Toán Lớp 7: Cho $\frac{a}{b}$ =$\frac{c}{d}$ và a, b, c, d khác 0. Chứng minh rằng
1.$\frac{a+b}{b}$ =$\frac{c+d}{d}$
2.$\frac{a-c}{c}$ =$\frac{b-d}{d}$
3.$\frac{a-b}{a}$ =$\frac{x}{y}$ $\frac{c-d}{d}$
4.$\frac{a^2+c^2}{b^2+d^2}$ =$\frac{ac}{bd}$
5.Với a^2=bc thì $\frac{â+b}{a-b}$ =$\frac{c+a}{c-a}$
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Comments ( 2 )
$\frac{1.}{}$
$\frac{a}{b}$=$\frac{c}{d}$=$\frac{a}{b}$+1=$\frac{c}{d}$+1=$\frac{a+b}{b}$=$\frac{c+d}{d}$
$\frac{2.}{}$
$\frac{a}{b}$=$\frac{c}{d}$=$\frac{a}{c}$=$\frac{b}{d}$
=$\frac{a}{c}$-1=$\frac{b}{d}$-1
=$\frac{a-c}{c}$=$\frac{b-d}{d}$
3.
$\frac{4.}{}$
Đặt $\frac{a}{b}$=$\frac{c}{d}$=k
=>a=bk;c=dk
Ta có:$\frac{ac}{bd}$=$\frac{bk.dk}{bd}$=$\frac{bd.k^2}{bd}$=$k^{2}$ (1)
$\frac{a^2+c^2}{b^2+d^2}$^$\frac{(bk)^2+(dk)^2}{b^2+d^2}$
=$\frac{b^2k^2+d^2k^2}{b^2+d^2}$=$\frac{(b^2+d^2).k^2}{b^2+d^2}$^=$k^{2}$(2)
Từ (1)và(2)=>
$\frac{ac}{bd}$=$\frac{a^2+c^2}{b^2+d^2}$
Mình ko bt lm bài 3 sorry bn nhé bn cho mình 5 sao cũng đc