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Đặt \(\frac{a}{b}=\frac{c}{d}=k\) ,ta có:
\(a=bk,c=dk\)
\(\Rightarrow\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{\left[b.\left(k+1\right)\right]^2}{\left[d.\left(k+1\right)\right]^2}=\frac{b^2.\left(k+1\right)^2}{d^2.\left(k+1\right)^2}=\frac{b^2}{d^2}\)(1)
\(\frac{ab}{cd}=\frac{bkb}{dkd}=\frac{b^2}{d^2}\)(2)
Từ (1) và (2) suy ra:
\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{ab}{cd}\)(đpcm)
Ta có: \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}=\frac{a^2+b^2+2ab}{c^2+d^2+2cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\left(1\right)\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{2ab}{2cd}=\frac{a^2+b^2-2ab}{c^2+d^2-2cd}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\left(\frac{a-b}{c-d}\right)^2\left(2\right)\)
Từ (1) và (2) \(\left(\frac{a+b}{c+d}\right)^2=\left(\frac{a-b}{c-d}\right)^2\)
TH1: \(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a+b+a-b}{c+d+c-d}=\frac{2a}{2c}=\frac{a}{c}\left(3\right)\)
\(\frac{a+b}{c+d}=\frac{a-b}{c-d}=\frac{a+b-a+b}{c+d-c+d}=\frac{2b}{2d}=\frac{b}{d}\left(4\right)\)
Từ (3) và (4) => \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}\)
TH2: \(\frac{a+b}{c+d}=\frac{b-a}{d-c}=\frac{a+b+b-a}{c+d+d-c}=\frac{2b}{2d}=\frac{b}{d}\left(5\right)\)
\(\frac{a+b}{c+d}=\frac{b-a}{d-c}=\frac{a+b-b+a}{c+d-d+c}=\frac{2a}{2c}=\frac{a}{c}\left(6\right)\)
Từ (5) và (6) => \(\frac{b}{c}=\frac{a}{d}\Rightarrow\frac{a}{b}=\frac{d}{c}\)
Vậy nếu \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\) thì \(\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\)
Ta có\(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
<=> cd(a2 + b2) = ab(c2 + d2)
<=> a2cd + b2cd - abc2 - abd2 = 0
<=> (a2cd - abc2) + (b2cd - abd2) = 0
<=> ac(ad - bc) + bd(bc - ad) = 0
<=> ac(ad - bc) - bd(ad - bc) = 0
<=> (ac - bd)(ad - bc) = 0
<=> \(\orbr{\begin{cases}ac-bd=0\\ad-bc=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}ac=bd\\ad=bc\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{a}{d}=\frac{b}{c}\\\frac{a}{b}=\frac{c}{d}\end{cases}}\left(\text{đpcm}\right)\)
a
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}-1=\frac{c}{d}-1\Rightarrow\frac{a-b}{b}=\frac{c-d}{d}\)
b
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{ab}{cd}\)
c
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{5a^2}{5b^2}=\frac{3c^2}{3d^2}=\frac{5a^2+3c^2}{3d^2+5b^2}\)
ta có: a/b = c/d
=> a/c = b/d = (a+b)/(c+d) = (a-b)/(c-d)
=> (a+b)/(a-b) = (c+d)/(c-d) ( đpcm)
ta có: a/b = c/d
=> a/c = b/d = (a+b)/(c+d) = (a-b)/(c-d)
=> (a+b)/(a-b) = (c+d)/(c-d) ( đpcm)
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Tôi chỉ gợi ý thôi. Bạn đặt tỉ lệ thức đã cho bằng 1 số k nào đó
Ta có : \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)
\(\Leftrightarrow\left(a^2+b^2\right)cd=ab\left(c^2+d^2\right)\)
\(\Leftrightarrow a^2cd+b^2cd=abc^2+abd^2\)
\(\Leftrightarrow\left(a^2cd-abd^2\right)+\left(b^2cd-abc^2\right)=0\)
\(\Leftrightarrow ad\left(ac-bd\right)-bc\left(ac-bd\right)=0\)
\(\Leftrightarrow\left(ac-bd\right)\left(ad-bc\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}ac=bd\\ad=bc\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{a}{b}=\frac{c}{d}\\\frac{a}{b}=\frac{d}{c}\end{cases}}\) (đpcm)