - d)
=> 9ac + 3ad - 3bc - bd = 9ac - 3ad + 3bc - bd
=> 3ad - 3bc = -3ad + 3bc
=> 3ad + 3ad = 3bc + 3bc
=> 6ad = 6bc
=> ad = bc
=> \(\frac{a}{b}=\frac{c}{d}\left(\text{đpcm}\right)\)
b) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó \(\frac{b^2+d^2}{a^2+c^2}=\frac{b^2+d^2}{\left(bk\right)^2+\left(dk\right)^2}=\frac{b^2+d^2}{d^2k^2+d^2k^2}=\frac{b^2+d^2}{k^2\left(b^2+d^2\right)}=\frac{1}{k^2}\)(1);
\(\frac{bd}{ac}=\frac{bd}{bkdk}=\frac{1}{k^2}\left(2\right)\)
Từ (1)(2) => \(\frac{b^2+d^2}{a^2+c^2}=\frac{bd}{ac}\)(đpcm)
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Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=kb;c=kd\)
Ta có:\(\frac{a^2+ac}{c^2-ac}=\frac{b^2k^2+bk.dk}{d^2k^2-bk.dk}=\frac{bk^2\left(b+d\right)}{dk^2\left(d-b\right)}=\frac{b\left(b+d\right)}{d\left(d-b\right)}\)(1)
\(\frac{b^2+bd}{d^2-bd}=\frac{b\left(b+d\right)}{d\left(d-b\right)}\)(2)
Từ 1 và 2 =>\(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\)
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Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=kb\\c=kd\end{matrix}\right.\)
Ta có: \(\frac{a^2+ac}{c^2-ac}=\frac{b^2.k^2+bk.dk}{d^2.k^2-bk.dk}=\frac{bk^2.\left(b+d\right)}{dk^2.\left(d-b\right)}=\frac{b.\left(b+d\right)}{d.\left(d-b\right)}\) (1)
\(\frac{b^2+bd}{d^2-bd}=\frac{b.\left(b+d\right)}{d.\left(d-b\right)}\) (2)
Từ (1) và (2) => \(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\left(đpcm\right).\)
Chúc bạn học tốt!
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Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=kb\\c=kd\end{cases}}\)
a) \(VT=\frac{a}{a+c}=\frac{kb}{kb+kd}=\frac{kb}{k\left(b+d\right)}=\frac{b}{b+d}=VP\)
=> đpcm
b) \(VT=\frac{a^2+c^2}{b^2+d^2}=\frac{\left(kb\right)^2+\left(kd\right)^2}{b^2+d^2}=\frac{k^2b^2+k^2d^2}{b^2+d^2}=\frac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)(1)
\(VP=\frac{ac}{bd}=\frac{kb\cdot kd}{bd}=\frac{k^2bd}{bd}=k^2\)(2)
Từ (1) và (2) => VT = VP => đpcm
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Tham khảo:Chứng minh a/b=c/d hoặc a/b=d/c biết (a^2+b^2)/(c^2+d^2)=ab/cd - An Nhiên
\(\text{Cho }\dfrac{a}{b}=\dfrac{d}{c}\text{ và }b,d\notin0\text{.CMR:}\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
\(\text{Ta có:}\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\text{Lại có:}\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=\dfrac{\left(bd\right).k^2}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{\left(b^2+d^2\right).k^2}{b^2+d^2}=k^2\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{b}.\frac{c}{d}=\frac{a}{b}.\frac{a}{b}=\frac{c}{d}.\frac{c}{d}\Leftrightarrow\frac{ac}{bd}=\frac{a^2}{b^2}=\frac{c^2}{d^2}\)
Áp dụng t/c dãy tỉ số bằng nhau: \(\frac{ac}{bd}=\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)
Ta có đpcm