Question

Bài 1: Cho \(\frac{a}{b}=\frac{c}{d}\) .CM:

a) \(\frac{a^2}{a^2+b^2}=\frac{c^2}{c^2+d^2}\) b) \(\left(\frac{a+c}{b+d}\right)^2=\frac{a^2+c^2}{b^2+d^2}\)

Bài 2: Cho 3 số a,b,c\(\ne\)0, sao cho a\(^2\)=bc. CM:

a) \(\frac{a^2+c^2}{b^2+a^2}=\frac{c}{b}\) b)\(\left(\frac{c+2019a}{a+2019b}\right)^2=\frac{c}{b}\)

Bài 4: Cho a,b,c,d khác 0 sao cho b²=ac, c²=bd.CM: \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)

Answer 1

Bài 1:

a) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)

\(\Rightarrow\frac{b}{a}=\frac{d}{c}\)

\(\Rightarrow\frac{b^2}{a^2}=\frac{d^2}{c^2}.\)

\(\Rightarrow\frac{b^2}{a^2}+1=\frac{d^2}{c^2}+1\)

\(\Rightarrow\frac{b^2}{a^2}+\frac{a^2}{a^2}=\frac{d^2}{c^2}+\frac{c^2}{c^2}.\)

\(\Rightarrow\frac{b^2+a^2}{a^2}=\frac{d^2+c^2}{c^2}\)

\(\Rightarrow\frac{a^2}{a^2+b^2}=\frac{c^2}{c^2+d^2}\left(đpcm\right).\)

Bài 4:

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Answer 2

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)

Answer 3

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.

Answer 4

Bài 4:
Từ điều kiện $b^2=ac; c^2=bd\Rightarrow \frac{a}{b}=\frac{b}{c}=\frac{c}{d}$

Đặt $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=t$

$\Rightarrow a=bt; b=ct; c=dt$. Khi đó:

$\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{(bt)^3+(ct)^3+(dt)^3}{b^3+c^3+d^3}=\frac{t^3(b^3+c^3+d^3)}{b^3+c^3+d^3}=t^3(1)$

Và:

$\frac{a}{d}=\frac{bt}{d}=\frac{ct.t}{d}=\frac{dt.t.t}{d}=t^3(2)$

Từ $(1);(2)$ ta có đpcm.