Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(a=bk;c=dk\)

1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)

\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)

Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)

2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)

\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)

Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)

3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)

4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)

\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)

Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)

có thể chứng minh mà ko phải đặt k ko

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{4a^4}{4c^4}=\frac{5b^4}{5d^4}\)

áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{4a^4}{4b^4}=\frac{5b^4}{5d^4}=\frac{4a^4+5b^4}{4b^4+5d^4}\)

\(\frac{4a^4}{4b^4}=\frac{a^4}{b^4}\)

vì \(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^4}{c^4}=\frac{a}{c}\cdot\frac{b}{d}\cdot\frac{a}{c}\cdot\frac{b}{d}=\frac{a^2}{c^2}\cdot\frac{b^2}{d^2}\)

\(\frac{a^4}{c^4}=\frac{a^2}{c^2}\cdot\frac{b^2}{d^2}=\frac{4a^4+5b^4}{4c^4+5d^4}\left(đpcm\right)\)

a.Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) => \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

=> \(\dfrac{4\left(bk\right)^4+5b^4}{4\left(dk\right)^4+5d^4}=\dfrac{b^4\left(4k^4+5\right)}{d^4\left(4k^4+5\right)}=\dfrac{b^4}{d^4}\)(1)

\(\dfrac{a^2b^2}{c^2d^2}=\dfrac{k^2b^2b^2}{k^2d^2d^2}=\dfrac{b^4}{d^4}\)(2)

Từ (1) và (2) suy ra: \(\dfrac{4a^4+5b^4}{4c^4+5d^4}=\dfrac{a^2b^2}{c^2d^2}\)

b.Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) => \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

=> \(\dfrac{\left(bk\right)^{2004}-b^{2004}}{\left(bk\right)^{2004}+b^{2004}}=\dfrac{b^{2004}\left(k^{2004}-1\right)}{b^{2004}\left(k^{2004}+1\right)}=\dfrac{k^{2004}-1}{k^{2004}+1}\) (1)

\(\dfrac{\left(dk\right)^{2004}-d^{2004}}{\left(dk\right)^{2004}+d^{2004}}=\dfrac{d^{2004}\left(k^{2004}-1\right)}{d^{2004}\left(k^{2004}+1\right)}=\dfrac{k^{2004}-1}{k^{2004}+1}\) (2)

Từ (1) và (2) suy ra: \(\dfrac{a^{2004}-b^{2004}}{a^{2004}+b^{2004}}=\dfrac{c^{2004}-d^{2004}}{c^{2004}+d^{2004}}\)

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{4a^4+5b^4}{4c^4+5d^4}=\dfrac{4b^4k^4+5b^4}{4d^4k^4+5d^4}=\dfrac{b^4\left(4k^4+5\right)}{d^4\left(k^4+5\right)}=\dfrac{b^4}{d^4}\\\dfrac{a^2b^2}{c^2d^2}=\dfrac{bk^2b^2}{dk^2d^2}=\dfrac{k^2b^4}{k^2d^4}=\dfrac{b^4}{d^4}\end{matrix}\right.\)

Vậy.....

\(\left\{{}\begin{matrix}\dfrac{a^{2004}-b^{2004}}{a^{2004}+b^{2004}}=\dfrac{b^{2004}k^{2004}-b^{2004}}{b^{2004}k^{2004}+b^{2004}}=\dfrac{b^{2004}\left(k^{2004}-1\right)}{b^{2004}\left(k^{2004}+1\right)}=\dfrac{k^{2004}-1}{k^{2004}+1}\\\dfrac{c^{2004}-d^{2004}}{c^{2004}+d^{2004}}=\dfrac{d^{2004}k^{2004}-d^{2004}}{d^{2004}k^{2004}+d^{2004}}=\dfrac{d^{2004}\left(k^{2004}-1\right)}{d^{2004}\left(k^{2004}+1\right)}=\dfrac{k^{2004}-1}{k^{2004}+1}\end{matrix}\right.\)

Vậy....

Theo đề bài, ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a^4}{c^4}=\dfrac{b^4}{d^4}=\dfrac{4a^4}{4c^4}=\dfrac{5b^4}{5d^4}=\dfrac{4a^4+5b^4}{4c^4+5d^4}\left(1\right)\)
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{a^2b^2}{b^4}=\dfrac{c^2d^2}{d^4}=\dfrac{a^2b^2}{c^2d^2}=\dfrac{b^4}{d^4}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{4a^4+5b^4}{4c^4+5d^4}=\dfrac{a^2b^2}{c^2d^2}\)(đpcm)
b/ Theo đề bài, ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a^{2004}}{c^{2004}}=\dfrac{b^{2004}}{d^{2004}}=\dfrac{a^{2004}+b^{2004}}{c^{2004}+d^{2004}}\left(1\right)\)
\(\Rightarrow\dfrac{a^{2004}}{c^{2004}}=\dfrac{b^{2004}}{d^{2004}}=\dfrac{a^{2004}-b^{2004}}{c^{2004}-d^{2004}}\left(2\right)\)
Từ (1) và (2)\(\Rightarrow\dfrac{a^{2004}+b^{2004}}{c^{2004}+d^{2004}}=\dfrac{a^{2004}-b^{2004}}{c^{2004}-d^{2004}}=\dfrac{a^{2004}-b^{2004}}{a^{2004}+b^{2004}}=\dfrac{c^{2004}-d^{2004}}{c^{2004}+d^{2004}}\left(đpcm\right)\)

1: Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{a}{a+c}=\dfrac{bk}{bk+dk}=\dfrac{b}{b+d}\)

2: Ta có: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)

nên \(\dfrac{a+c}{a}=\dfrac{b+d}{b}\)

3: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k\)

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

Do đó: \(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}\)

4: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)

=>a/c=a+b/c+d

Theo tính chất dãy tỉ số bằng nhau:

\(x=\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}=\frac{1}{2}\)

ta có: \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\)

áp dụng tc của dãy tỉ số bằng nhau ta có:

 \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{b+c+c+a+a+b}=\frac{a+b+c}{2.\left(a+b+c\right)}=\frac{1}{2}\)

\(\Rightarrow x=\frac{1}{2}\)

Vậy ...

day la cac tinh chat ma

a/

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

Áp dụng tỉ lệ thức ta có:

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

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

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

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