Giải thích các bước giải:
Bài 1:
Ta có: $\dfrac{a}{b}=\dfrac{c}{d}=k$
$\to a=bk, c=dk$
$\to \dfrac{a-b}{c-d}=\dfrac{bk-b}{dk-d}=\dfrac{b(k-1)}{d(k-1)}=\dfrac{b}d$
Ta có: $\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k(b+d)}{b+d}=k$
Do $\dfrac{b}d$ không chắc chắn bằng $k$
$\to \dfrac{a-b}{c-d}=\dfrac{a+c}{b+d}$ là sai
$\to $Đề sai
Bài 2:
Ta có:
$\dfrac{2a+b}{a-2b}=\dfrac{2c+d}{c-2d}$
$\to \dfrac{2a+b}{a-2b}-2=\dfrac{2c+d}{c-2d}-2$
$\to \dfrac{5b}{a-2b}=\dfrac{5d}{c-2d}$
$\to \dfrac{b}{a-2b}=\dfrac{d}{c-2d}$
$\to \dfrac{a-2b}{b}=\dfrac{c-2d}d$
$\to \dfrac ab-2=\dfrac cd-2$
$\to \dfrac ab=\dfrac cd$
$\to đpcm$
Bài 3:
Ta có:
$\dfrac ab=\dfrac cd$
$\to\dfrac ac=\dfrac bd=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}$(dãy tỉ số bằng nhau)
$\to đpcm$
Bài 4:
Ta có:
$\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}$
$\to (a+3c)(b+d)=(b+3d)(a+c)$
$\to ab+ad+3bc+3cd=ab+bc+3ad+3cd$
$\to 2bc=2ad$
$\to bc=ad$
$\to \dfrac{a}b=\dfrac cd$
Bài 5:
Ta có:
$\dfrac{a+b}{b+c}=\dfrac{c+d}{d+a}$
$\to (a+b)(d+a)=(c+d)(b+c)$
$\to ad+a^2+bd+ab=bc+c^2+bd+cd$
$\to ad+a^2+ab=bc+c^2+cd$
$\to ad+a^2+ab-bc-c^2-cd=0$
$\to (ad-cd)+(a^2-c^2)+(ab-bc)=0$
$\to d(a-c)+(a-c)(a+c)+b(a-c)=0$
$\to (a-c)(d+a+c+b)=0$
$\to a-c=0$ hoặc $d+a+c+b=0$
$\to a=c$ hoặc $a+b+c+d=0$
