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

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

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

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

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

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

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

câu cuối lm tương tự

Ta có: \(\frac{a}{b}=\frac{c}{d}\)

=> ad = bc 

=> 3ac + ad = 3ac + bc

=> a(3c + d) = c(3a + b)

=> \(\frac{a}{3a+b}=\frac{c}{3a+d}\) (ĐPCM)

b) Ta có:

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

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

=> a = c.k; b = d.k

=> a² = c².k²; b² = d².k²

=> \(\frac{a^2+c^2}{b^2+d^2}=\frac{\left(c^2.k^2\right)+c^2}{\left(d^2.k^2\right)+d^2}\)= \(\frac{c^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}\)=\(\frac{c^2}{d^2}=\frac{a^2}{b^2}=\frac{ac}{bd}\)

=> ĐPCM

Câu c tương tự câu b

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow\)a=bk , c=dk

Ta có:

\(\left(\frac{a+b}{c+d}\right)^2=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\)\(\frac{\left(b\left(k+1\right)\right)^2}{\left(d\left(k+1\right)\right)^2}=\frac{b^2\times\left(k+1\right)^2}{d^2\times\left(k+1\right)^2}=\frac{b^2}{d^2}\)( 1 )

\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2\times k^2+b^2}{d^2\times k^2+d^2}\)= \(\frac{b^2\times\left(k^2+1\right)}{d^2\times\left(k^2+1\right)}=\frac{b^2}{d^2}\)( 2 )

Từ ( 1 ) và ( 2 ) \(\Rightarrow\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)(dpcm)

* Giả sử tất cả các tỷ lệ thức đều có nghĩa.

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

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

a) \(\frac{a}{b}=\frac{c}{d}\)

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

áp dụng t/c dãy tỉ số = nhau : 

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

mấy bài kia cũng tương tự em ạ !

gợi ý: đặt chung cho cả 4 phần a/b = c/d = k( k khác 0)

                                               => a=bk; c=dk

rồi thay vào các biểu thức

nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

*a/b=c/d=k=>a=bk;c=dk

Thay a=bk vào 2a+3b/2a-3b=2bk+3b/2bk-3b=2k+3/2k-3

Tương tự thay c=dk vào 2c+3d/2c-3d=2dk+3d/2dk-3d=2k+3/2k-3

=>2a+3b/2a-3b=2c+3d/2c-3d

*a/b=c/d=>a/c=b/d=k

=>k^2=a^2/c^2=c^2/d^2=a^2-b^2/c^2-d^2 (1)

k^2=a/c.b/d=ab/cd (2)

Từ (1) và (2)=>ab/cd=a^2-b^2/c^2-d^2

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

=>k^2=(a+b/c+d)^2 

k^2=a^2/c^2=b^2/d^2=a^2+b^2/c^2+d^2

=>(a+b/c+d)^2=a^2+b^2/c^2+d^2

Đặt \(\frac{a}{b}=\frac{c}{d}=k\left(k\in R\right)\)thì a = bk ; c = dk .Ta có :

 \(\frac{2a+3b}{2a-3b}=\frac{2bk+3b}{2bk-3b}=\frac{b\left(2k+3\right)}{b\left(2k-3\right)}=\frac{2k+3}{2k-3}\left(1\right)\)

 \(\frac{2c+3d}{2c-3d}=\frac{2dk+3d}{2dk-3d}=\frac{d\left(2k+3\right)}{d\left(2k-3\right)}=\frac{2k+3}{2k-3}\left(2\right)\)

 \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\left(3\right)\); \(\frac{a^2-b^2}{c^2-d^2}=\frac{b^2k^2-b^2}{d^2k^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\left(4\right)\)

\(\left(\frac{a+b}{c+d}\right)^2=\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\frac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\frac{b^2}{d^2}\left(5\right)\)

\(\frac{a^2+b^2}{c^2+d^2}=\frac{b^2k^2+b^2}{d^2k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\left(6\right)\)

Từ (1) và (2) , (3) và (4) , (5) và (6) , ta suy ra 3 tỉ lệ thức cần chứng minh từ tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\)

chỉ cần thừa nhận không cần chứng minh

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

\(\Rightarrow\hept{\begin{cases}c+d=\left(d+a\right)k\\a+b=\left(b+c\right)k\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}c+d=dk+ak\\a+b=bk+ck\end{cases}}\)

\(\Rightarrow a+b+c+d=bk+ck+dk+ak\)

\(\Rightarrow a+b+c+d=\left(a+b+c+d\right)k\)

\(\Rightarrow k=1\)

\(\Rightarrow\hept{\begin{cases}c+d=d+a\\a+b=b+c\end{cases}}\)

\(\Rightarrow c+d-d-a=0\)

\(\Rightarrow c-a=0\)

\(\Rightarrow c=a\)