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Do a/b=c/d ⇔ ad=bc
1) Ta có: (a+c)b=ab+bc
(b+d)a=ab+ad
Do bc=ad nên ab+ad=ab+bc
Suy ra (a+c)b=(b+d)a (đpcm)
2) Ta có: (b+d)c=bc+dc
(a+c)d=ad+cd
Do bc=ad nên bc+dc=ad+cd
Suy ra (b+d)c=(b+d)c (đpcm)
3)Ta có:(a+b)(c-d)=ac-ad+bc-bd=(ac-bd)-(ad-bc)
(a-b)(c+d)=ac+ad-bc-bd=(ac-bd)+(ad-bc)
Do ad=bc ⇔ ad-bc=0 nên (ac-bd)-(ad-bc)=(ac-bd)+(ad-bc)
⇔(a+b)(c-d)= (a-b)(c+d) (đpcm)
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Đặt \(\frac{a}{b}=\frac{c}{d}=k\) => a = bk ; c = dk
\(\frac{a^2+ac}{c^2-ac}=\frac{\left(bk\right)^2+bk.dk}{\left(dk\right)^2-bk.dk}=\frac{b^2.k^2+k^2bd}{d^2k^2-k^2bd}=\frac{k^2\left(b^2+bd\right)}{k^2\left(d^2-bd\right)}=\frac{b^2+bd}{d^2-bd}\) (đpcm)
Vậy \(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\)
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Ta có :
\(\frac{a}{b}=\frac{c}{d}=\frac{a-b}{c-d}\Rightarrow\left(\frac{a}{b}\right)^2=\left(\frac{a-b}{c-d}\right)^2=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\) (1)
Lại có \(\left(\frac{a}{b}\right)^2=\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{c}{d}=\frac{a.b}{c.d}\left(\text{ do }\frac{a}{b}=\frac{c}{d}\right)\)(2)
Từ (1) và (2) => \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{a.b}{c.d}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Từ \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\)(1)
Từ \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{a+b}{c+d}\right)^2\)(2)
Từ (1) và (2) \(\Rightarrow\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}=\left(\frac{a}{c}\right)^2\left(đpcm\right)\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
\(\dfrac{a-b}{2\left(c-d\right)}=\dfrac{bk-b}{2\left(dk-d\right)}=\dfrac{b\left(k-1\right)}{2d\left(k-1\right)}=\dfrac{b}{2d}\)
\(\dfrac{a+b}{2\left(c+d\right)}=\dfrac{bk+b}{2\left(dk+d\right)}=\dfrac{b\left(k+1\right)}{2d\left(k+1\right)}=\dfrac{b}{2d}\)
Do đó: \(\dfrac{a-b}{2\left(c-d\right)}=\dfrac{a+b}{2\left(c+d\right)}\)