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Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=bk;c=dk\)
\(\Rightarrow VT=\frac{bk}{bk-b}=\frac{bk}{b\left(k-1\right)}=\frac{k}{k-1}\left(1\right)\)
\(\Rightarrow VP=\frac{c}{c-d}=\frac{dk}{dk-d}=\frac{dk}{d\left(k-1\right)}=\frac{k}{k-1}\left(2\right)\)
Từ (1) và (2) =>Đpcm
a) Từ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có: \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)
\(\frac{a}{c}=\frac{a-b}{c-d}\Rightarrow\frac{a}{a-b}=\frac{c}{c-d}\)
b) \(\frac{a}{b}=\frac{c}{d}\Rightarrow1+\frac{a}{b}=1+\frac{c}{d}\Rightarrow\frac{b+a}{b}=\frac{d+c}{d}\)
vậy \(\frac{a+b}{b}=\frac{c+d}{d}\)
a) Ta có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{b}{a}=\frac{d}{c}\Rightarrow1-\frac{b}{a}=1-\frac{d}{c}\Rightarrow\frac{a-b}{a}=\frac{c-d}{c}\)
b) Ta có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{b}{a}=\frac{d}{c}\Rightarrow1+\frac{b}{a}=1+\frac{d}{c}\Rightarrow\frac{a+b}{a}=\frac{c+d}{c}\)
a) ta có: \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow\frac{a}{b}=k\Rightarrow a=bk\)
\(\frac{c}{d}=k\Rightarrow c=dk\)
thay vào \(\frac{a^2-b^2}{ab}=\frac{\left(bk^2\right)-b^2}{bkb}=\frac{bkbk-bb}{bkb}=\frac{bb\times\left(kk-1\right)}{bbk}=\frac{kk-1}{k}\)
\(\frac{c^2-d^2}{cd}=\frac{\left(dk^2\right)-d^2}{dkd}=\frac{dkdk-dd}{dkd}=\frac{dd\times\left(kk-1\right)}{ddk}=\frac{kk-1}{k}\)
\(\Rightarrow\frac{a^2-b^2}{ab}=\frac{c^2-d^2}{cd}\left(=\frac{kk-1}{k}\right)\)
b) ta có \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow\frac{a}{b}=k\Rightarrow a=bk\)
\(\Rightarrow\frac{c}{d}=k\Rightarrow c=dk\)
thay vào \(\frac{\left(a+b\right)^2}{a^2+b^2}=\frac{\left(bk+b\right)^2}{bkbk+bb}=\frac{b\left(k+1\right)\times b\left(k+1\right)}{bb\left(kk+1\right)}=\frac{bb\left(k+1\right)\left(k+1\right)}{bb\left(kk+1\right)}=\frac{\left(k+1\right)\left(k+1\right)}{kk+1}\)
\(\frac{\left(c+d\right)^2}{c^2+d^2}=\frac{\left(dk+d\right)^2}{dkdk+dd}=\frac{\left(d\left(k+1\right)\right)^2}{dd\left(kk+1\right)}=\frac{d\left(k+1\right)\times d\left(k+1\right)}{dd\left(kk+1\right)}=\frac{dd\left(k+1\right)\left(k+1\right)}{dd\left(kk+1\right)}=\frac{\left(k+1\right)\left(k+1\right)}{kk+1}\)
\(\Rightarrow\frac{\left(a+b\right)^2}{a^2+b^2}=\frac{\left(c+d\right)^2}{c^2+d^2}\left(=\frac{\left(k+1\right)\left(k+1\right)}{kk+1}\right)\)
(a² + b²) / (c² + d²) = ab/cd
<=> (a² + b²)cd = ab(c² + d²)
<=> a²cd + b²cd = abc² + abd²
<=> a²cd - abc² - abd² + b²cd = 0
<=> ac(ad - bc) - bd(ad - bc) = 0
<=> (ac - bd)(ad - bc) = 0
<=> ac - bd = 0 hoặc ad - bc = 0
<=> ac = bd hoặc ad = bc
<=> a/b = d/c hoặc a/b = c/d (đpcm)
Lời giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=t(t\neq \pm 1)\) \(\Rightarrow a=bt;c=dt\)
Khi đó:
\(\frac{a+b}{a-b}=\frac{bt+b}{bt-b}=\frac{b(t+1)}{b(t-1)}=\frac{t+1}{t-1}\)
\(\frac{c+d}{c-d}=\frac{dt+d}{dt-d}=\frac{d(t+1)}{d(t-1)}=\frac{t+1}{t-1}\)
\(\Rightarrow \frac{a+b}{a-b}=\frac{c+d}{c-d}\) (đpcm)
Cách khác:
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau,ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a-b}{c-d}\Rightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\left(đpcm\right)\)
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