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Dễ mà
Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Áp dụng t/c dãy tỉ số bằng nhau:
Ta có: \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}=\frac{a-b}{c-d}\)(1)
Từ (1),
Ta có: \(\frac{a+b}{c+d}\cdot\frac{a+b}{c+d}=\frac{a+b}{c+d}\cdot\frac{a-b}{c-d}\)(nhân mỗi vế với \(\frac{a+b}{c+d}\))
Vậy \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(a+b\right)\left(a-b\right)}{\left(c+d\right)\left(c-d\right)}=\frac{a^2-b^2}{c^2-d^2}\)(đpcm)
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+) Chứng minh: \(\frac{a}{b}>\frac{a+c}{b+d}\) (1)
Xét hiệu: \(\frac{a}{b}-\frac{a+c}{b+d}=\frac{a\left(b+d\right)-b\left(a+c\right)}{b\left(b+d\right)}=\frac{ab+ad-ab-bc}{b\left(b+d\right)}=\frac{ad-bc}{b\left(b+d\right)}\)
Vì a/b > c/d ; b; d > 0 => ad > bc => ad - bc > 0 .T a có b(b +d) > 0 nên Hiệu trên > 0 => \(\frac{a}{b}>\frac{a+c}{b+d}\)
+) Chứng minh: \(\frac{a+c}{b+d}>\frac{c}{d}\)
Xét hiệu: \(\frac{a+c}{b+d}-\frac{c}{d}=\frac{\left(a+c\right)d-c\left(b+d\right)}{b\left(b+d\right)}=\frac{ad-bc}{b.\left(b+d\right)}>0\)
=> \(\frac{a+c}{b+d}>\frac{c}{d}\) (2)
Từ (1)(2 ta có đpcm
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\(\frac{a+b}{c+d}=\frac{b+c}{a+d}\)
\(\Rightarrow\left(a+b\right)\left(a+d\right)=\left(b+c\right)\left(c+d\right)\)
\(\Rightarrow\left(a+b\right)a+\left(a+b\right)d=\left(b+c\right)c+\left(b+c\right)d\)
\(\Rightarrow a^2+ab+ad+bd=bc+c^2+bd+cd\)
\(\Rightarrow\left(a^2+ab+ad\right)+bd=\left(c^2+bc+cd\right)+bd\)
\(\Rightarrow a.\left(a+b+d\right)=c.\left(c+b+d\right)\)
xét a< c =>a.(a+b+d)<c(c+b+d)
xét a>c =>a.(a+b+d)>c(c+b+d)
=>a=c
=>đpcm
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Bài 1:
Đặt \(\frac{a}{b}=\frac{c}{d}=t\Rightarrow a=bt; c=dt\). Khi đó:
a)
\(\frac{a^2}{a^2+b^2}=\frac{(bt)^2}{(bt)^2+b^2}=\frac{b^2t^2}{b^2(t^2+1)}=\frac{t^2}{t^2+1}(1)\)
\(\frac{c^2}{c^2+d^2}=\frac{(dt)^2}{(dt)^2+d^2}=\frac{d^2t^2}{d^2(t^2+1)}=\frac{t^2}{t^2+1}(2)\)
Từ $(1);(2)$ suy ra đpcm.
b)
\(\left(\frac{a+c}{b+d}\right)^2=\left(\frac{bt+dt}{b+d}\right)^2=\left(\frac{t(b+d)}{b+d}\right)^2=t^2(3)\)
\(\frac{a^2+c^2}{b^2+d^2}=\frac{(bt)^2+(dt)^2}{b^2+d^2}=\frac{t^2(b^2+d^2)}{b^2+d^2}=t^2(4)\)
Từ $(3);(4)\Rightarrow \left(\frac{a+c}{b+d}\right)^2=\frac{a^2+c^2}{b^2+d^2}$ (đpcm)
Bài 2:
Từ $a^2=bc\Rightarrow \frac{a}{c}=\frac{b}{a}$
Đặt $\frac{a}{c}=\frac{b}{a}=t\Rightarrow a=ct; b=at$. Khi đó:
a)
$\frac{a^2+c^2}{b^2+a^2}=\frac{(ct)^2+c^2}{(at)^2+a^2}=\frac{c^2(t^2+1)}{a^2(t^2+1)}=\frac{c^2}{a^2}=(\frac{c}{a})^2=\frac{1}{t^2}(1)$
Và:
$\frac{c}{b}=\frac{a}{tb}=\frac{a}{t.at}=\frac{1}{t^2}(2)$
Từ $(1);(2)$ suy ra đpcm.
b)
$\left(\frac{c+2019a}{a+2019b}\right)^2=\left(\frac{c+2019a}{ct+2019at}\right)^2=\left(\frac{c+2019a}{t(c+2019a)}\right)^2=\frac{1}{t^2}(3)$
Từ $(2);(3)$ suy ra đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2-c^2}{b^2-d^2}\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{a}{b}\cdot\frac{a}{b}=\frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}\left(dpcm\right)\)
Có \(\frac{a}{b}=\frac{c}{d}\left(b,d\ne0\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{b}=\frac{c}{d}=\frac{a^2}{b^2}=\frac{a}{b}.\frac{c}{d}=\frac{a^2-c^2}{b^2-d^2}=\frac{ac}{bd}\)
Vậy \(\frac{ac}{bd}=\frac{a^2-c^2}{b^2-d^2}\)( đpcm )
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\(\frac{a-b}{a}=\frac{c-d}{c}\Leftrightarrow1-\frac{b}{a}=1-\frac{d}{c}\Leftrightarrow\frac{b}{a}=\frac{d}{c}\Leftrightarrow\frac{a}{b}=\frac{c}{d}\) ( Đúng với gt)