Cách 1: Sử dụng t/c dãy tỉ số bằng nhau ta được

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

Cách 2: 

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

\(\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\) thay vào ta được:

\(\frac{a+2c}{b+2d}=\frac{bk+2dk}{b+2d}=\frac{k\left(b+2d\right)}{b+2d}=k\)

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

=> đpcm

cách 1

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

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

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

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

=> \(\frac{a+2c}{b+2d}=\frac{a-c}{b-d}\)

cách 2:

đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=b.k;c=d.k\)

\(\frac{a+2c}{b+2d}=\frac{bk+2dk}{b+2d}=\frac{k\left(b+2d\right)}{b+2d}=k\)

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

=> \(\frac{a+2c}{b+2d}=\frac{a-c}{b-d}\)

Tìm hai số hữu tỉ a và b biết: a-b=a:b=2*(a+b)

Cảm ơn mn:vvv

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

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

b ) \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{5a}{5c}=\frac{3b}{3d}=\frac{3a}{3c}=\frac{2b}{2d}=\frac{5a-3b}{5c-3d}=\frac{3a+2b}{3c+2d}\)

\(\Rightarrow\frac{5a-3b}{3a+2b}=\frac{5c-3d}{3c+2d}\) ( đpcm )

dat x/5=y/7=k

=)x=5k

   y=7k(1)

Thay 1 vao bthuc xy=140 ta duoc

5k.7k=140

=)35.k^2=140

=)k^2=4

=)k=2 hoac k=-2

thay k=2 vao 1 ta duoc

x=5.2=10

y=7.2=14

thay k=-2 vao 1 ta dc

x=5.(-2)=-10

y=7.(-2)=-14

1) Ta có:
\(\dfrac{a}{a+b}\)=\(\dfrac{c}{c+d}\)
=>a.(c+d) = c.(a+b)
a.c+a.d = a.c+b.d
Do đó a.d=b.d
=>\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)( đpcm)

Câu 2: 

Đặt a/b=c/d=k

=>a=bk; c=dk

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

\(\dfrac{-5a+3c}{-5b+3d}=\dfrac{-5bk+3dk}{-5b+3d}=k\)

=>\(\dfrac{3a+2c}{3b+2d}=\dfrac{-5a+3c}{-5b+3d}\)

b: \(\dfrac{a^2}{b^2}=\dfrac{b^2k^2}{b^2}=k^2\)

\(\dfrac{2c^2-ac}{2d^2-bd}=\dfrac{c\left(2c-a\right)}{d\left(2d-b\right)}=\dfrac{dk}{d}\cdot\dfrac{2dk-bk}{2d-b}=k^2\)

=>\(\dfrac{a^2}{b^2}=\dfrac{2c^2-ac}{2d^2-bd}\)

Theo bài ra ta có : 

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

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

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

Nếu a + b + c + d = 0

\(\Rightarrow\frac{0}{a}=\frac{0}{b}=\frac{0}{c}=\frac{0}{d}\)

\(\Rightarrow\orbr{\begin{cases}a=b=c=d\\a\ne b\ne c\ne d\end{cases}}\)(loại) 

Nếu a + b + c + d \(\ne\)0

=> \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=\frac{1}{d}\)

=> a = b = c = d (đpcm)