Question

[NHN14/09]

Ai giúp mk với

Answer 1

[Akai Haruma]

Lời giải:

Đặt $\frac{a}{c}=\frac{c}{b}=k\Rightarrow a=ck; c=bk$.

Khi đó:

a.

$\frac{a^2+c^2}{b^2+c^2}=\frac{(ck)^2+c^2}{b^2+(bk)^2}=\frac{c^2(k^2+1)}{b^2(k^2+1)}=\frac{c^2}{b^2}=\frac{(bk)^2}{b^2}=k^2(1)$

$\frac{a}{b}=\frac{ck}{b}=\frac{bk.k}{b}=k^2(2)$

Từ $(1); (2)\Rightarrow \frac{a^2+c^2}{b^2+c^2}=\frac{a}{b}$

b.

$\frac{b^2-a^2}{a^2+c^2}=\frac{b^2-(ck)^2}{(ck)^2+c^2}$

$=\frac{b^2-c^2k^2}{c^2(k^2+1)}=\frac{b^2-b^2k^4}{c^2(k^2+1)}$

$=\frac{b^2(1-k^4)}{c^2(k^2+1)}=\frac{b^2(1-k^2)(1+k^2)}{c^2(k^2+1)}$

$=\frac{b^2(1-k^2)}{c^2}=\frac{b^2(1-k^2)}{b^2k^2}=\frac{1-k^2}{k^2}(3)$
Và:

$\frac{b-a}{a}=\frac{b-ck}{ck}=\frac{b-bk^2}{bk^2}=\frac{b(1-k^2)}{bk^2}=\frac{1-k^2}{k^2}(4)$

Từ $(3); (4)$ suy ra $\frac{b^2-a^2}{a^2+c^2}=\frac{b-a}{a}$

Answer 2

[NHN14/09]

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