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

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 )

Sửa lại tí:

\(\frac{a}{b}=\frac{c}{d}=\frac{a^2}{b^2}=\frac{a}{b}.\frac{c}{d}=\frac{a-c}{b-d}=\frac{a^2-c^2}{b^2-d^2}=\frac{\left(a-c\right)^2}{\left(b-d\right)^2}=\frac{ac}{bd}\)

Ta có :

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

\(\Rightarrow\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\) (đpcm)

a) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\begin{cases}a=kb\\c=kd\end{cases}\)

=> \(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(kb\right)^2+b^2}{\left(kd\right)^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\) (1)

\(\frac{ab}{cd}=\frac{kbb}{kdd}=\frac{k.b^2}{k.d^2}=\frac{b^2}{d^2}\) (1)

Từ (1) và (2) => \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}\)

b) Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\)

Ta có: \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=k^3\)

Mà: \(k^3=\frac{a}{d}\) => \(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)

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

\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a}{c}\cdot\frac{b}{d}=\frac{ab}{cd}\)

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

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

\(\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{c}\right)^3=\left(\frac{c}{d}\right)^3=\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}\)

\(\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\)

Mà \(\left(\frac{a}{b}\right)^3=\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{d}=\frac{a}{d}=\frac{a^3}{b^3}\)

\(\Rightarrow\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)

\(\Rightarrow\frac{a^2+ac}{c^2-ac}=\frac{\left(bk\right)^2+\left(bk\right)\left(dk\right)}{\left(dk\right)^2-\left(bk\right)\left(dk\right)}=\frac{k^2\left(b^2+bd\right)}{k^2\left(d^2-bd\right)}=\frac{b^2+bd}{d^2-bd}\) (đpcm)

Vậy \(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\)

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Konasuba

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)\(\Rightarrow a=bk;c=dk\)

1)Xét \(VT=\frac{\left(bk\right)^2+bkdk}{\left(dk\right)^2-bkdk}=\frac{b^2k^2+bdk^2}{d^2k^2-bdk^2}=\frac{k^2\left(b^2+bd\right)}{k^2\left(d^2-bd\right)}=\frac{b^2+bd}{d^2-bd}=VP\)

Suy ra Đpcm

2)Xét \(VT=\frac{3\left(bk\right)^2+\left(dk\right)^2}{3b^2+d^2}=\frac{3b^2k^2+d^2k^2}{3b^2+d^2}=\frac{k^2\left(3b^2+d^2\right)}{3b^2+d^2}=k^2\left(1\right)\)

Xét \(VP=\frac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\frac{k^2\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\left(2\right)\)

Từ (1) và (2) suy ra Đpcm

Theo đề bài ta được:

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

Ta có:

\(\dfrac{a^2+ac}{c^2-ac}=\dfrac{a\left(a+c\right)}{c\left(c-a\right)}=\dfrac{bk\left(bk+dk\right)}{dk\left(dk-bk\right)}=\dfrac{bk\left[k\left(b+d\right)\right]}{dk\left[k\left(d-b\right)\right]}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\left(1\right)\)

\(\dfrac{b^2+bd}{d^2-bd}=\dfrac{b\left(b+d\right)}{d\left(d-b\right)}\left(2\right)\)

Từ (1) và (2) suy ra:\(\dfrac{a^2+ac}{c^2-ac}=\dfrac{b^2+bd}{d^2-bd}\)