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Đặt \(\frac{a}{b}=\frac{c}{d}=k\)=> \(\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó, ta có: \(\frac{\left(bk\right)^2+\left(bk\right)\left(dk\right)}{\left(dk^2\right)-\left(bk\right)\left(dk\right)}=\frac{b^2.k^2+b.d.k^2}{d^2.k^2-b.d.k^2}=\frac{\left(b^2+bd\right).k^2}{\left(d^2-bd\right).k^2}=\frac{b^2+bd}{d^2-bd}\)
Vậy ...
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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.
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a)\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{ac}{bd}=\frac{a^2}{b^2}=\frac{c^2}{d^2}\)
Áp dụng t/c dãy tỉ số bằng nhau: \(\frac{ac}{bd}=\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}\)(đpcm)
b)\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{b}+2=\frac{c}{d}+2\Leftrightarrow\frac{a+2b}{b}=\frac{c+2d}{d}\)(đpcm)
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\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}\)
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 )
\(\frac{a^2+ac}{ac+c^2}=\frac{a\left(a+c\right)}{c\left(a+c\right)}=\frac{a}{c}\)
=> điều phải chứng minh